In our place we&#39;re split between using mysqli and PDO for stuff like prepared statements and transaction support. Some projects use one, some the other. There is little realistic likelihood of us ever moving to another RDBMS.

I prefer PDO for the single reason that it allows named parameters for prepared statements, and as far as I am aware mysqli does not.

Are there any other pros and cons to choosing one over the other as a standard as we consolidate our projects to use just one approach?

mysqli or PDO - what are the pros and cons?

Which is better, MySQL or MySQLi? And why? Which should I use?

I mean better not just in terms of performance, but any other relevant feature.


MySQL vs MySQLi when using PHP

In my local/development environment, the MySQLi query is performing OK. However, when I upload it on my web host environment, I get this error:

&gt; Fatal error: Call to a member function bind_param() on a non-object in...

Here is the code:

    global $mysqli;
    $stmt = $mysqli-&gt;prepare(&quot;SELECT id, description FROM tbl_page_answer_category WHERE cur_own_id = ?&quot;);
    $stmt-&gt;bind_param(&#39;i&#39;, $cur_id);
    $stmt-&gt;execute();
    $stmt-&gt;bind_result($uid, $desc);

To check my query, I tried to execute the query via control panel [phpMyAdmin][1] and the result is OK.

  [1]: https://en.wikipedia.org/wiki/PhpMyAdmin


How to deal with mysqli problems? mysqli_fetch_array(): Argument #1 must be of type mysqli_result

What is the difference between MySQL, [MySQLi][1] and [PDO][2]?

Which one is the best suited to use with PHP-MySQL?

  [1]: http://en.wikipedia.org/wiki/MySQLi
  [2]: https://en.wikipedia.org/wiki/PHP#Development_and_community


What is the difference between MySQL, MySQLi and PDO?

Here&#39;s  my code:

    include &#39;conn.php&#39;;
    $conn = new Connection();
    $query = &#39;SELECT EmailVerified, Blocked FROM users WHERE Email = ? AND SLA = ? AND `Password` = ?&#39;;
    $stmt = $conn-&gt;mysqli-&gt;prepare($query);
    $stmt-&gt;bind_param(&#39;sss&#39;, $_POST[&#39;EmailID&#39;], $_POST[&#39;SLA&#39;], $_POST[&#39;Password&#39;]);
    $stmt-&gt;execute();
    $result = $stmt-&gt;get_result();

I get the error on last line as: **Call to undefined method mysqli_stmt::get_result()**

Here is the code for conn.php:

    define(&#39;SERVER&#39;, &#39;localhost&#39;);
    define(&#39;USER&#39;, &#39;root&#39;);
    define(&#39;PASS&#39;, &#39;xxxx&#39;);
    define(&#39;DB&#39;, &#39;xxxx&#39;);
    class Connection{
        /**
         * @var Resource 
         */
        var $mysqli = null;
    
        function __construct(){
            try{
                if(!$this-&gt;mysqli){
                    $this-&gt;mysqli = new MySQLi(SERVER, USER, PASS, DB);
                    if(!$this-&gt;mysqli)
                        throw new Exception(&#39;Could not create connection using MySQLi&#39;, &#39;NO_CONNECTION&#39;);
                }
            }
            catch(Exception $ex){
                echo &quot;ERROR: &quot;.$e-&gt;getMessage();
            }
        }
    }

If I write this line: 

    if(!stmt) echo &#39;Statement prepared&#39;; else echo &#39;Statement NOT prepared&#39;;

It prints **&#39;Statement NOT prepared&#39;**. If I run the query directly in the IDE replacing ? marks with values, it works fine. Please note that $conn object works fine in other queries in the project.

Any help please.......

Call to undefined method mysqli_stmt::get_result

I am doing a tutorial and am getting this error:

&gt; Fatal error: Class &#39;MySQLi&#39; not found (LONG URL) on line 8

The code on line 8 is:

    $mysqli = new MySQLi($db_server, $db_user, $db_pass, $db_name);

I saw online someone said to see if it was turned on in my phpinfo(), but there wasn&#39;t anything listed in there under for &quot;mysqli&quot;.

Also, I am running PHP version 5.2.5

 


How to solve &quot;Fatal error: Class &#39;MySQLi&#39; not found&quot;?

I have searched the net and so far what I have seen is that you can use `mysql_` and `mysqli_` together meaning:

    &lt;?php
    $con=mysqli_connect(&quot;localhost&quot;, &quot;root&quot; ,&quot;&quot; ,&quot;mysql&quot;);

    if( mysqli_connect_errno( $con ) ) {
        echo &quot;failed to connect&quot;;
    }else{
        echo &quot;connected&quot;;
    }
    mysql_close($con);
    echo &quot;Done&quot;;
    ?&gt;
or 

    &lt;?php
    $con=mysql_connect(&quot;localhost&quot;, &quot;root&quot; ,&quot;&quot; ,&quot;mysql&quot;);
    if( mysqli_connect_errno( $con ) ) {
        echo &quot;failed to connect&quot;;
    }else{
        echo &quot;connected&quot;;
    }
    mysqli_close($con);
    echo &quot;Done&quot;;
    ?&gt;

Are valid but when I use this code what I get is:

    Connected
    Warning: mysql_close() expects parameter 1 to be resource, object given in D:\************.php on line 9
    Done

For the first and the same except with `mysqli_close()`. For the second one.

