How to add new column to MYSQL table?
PhpMysqlAddAlterPhp Problem Overview
I am trying to add a new column to my MYSQL table using PHP. I am unsure how to alter my table so that the new column is created. In my assessment table I have:
assessmentid | q1 | q2 | q3 | q4 | q5
Say I have a page with a textbox and I type q6 in to the textbox and press a button then the table is updated to:
assessmentid | q1 | q2 | q3 | q4 | q5 | q6
My code:
<?php
mysql_query("ALTER TABLE `assessment` ADD newq INT(1) NOT NULL AFTER `q10`");
?>
<form method="post" action="">
<input type="text" name="newq" size="20">
<input type="submit" name="submit" value="Submit">
Php Solutions
Solution 1 - Php
your table:
q1 | q2 | q3 | q4 | q5
you can also do
ALTER TABLE yourtable ADD q6 VARCHAR( 255 ) after q5
Solution 2 - Php
$table = 'your table name';
$column = 'q6'
$add = mysql_query("ALTER TABLE $table ADD $column VARCHAR( 255 ) NOT NULL");
you can change VARCHAR( 255 ) NOT NULL into what ever datatype you want.
Solution 3 - Php
-
You can add a new column at the end of your table
ALTER TABLE assessment ADD q6 VARCHAR( 255 ) -
Add column to the begining of table
ALTER TABLE assessment ADD q6 VARCHAR( 255 ) FIRST -
Add column next to a specified column
ALTER TABLE assessment ADD q6 VARCHAR( 255 ) after q5
and more options here
Solution 4 - Php
Something like:
$db = mysqli_connect("localhost", "user", "password", "database");
$name = $db->mysqli_real_escape_string($name);
$query = 'ALTER TABLE assesment ADD ' . $name . ' TINYINT NOT NULL DEFAULT \'0\'';
if($db->query($query)) {
echo "It worked";
}
Haven't tested it but should work.
Solution 5 - Php
You should look into normalizing your database to avoid creating columns at runtime.
Make 3 tables:
- assessment
- question
- assessment_question (columns assessmentId, questionId)
Put questions and assessments in their respective tables and link them together through assessment_question using foreign keys.
Solution 6 - Php
Based on your comment it looks like your'e only adding the new column if: mysql_query("SELECT * FROM assessment"); returns false. That's probably not what you wanted. Try removing the '!' on front of $sql in the first 'if' statement. So your code will look like:
$sql=mysql_query("SELECT * FROM assessment");
if ($sql) {
mysql_query("ALTER TABLE assessment ADD q6 INT(1) NOT NULL AFTER q5");
echo 'Q6 created';
}else...
Solution 7 - Php
for WORDPRESS:
global $wpdb;
$your_table = $wpdb->prefix. 'My_Table_Name';
$your_column = 'My_Column_Name';
if (!in_array($your_column, $wpdb->get_col( "DESC " . $your_table, 0 ) )){ $result= $wpdb->query(
"ALTER TABLE $your_table ADD $your_column VARCHAR(100) CHARACTER SET utf8 NOT NULL " //you can add positioning phraze: "AFTER My_another_column"
);}
Solution 8 - Php
ALTER TABLE `stor` ADD `buy_price` INT(20) NOT NULL ;
Solution 9 - Php
The problem with the ALTER TABLE in the PHP code is in this line:
mysql_query("ALTER TABLE assessment ADD newq INT(1) NOT NULL AFTER q10");
It should be AFTER q5 since there is no q10 in your table sample. So, it becomes ALTER TABLE assessment ADD newq INT(1) NOT NULL AFTER q5;
Tried the same logic in my own table in [Skyvia] and it should work as seen below. I added the description column after the To column in the Title table.
[Skyvia] - the link to https://skyvia.com/query/online-mysql-query-builder
[image alter-table-add-column]