What is the problem? Can&#39;t I use `mysql_` and `mysqli` together? Or is it normal? Is the way I can check if the connections are valid at all? (the `if(mysq...)`)





Can I mix MySQL APIs in PHP?

I am trying to execute my PHP code, which calls two MySQL queries via mysqli, and get the error &quot;Commands out of sync; you can&#39;t run this command now&quot;.

Here is the code I am using

    &lt;?php
    $con = mysqli_connect(&quot;localhost&quot;, &quot;user&quot;, &quot;password&quot;, &quot;db&quot;);
    if (!$con) {
    	echo &quot;Can&#39;t connect to MySQL Server. Errorcode: %s\n&quot;. Mysqli_connect_error();
    	exit;
    }
    $con-&gt;query(&quot;SET NAMES &#39;utf8&#39;&quot;);
    $brand =&quot;o&quot;;
    $countQuery = &quot;SELECT ARTICLE_NO FROM AUCTIONS WHERE upper(ARTICLE_NAME) LIKE % ? %&quot;;
    if ($numRecords = $con-&gt;prepare($countQuery)) {
    	$numRecords-&gt;bind_param(&quot;s&quot;, $brand);
    	$numRecords-&gt;execute();
    	$data = $con-&gt;query($countQuery) or die(print_r($con-&gt;error));
    	$rowcount = $data-&gt;num_rows;
    	$rows = getRowsByArticleSearch(&quot;test&quot;, &quot;Auctions&quot;, &quot; &quot;);
    	$last = ceil($rowcount/$page_rows);
    }  else {

    print_r($con-&gt;error);
    }
    foreach ($rows as $row) {
    	$pk = $row[&#39;ARTICLE_NO&#39;];
    	echo &#39;&lt;tr&gt;&#39; . &quot;\n&quot;;
    	echo &#39;&lt;td&gt;&lt;a href=&quot;#&quot; onclick=&quot;updateByPk(\&#39;Layer2\&#39;, \&#39;&#39; . $pk . &#39;\&#39;)&quot;&gt;&#39;.$row[&#39;USERNAME&#39;].&#39;&lt;/a&gt;&lt;/td&gt;&#39; . &quot;\n&quot;;
    	echo &#39;&lt;td&gt;&lt;a href=&quot;#&quot; onclick=&quot;updateByPk(\&#39;Layer2\&#39;, \&#39;&#39; . $pk . &#39;\&#39;)&quot;&gt;&#39;.$row[&#39;shortDate&#39;].&#39;&lt;/a&gt;&lt;/td&gt;&#39; . &quot;\n&quot;;
    	echo &#39;&lt;td&gt;&lt;a href=&quot;#&quot; onclick=&quot;deleterec(\&#39;Layer2\&#39;, \&#39;&#39; . $pk . &#39;\&#39;)&quot;&gt;DELETE RECORD&lt;/a&gt;&lt;/td&gt;&#39; . &quot;\n&quot;;
    	echo &#39;&lt;/tr&gt;&#39; . &quot;\n&quot;;
    }
    function getRowsByArticleSearch($searchString, $table, $max) {
    	$con = mysqli_connect(&quot;localhost&quot;, &quot;user&quot;, &quot;password&quot;, &quot;db&quot;);
    	$recordsQuery = &quot;SELECT ARTICLE_NO, USERNAME, ACCESSSTARTS, ARTICLE_NAME, date_format(str_to_date(ACCESSSTARTS, &#39;%d/%m/%Y %k:%i:%s&#39;), &#39;%d %m %Y&#39; ) AS shortDate FROM AUCTIONS WHERE upper(ARTICLE_NAME) LIKE &#39;%?%&#39; ORDER BY str_to_date(ACCESSSTARTS, &#39;%d/%m/%Y %k:%i:%s&#39;)&quot; . $max;
    	if ($getRecords = $con-&gt;prepare($recordsQuery)) {
    		$getRecords-&gt;bind_param(&quot;s&quot;, $searchString);
    		$getRecords-&gt;execute();
    		$getRecords-&gt;bind_result($ARTICLE_NO, $USERNAME, $ACCESSSTARTS, $ARTICLE_NAME, $shortDate);
    		while ($getRecords-&gt;fetch()) {
    			$result = $con-&gt;query($recordsQuery);
    			$rows = array();
    			while($row = $result-&gt;fetch_assoc()) {
    				$rows[] = $row;
    			}
    			return $rows;
    		}
    	}
    }

I have tried reading up on this, but I am unsure of what to do. I have read about store result and free result, however these have made no difference when using them. I am unsure at exactly which point this error is being caused, and would like to know why it is being caused, and how to fix it.

Going by my debug statements, the first if loop for countQuery is not even being entered, because of an error in my sql syntax near near `&#39;% ? %&#39;`. However if I just select `*` instead of trying to limit based on a LIKE clause, I still get the command out of sync error.

Commands out of sync; you can&#39;t run this command now

I understand that you should NEVER trust user input from a form, mainly due to the chance of SQL injection. 

However, does this also apply to a form where the only input is from a dropdown(s) (see below)?

I&#39;m saving the `$_POST[&#39;size&#39;]` to a Session which is then used throughout the site to query the various databases (with a `mysqli` Select query) and any SQL injection would definitely harm (possibly drop) them. 

There is no area for typed user input to query the databases, only dropdown(s).  

    &lt;form action=&quot;welcome.php&quot; method=&quot;post&quot;&gt;
    &lt;select name=&quot;size&quot;&gt;
      &lt;option value=&quot;All&quot;&gt;Select Size&lt;/option&gt; 
      &lt;option value=&quot;Large&quot;&gt;Large&lt;/option&gt;
      &lt;option value=&quot;Medium&quot;&gt;Medium&lt;/option&gt;
      &lt;option value=&quot;Small&quot;&gt;Small&lt;/option&gt;
    &lt;/select&gt;
    &lt;input type=&quot;submit&quot;&gt;
    &lt;/form&gt;
  

Do I have to guard against SQL injection if I used a dropdown?

    mysqli_real_connect(): (HY000/2002): No such file or directory

PhpMyAdmin error on MacOS. I want answer I really have no idea what I need to do to resolve this.

mysqli_real_connect(): (HY000/2002): No such file or directory

I would like to see an example of how to call using `bind_result` vs. `get_result` and what would be the purpose of using one over the other.

Also the pro and cons of using each.

What is the limitation of using either and is there a difference.

Example of how to use bind_result vs get_result

In PHP I&#39;m getting the following warning whenever I try to connect to a database (via `mysql_connect`)

&gt;Warning: mysql_connect(): Headers and client library minor version mismatch. Headers:50162 Library:50524

In my `php -i` output I have the following values listed under mysqli

&gt; Client API library version =&gt; 5.5.24
&gt;
&gt; Client API header version =&gt; 5.1.62

I&#39;ve tried updating php5-mysql *and* php but I&#39;m already at the latest version of both of them. How do I go about updating the header version so I stop seeing this warning?

**EDIT**

My MySQL files should all be updated to be the latest version:

    $ apt-get install mysql.*5.5
    . . .
    mysql-client-5.5 is already the newest version.
    mysql-server-core-5.5 is already the newest version.
    mysql-server-5.5 is already the newest version.
    mysql-testsuite-5.5 is already the newest version.
    mysql-source-5.5 is already the newest version.

Removing old versions

    $ apt-get remove mysql.*5.1
    . . .
    Package handlersocket-mysql-5.1 is not installed, so not removed
    Package mysql-cluster-client-5.1 is not installed, so not removed
    Package mysql-cluster-server-5.1 is not installed, so not removed
    Package mysql-client-5.1 is not installed, so not removed
    Package mysql-client-core-5.1 is not installed, so not removed
    Package mysql-server-5.1 is not installed, so not removed
    Package mysql-server-core-5.1 is not installed, so not removed
    Package mysql-source-5.1 is not installed, so not removed

Headers and client library minor version mismatch

I have installed PHP 7 and MySQL 5.5.47 on Ubuntu&amp;nbsp;14.04 (Trusty Tahr).

I have checked installed extension using:

    sudo apt-cache search php7-*

It outputs:

    php7.0-common - Common files for packages built from the PHP source
    libapache2-mod-php7.0 - server-side, HTML-embedded scripting language (Apache 2 module)
    php7.0-cgi - server-side, HTML-embedded scripting language (CGI binary)
    php7.0-cli - command-line interpreter for the PHP scripting language
    php7.0-phpdbg - server-side, HTML-embedded scripting language (PHPDBG binary)
    php7.0-fpm - server-side, HTML-embedded scripting language (FPM-CGI binary)
    libphp7.0-embed - HTML-embedded scripting language (Embedded SAPI library)
    php7.0-dev - Files for PHP7.0 module development
    php7.0-dbg - Debug symbols for PHP7.0
    php7.0-curl - CURL module for PHP
    php7.0-enchant - Enchant module for PHP
    php7.0-gd - GD module for PHP
    php7.0-gmp - GMP module for PHP
    php7.0-imap - IMAP module for PHP
    php7.0-interbase - Interbase module for PHP
    php7.0-intl - Internationalisation module for PHP
    php7.0-ldap - LDAP module for PHP
    php7.0-mcrypt - libmcrypt module for PHP
    php7.0-readline - readline module for PHP
    php7.0-odbc - ODBC module for PHP
    php7.0-pgsql - PostgreSQL module for PHP
    php7.0-pspell - pspell module for PHP
    php7.0-recode - recode module for PHP
    php7.0-snmp - SNMP module for PHP
    php7.0-tidy - tidy module for PHP
    php7.0-xmlrpc - XMLRPC-EPI module for PHP
    php7.0-xsl - XSL module for PHP
    php7.0 - server-side, HTML-embedded scripting language (metapackage)
    php7.0-json - JSON module for PHP
    php-all-dev - package depending on all supported PHP development packages
    php7.0-sybase - Sybase module for PHP
    php7.0-sqlite3 - SQLite3 module for PHP
    php7.0-mysql - MySQL module for PHP
    php7.0-opcache - Zend OpCache module for PHP
    php7.0-bz2 - bzip2 module for PHP

I am not able to see the MySQLi extension using phpinfo() either. How can I enable/install MySQLi extension in PHP&amp;nbsp;7?

That&#39;s why I cannot use phpMyAdmin. It says &quot;The mysqli extension is missing.&quot;





How can I enable the MySQLi extension in PHP 7?

I&#39;m trying to get my head around MySQli and I&#39;m confused by the error reporting.
I am using the return value of the MySQLi &#39;prepare&#39; statement to detect errors when executing SQL, like this:

    $stmt_test =  $mysqliDatabaseConnection-&gt;stmt_init();
    if($stmt_test-&gt;prepare(&quot;INSERT INTO testtable VALUES (23,44,56)&quot;))
    {
     $stmt_test-&gt;execute();
     $stmt_test-&gt;close();
    }
    else echo(&quot;Statement failed: &quot;. $stmt_test-&gt;error . &quot;&lt;br&gt;&quot;);

But, is the return value of the prepare statement only detecting if there is an error in the preperation of the SQL statement and not detecting execution errors? If so should I therefore change my execute line to flag errors as well like this:

    if($stmt_test-&gt;execute()) $errorflag=true;

And then just to be safe should I also do the following after the statement has executed:

    if($stmt_test-&gt;errno) {$errorflag=true;}

...Or was I OK to start with  and the return value on the MySQLi prepare&#39; statement captures all errors associated with the complete execution of the query it defines?

Thanks 
C

MySQLi prepared statements error reporting

Based on this code below I use for regular mysql, how could I convert it to use mysqli?

**Is it as simple as changing `mysql_query($sql);` to `mysqli_query($sql);`?**

    &lt;?PHP
    
    //in my header file that is included on every page I have this
    $DB[&quot;dbName&quot;] = &quot;emails&quot;;
    $DB[&quot;host&quot;] = &quot;localhost&quot;;
    $DB[&quot;user&quot;] = &quot;root&quot;;
    $DB[&quot;pass&quot;] = &quot;&quot;;
    $link = mysql_connect($DB[&#39;host&#39;], $DB[&#39;user&#39;], $DB[&#39;pass&#39;]) or die(&quot;&lt;center&gt;An Internal Error has Occured. Please report following error to the webmaster.&lt;br&gt;&lt;br&gt;&quot;.mysql_error().&quot;&#39;&lt;/center&gt;&quot;);
    mysql_select_db($DB[&#39;dbName&#39;]);
    // end header connection part
    
    // function from a functions file that I run a mysql query through in any page.
    function executeQuery($sql) {
    	$result = mysql_query($sql);
    	if (mysql_error()) {
    		$error = &#39;&lt;BR&gt;&lt;center&gt;&lt;font size=&quot;+1&quot; face=&quot;arial&quot; color=&quot;red&quot;&gt;An Internal Error has Occured.&lt;BR&gt; The error has been recorded for review&lt;/font&gt;&lt;/center&gt;&lt;br&gt;&#39;;
    		if ($_SESSION[&#39;auto_id&#39;] == 1) {
    			$sql_formatted = highlight_string(stripslashes($sql), true);
    			$error .= &#39;&lt;b&gt;The MySQL Syntax Used&lt;/b&gt;&lt;br&gt;&#39; . $sql_formatted . &#39;&lt;br&gt;&lt;br&gt;&lt;b&gt;The MySQL Error Returned&lt;/b&gt;&lt;br&gt;&#39; . mysql_error();
    		}
    		die($error);
    	}
    	return $result;
    }
    
    // example query ran on anypage of the site using executeQuery function
    $sql=&#39;SELECT auto_id FROM friend_reg_user WHERE auto_id=&#39; .$info[&#39;auto_id&#39;];
    $result_member=executequery($sql);
    if($line_member=mysql_fetch_array($result_member)){
    	extract($line_member);
    } else {
    	header(&quot;location: index.php&quot;);
    	exit;
    }
    ?&gt;

How to change mysql to mysqli?

I just need help on this PHP error which I do not quite understand:

&gt; Fatal error: Cannot pass parameter 2 by reference in /web/stud/openup/inactivatesession.php on line 13

    &lt;?php

    error_reporting(E_ALL);

    include(&#39;connect.php&#39;);

    $createDate = mktime(0,0,0,09,05,date(&quot;Y&quot;));
    $selectedDate =  date(&#39;d-m-Y&#39;, ($createDate));

    $sql = &quot;UPDATE Session SET Active = ? WHERE DATE_FORMAT(SessionDate,&#39;%Y-%m-%d&#39; ) &lt;= ?&quot;;										    
    $update = $mysqli-&gt;prepare($sql);
    $update-&gt;bind_param(&quot;is&quot;, 0, $selectedDate);  //LINE 13
    $update-&gt;execute();

    ?&gt;

What does this error mean? How can this error be fixed?

PHP error: &quot;Cannot pass parameter 2 by reference&quot;

I&#39;m trying to install vanilla forums on my Mac, and for this I just created a database and a user from the MySQL command line:
    
    mysql&gt; CREATE DATABASE vanilla;
    Query OK, 1 row affected (0.00 sec)
    
    mysql&gt; create user &#39;vanilla_user3&#39;@&#39;localhost&#39; IDENTIFIED BY &#39;vanilla_password&#39;;
    Query OK, 0 rows affected (0.00 sec)
    
    mysql&gt; GRANT ALL PRIVILEGES ON * . * TO &#39;vanilla_user3&#39;@&#39;localhost&#39;;
    Query OK, 0 rows affected (0.00 sec)
    
    mysql&gt; FLUSH PRIVILEGES;
    Query OK, 0 rows affected (0.00 sec)

So I try to connect using the following code:

    $con=mysqli_connect(&quot;localhost&quot;,&quot;vanilla_user3&quot;,&quot;vanilla_password&quot;,&quot;vanilla&quot;);
    if (mysqli_connect_errno($con)) {
      echo &quot;Failed to connect to MySQL: &quot; . mysqli_connect_error();
    }

but unfortunately, I get an error saying 

&gt;Warning: mysqli_connect(): (HY000/2002): No such file or directory in /Users/kramer65/Sites/vanilla/info.php on line 3
Failed to connect to MySQL: No such file or directory

Any idea where I&#39;m going wrong?


Warning: mysqli_connect(): (HY000/2002): No such file or directory

For 2 days now I&#39;m trying to solve this, but unfortunately no result. Let me tell you my story about the problem. I&#39;ve bulid an application on a site, and the application deals with the reviews. But, I&#39;m trying to put it on another site, and I copyed the php files, the sql file from the old site, and moved them to the new site (they are on different FTP servers). When I&#39;m trying to go to the pages from the application, I receive this FATAL ERROR:

&gt; Fatal error: Call to undefined function mysqli_connect()

The code that I wrote to connect to the database is this (with hidden credentials):

    $con = mysqli_connect(&quot;&quot;,&quot;*the_name*&quot;,&quot;*the_pass*&quot;,&quot;*the_database*&quot;);
	if (mysqli_connect_errno()) {
		echo &quot;Failed to connect to MySQL: &quot; . mysqli_connect_error();
	}

Why do I get the error? It works on the old server, and the code I think it&#39;s not the problem, because it works on localhost, and on the new server it doesn&#39;t. Can anyone help me?


Fatal error: Call to undefined function mysqli_connect()

I am using the following script to process a form to add info to my website. The problem I am having is when I submit the form nothing gets submitted to the database, and there are no errors. How can I add error reporting to my query?

    &lt;?php
    if (isset($_POST[&#39;itemdescription&#39;])) {$itemdescription = $_POST[&#39;itemdescription&#39;];}else {$itemdescription = &#39;&#39;;}
    if (isset($_POST[&#39;itemnumber&#39;])) {$itemnumber = $_POST[&#39;itemnumber&#39;];}else {$itemnumber = &#39;&#39;;}
    if (isset($_POST[&#39;sellerid&#39;])) {$sellerid = $_POST[&#39;sellerid&#39;];}else {$sellerid = &#39;&#39;;}
    if (isset($_POST[&#39;purchasedate&#39;])) {$purchasedatepre = $_POST[&#39;purchasedate&#39;];$date = DateTime::createFromFormat(&quot;D F d, Y&quot;, $purchasedatepre);$purchasedate = date(&#39;Y-m-d&#39;,strtotime($purchasedatepre));}else {$purchasedatepre = &#39;&#39;; $purchasedate = &#39;&#39;;}
    if (isset($_POST[&#39;otherinfo&#39;])) {$otherinfo = $_POST[&#39;otherinfo&#39;];}else {$otherinfo = &#39;&#39;;}
    if (isset($_POST[&#39;numberofitems&#39;])) {$numberofitems = $_POST[&#39;numberofitems&#39;];}else {$numberofitems = &#39;&#39;;}
    if (isset($_POST[&#39;numberofitemsused&#39;])) {$numberofitemsused = $_POST[&#39;numberofitemsused&#39;];}else {$numberofitemsused = &#39;&#39;;}
    if (isset($_POST[&#39;isitdelivered&#39;])) {$isitdelivered = $_POST[&#39;isitdelivered&#39;];}else {$isitdelivered = &#39;&#39;;}
    if (isset($_POST[&#39;price&#39;])) {$price = $_POST[&#39;price&#39;];}else {$price = &#39;&#39;;}
    
    $itemdescription = str_replace(&quot;&#39;&quot;, &quot;&quot;, &quot;$itemdescription&quot;);
    $itemnumber = str_replace(&quot;&#39;&quot;, &quot;&quot;, &quot;$itemnumber&quot;);
    $sellerid = str_replace(&quot;&#39;&quot;, &quot;&quot;, &quot;$sellerid&quot;);
    $otherinfo = str_replace(&quot;&#39;&quot;, &quot;&quot;, &quot;$otherinfo&quot;);
    
    include(&quot;connectmysqli.php&quot;); 
    
    mysqli_query($db,&quot;INSERT INTO stockdetails (`itemdescription`,`itemnumber`,`sellerid`,`purchasedate`,`otherinfo`,`numberofitems`,`isitdelivered`,`price`) VALUES (&#39;$itemdescription&#39;,&#39;$itemnumber&#39;,&#39;$sellerid&#39;,&#39;$purchasedate&#39;,&#39;$otherinfo&#39;,&#39;$numberofitems&#39;,&#39;$numberofitemsused&#39;,&#39;$isitdelivered&#39;,&#39;$price&#39;)&quot;);
    
    // header(&#39;Location: stockmanager.php?&amp;key=&#39;.$key);
    ?&gt;

How to display errors for my MySQLi query?

How do you insert data into a MySQL date or time column using PHP mysqli and bind_param?

Using Mysqli bind_param with date and time columns?

This is my code but it dosn&#39;t work:

    $param = &quot;%{$_POST[&#39;user&#39;]}%&quot;;
    $stmt = $db-&gt;prepare(&quot;SELECT id,Username FROM users WHERE Username LIKE ?&quot;);
    $stmt-&gt;bind_param(&quot;s&quot;, $param);
    $stmt-&gt;execute();
    $stmt-&gt;bind_result($id,$username);
    $stmt-&gt;fetch();


This code it doesn&#39;t seem to work. I have searched it a lot.
Also it may return more than 1 row.
So how can I get all the results even if it returns more than 1 row?


How can I with mysqli make a query with LIKE and get all results?

I keep getting the following errors with mysql connection through XAMPP and I don&#39;t know what to do:

[![enter image description here][1]][1]

That&#39;s the code in the config.inc.php

    &lt;?php
    /*
     * This is needed for cookie based authentication to encrypt password in
     * cookie
     */
    $cfg[&#39;blowfish_secret&#39;] = &#39;xampp&#39;; /* YOU SHOULD CHANGE THIS FOR A MORE 
    SECURE COOKIE AUTH! */

    /*
     * Servers configuration
     */
    $i = 0;

    /*
    * First server
    */
    $i++;

    /* Authentication type and info */
    $cfg[&#39;Servers&#39;][$i][&#39;auth_type&#39;] = &#39;config&#39;;
    $cfg[&#39;Servers&#39;][$i][&#39;user&#39;] = &#39;root&#39;;
    $cfg[&#39;Servers&#39;][$i][&#39;password&#39;] = &#39;&#39;;
    $cfg[&#39;Servers&#39;][$i][&#39;extension&#39;] = &#39;mysqli&#39;;
    $cfg[&#39;Servers&#39;][$i][&#39;AllowNoPassword&#39;] = true;
    $cfg[&#39;Lang&#39;] = &#39;&#39;;

    /* Bind to the localhost ipv4 address and tcp */
    $cfg[&#39;Servers&#39;][$i][&#39;host&#39;] = &#39;127.0.0.1&#39;;
    $cfg[&#39;Servers&#39;][$i][&#39;connect_type&#39;] = &#39;tcp&#39;;

    /* User for advanced features */
    $cfg[&#39;Servers&#39;][$i][&#39;controluser&#39;] = &#39;pma&#39;;
    $cfg[&#39;Servers&#39;][$i][&#39;controlpass&#39;] = &#39;&#39;;

    /* Advanced phpMyAdmin features */
    $cfg[&#39;Servers&#39;][$i][&#39;pmadb&#39;] = &#39;phpmyadmin&#39;;
    $cfg[&#39;Servers&#39;][$i][&#39;bookmarktable&#39;] = &#39;pma__bookmark&#39;;
    $cfg[&#39;Servers&#39;][$i][&#39;relation&#39;] = &#39;pma__relation&#39;;
    $cfg[&#39;Servers&#39;][$i][&#39;table_info&#39;] = &#39;pma__table_info&#39;;
    $cfg[&#39;Servers&#39;][$i][&#39;table_coords&#39;] = &#39;pma__table_coords&#39;;
    $cfg[&#39;Servers&#39;][$i][&#39;pdf_pages&#39;] = &#39;pma__pdf_pages&#39;;
    $cfg[&#39;Servers&#39;][$i][&#39;column_info&#39;] = &#39;pma__column_info&#39;;
    $cfg[&#39;Servers&#39;][$i][&#39;history&#39;] = &#39;pma__history&#39;;
    $cfg[&#39;Servers&#39;][$i][&#39;designer_coords&#39;] = &#39;pma__designer_coords&#39;;
    $cfg[&#39;Servers&#39;][$i][&#39;tracking&#39;] = &#39;pma__tracking&#39;;
    $cfg[&#39;Servers&#39;][$i][&#39;userconfig&#39;] = &#39;pma__userconfig&#39;;
    $cfg[&#39;Servers&#39;][$i][&#39;recent&#39;] = &#39;pma__recent&#39;;
    $cfg[&#39;Servers&#39;][$i][&#39;table_uiprefs&#39;] = &#39;pma__table_uiprefs&#39;;
    $cfg[&#39;Servers&#39;][$i][&#39;users&#39;] = &#39;pma__users&#39;;
    $cfg[&#39;Servers&#39;][$i][&#39;usergroups&#39;] = &#39;pma__usergroups&#39;;
    $cfg[&#39;Servers&#39;][$i][&#39;navigationhiding&#39;] = &#39;pma__navigationhiding&#39;;
    $cfg[&#39;Servers&#39;][$i][&#39;savedsearches&#39;] = &#39;pma__savedsearches&#39;;
    $cfg[&#39;Servers&#39;][$i][&#39;central_columns&#39;] = &#39;pma__central_columns&#39;;
    $cfg[&#39;Servers&#39;][$i][&#39;designer_settings&#39;] = &#39;pma__designer_settings&#39;;
    $cfg[&#39;Servers&#39;][$i][&#39;export_templates&#39;] = &#39;pma__export_templates&#39;;
    $cfg[&#39;Servers&#39;][$i][&#39;favorite&#39;] = &#39;pma__favorite&#39;;

    /*
     * End of servers configuration
     */

    ?&gt;


I&#39;ve already reset the password from MySql from the command line two times and added the changes to this code above and to the php.ini file, but at each time it goes back to the error. Any help?


  [1]: https://i.stack.imgur.com/nATTy.jpg

phpMyAdmin ERROR: mysqli_real_connect(): (HY000/1045): Access denied for user &#39;pma&#39;@&#39;localhost&#39; (using password: NO)

I&#39;d like to know if anyone has any first-hand experience with this dichotomy. A few blogs say the mysql extension is faster than mysqli.  Is this true?

And I&#39;m only asking about speed.  I know mysqli has features that are not present in the older extension.

Which is fastest in PHP- MySQL or MySQLi?

I cannot find instructions about installing MySQLi on a Mac. Just to be clear, MySQL is up to date and I am running PHP 5. How do I install it? Where do I even get it from? Thanks for your help. I&#39;ll be giving an up vote and a check mark to whoever answers this!

How to install MySQLi on MacOS

I have been using CI just fine using the MySQL driver. I want to use the MySQL driver instead, but as soon as I change it (just add the ‘i’ at the end of MySQL, and added the port number) I get the following error message

   &gt;A Database Error Occurred
    
   &gt;Unable to connect to your database server using the provided settings.
    
   &gt;Filename: core/Loader.php
    
   &gt;Line Number: 232 

my setting look like this:

    $db[&#39;default&#39;][&#39;hostname&#39;] = $hostname;
    $db[&#39;default&#39;][&#39;username&#39;] = $username;
    $db[&#39;default&#39;][&#39;password&#39;] = $password;
    $db[&#39;default&#39;][&#39;database&#39;] = $database;
    $db[&#39;default&#39;][&#39;dbdriver&#39;] = &#39;mysqli&#39;;
    $db[&#39;default&#39;][&#39;port&#39;]     = &quot;3306&quot;;  
    $db[&#39;default&#39;][&#39;dbprefix&#39;] = &#39;&#39;;
    $db[&#39;default&#39;][&#39;pconnect&#39;] = TRUE;
    $db[&#39;default&#39;][&#39;db_debug&#39;] = TRUE;
    $db[&#39;default&#39;][&#39;cache_on&#39;] = FALSE;
    $db[&#39;default&#39;][&#39;cachedir&#39;] = &#39;&#39;;
    $db[&#39;default&#39;][&#39;char_set&#39;] = &#39;utf8&#39;;
    $db[&#39;default&#39;][&#39;dbcollat&#39;] = &#39;utf8_general_ci&#39;;
    $db[&#39;default&#39;][&#39;swap_pre&#39;] = &#39;&#39;;
    $db[&#39;default&#39;][&#39;autoinit&#39;] = TRUE;
    $db[&#39;default&#39;][&#39;stricton&#39;] = FALSE; 
    
    where
    $hostname = &#39;localhost&#39;;
    $username = &#39;myusernamegoeshere&#39;;
    $password = &#39;mypasswordgoeshere&#39;;
    $database = &#39;mydatabasenamegoeshere&#39;; 

I&#39;m Using:
    &gt;CI 2.0.2 php 5.3.4 Apache/2.2.17 (Unix) mysql 5.5.13 mysql.default_port 3306

Am I doing anything wrong?

Thank you,



CodeIgniter: Unable to connect to your database server using the provided settings Error Message

I want retrieve the id of a inserted row in the database, but I don&#39;t know how to do this.

I tried to return using the SQL clause `RETURNING id`, but not works.

How I can return the id after the insertion of a row?

Using Prepared Statement, how I return the id of the inserted row?

I&#39;m playing around with MySQLi at the moment, trying to figure out how it all works. In my current projects I always like to echo out a query string while coding, just to make sure that everything is correct, and to quickly debug my code. But... how can I do this with a prepared MySQLi statement?

Example:

&lt;pre&gt;&lt;code&gt;$id = 1;
$baz = &#39;something&#39;;

if ($stmt = $mysqli-&gt;prepare(&quot;SELECT foo FROM bar WHERE id=? AND baz=?&quot;)) {
  $stmt-&gt;bind_param(&#39;is&#39;,$id,$baz);
  // how to preview this prepared query before acutally executing it?
  // $stmt-&gt;execute();
}&lt;/code&gt;&lt;/pre&gt;

I&#39;ve been going through this list (&lt;a href=&quot;http://www.php.net/mysqli&quot;&gt;http://www.php.net/mysqli&lt;/a&gt;) but without any luck.

&lt;br /&gt;&lt;b&gt;EDIT&lt;/b&gt;

Well, if it&#39;s not possible from within MySQLi, maybe I&#39;ll stick with something like this:

&lt;pre&gt;&lt;code&gt;function preparedQuery($sql,$params) {
  for ($i=0; $i&amp;lt;count($params); $i++) {
    $sql = preg_replace(&#39;/\?/&#39;,$params[$i],$sql,1);
  }
  return $sql;
}

$id = 1;
$baz = &#39;something&#39;;

$sql = &quot;SELECT foo FROM bar WHERE id=? AND baz=?&quot;;

echo preparedQuery($sql,array($id,$baz));

// outputs: SELECT foo FROM bar WHERE id=1 AND baz=something
&lt;/code&gt;&lt;/pre&gt;
Far from perfect obviously, since it&#39;s still pretty redundant &amp;mdash; something I wanted to prevent &amp;mdash; and it also doesn&#39;t give me an idea as to what&#39;s being done with the data by MySQLi. But I guess this way I can quickly see if all the data is present and in the right place, and it&#39;ll save me &lt;em&gt;some&lt;/em&gt; time compared to fitting in the variables manually into the query &amp;mdash; that can be a pain with many vars.

How to echo a MySQLi prepared statement?

Thanks to the answers I have figured out that I am unable to use `fetch_all()` because i am using `PHP 5.2.17` - `fetch_assoc` with `while` loop worked.
___

The function I am using `fetch_all` is coming back with this error: 

&gt; Fatal error: Call to undefined method mysqli_result::fetch_all() in

    $mysqli = new mysqli($host, $username, $password, $database);
    $query = &quot;LONG QUERY that works, tested in phpmyadmin&quot;
    $result = $mysqli-&gt;query($query);
    $result-&gt;fetch_all(); or  $mysqli-&gt;fetch_all() tried both
    mysqli_fetch_all() was already tried.
    $mysqli-&gt;close(); 

I am able to connect to the DB and I have pulled single rows. When I place the query in PHPMYADMIN I get 5 rows back. 

Does this function even work? Is there a way I can place my data into an assoc array on my own?

mysqli fetch_all() not a valid function?

With the following piece of code, how do i know that anything was inserted in to the db?

    if ($stmt = $connection-&gt;prepare(&quot;insert into table (blah) values (?)&quot;)) {
    $stmt-&gt;bind_param(&quot;s&quot;, $blah);	
    $stmt-&gt;execute();			
    $stmt-&gt;close();									
    }

I had thought adding the following line would have worked but apparently not.

    if($stmt-&gt;affected_rows==-1){$updateAdded=&quot;N&quot;; echo &quot;failed&quot;;}	

And then use the $updatedAdded=&quot;N&quot; to then skip other pieces of code further down the page that are dependent on the above insert being successful.

Any ideas?

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