From inside a Controller:

    $this-&gt;_helper-&gt;layout-&gt;setLayout(&#39;/path/to/your/layout_script&#39;);

(via [these docs][1])


  [1]: http://framework.zend.com/manual/en/zend.layout.quickstart.html#zend.layout.quickstart.mvc

EDIT: I should mention that the path is relative to whatever your layout directory is (by default, it&#39;s `application/layouts/scripts/`)

You can also use like this

    // Within controller
    Zend_Layout::getMvcInstance()-&gt;setLayout(&#39;layout_name&#39;);

    //Within view script

    &lt;?php $this-&gt;layout()-&gt;setLayout(&#39;layout_name&#39;); ?&gt;


Your layout must be in /layouts/scripts/ folder, otherwise you need to specify the path also. No need to write .phtml, just name of the layout

I have the following table

    DROP TABLE IF EXISTS `test`.`foo`;
    CREATE TABLE  `test`.`foo` (
      `id` int(10) unsigned NOT NULL auto_increment,
      `name` varchar(45) NOT NULL,
      PRIMARY KEY  (`id`)
    ) ENGINE=InnoDB DEFAULT CHARSET=latin1;

Then I try to get records based on the primary key

    SELECT * FROM foo f where f.id IN (2, 3, 1);

I then get the following result

    +----+--------+
    | id | name   |
    +----+--------+
    |  1 | first  |
    |  2 | second |
    |  3 | third  |
    +----+--------+
    3 rows in set (0.00 sec)

As one can see, the result is ordered by id. What I&#39;m trying to achieve is to get the results ordered in the sequence I&#39;m providing in the query. Given this example it should return

    +----+--------+
    | id | name   |
    +----+--------+
    |  2 | second |
    |  3 | third  |
    |  1 | first  |
    +----+--------+
    3 rows in set (0.00 sec)


Maintaining order in MySQL &quot;IN&quot; query

I read a book which referred to the .net **CLR as a virtual machine**? Can anyone justify this? What is the reason we need the concept of virtual machines on some development platforms?

Isn&#39;t it possible to develop a native framework [one without virtual machine] that is fully object oriented and as powerful as .net?

The book which refers to CLR as virtual machine is &quot;**Professional .Net Framework 2.0**&quot;.

Is the CLR a virtual machine?

I&#39;m sure it&#39;s a simple one-liner, but I can&#39;t seem to find it.

How can I use a different layout file for a particular action?

**Update:** This worked for me, thanks!

    // Within controller
    $this-&gt;_helper-&gt;_layout-&gt;setLayout(&#39;other-layout&#39;) //other-layout.phtml

    //Within view script
    &lt;?php $this-&gt;layout()-&gt;setLayout(&#39;other-layout&#39;); ?&gt;

How to switch layout files in Zend Framework?

I'm sure it's a simple one-liner, but I can't seem to find it.
How can I use a different layout file for a particular action?
Update: This worked for me, thanks!
<pre><code class="hljs language-php">// Within controller
$this->_helper->_layout->setLayout('other-layout') //other-layout.phtml

//Within view script
&#x3C;?php $this->layout()->setLayout('other-layout'); ?>
</code></pre>

If I had an array like:

    $array[&#39;foo&#39;] = 400;
    $array[&#39;bar&#39;] = &#39;xyz&#39;;

And I wanted to get the first item out of that array without knowing the key for it, how would I do that? Is there a function for this?

How to get the first item from an associative PHP array?

How can we round off a number to the nearest 10 in php?

Say I have `23`, what code would I use to round it off to `30`?

How to round up a number to nearest 10?

I saw different binaries for PHP, like non-thread or thread safe?

What does this mean? 

What is the difference between these packages?


What is thread safe or non-thread safe in PHP?

What are the best current libraries/methods to authenticate users without the use of a CMS or heavy framework?

Responses should include suggestions for anything you think should be considered the standard for new PHP development involving user authentication.

PHP best practices for user authentication and password security

This code produces the result as 56.

	function x ($y) {
		function y ($z) {
			return ($z*2);
	 	}

		return($y+3);
	}

	$y = 4;
	$y = x($y)*y($y);
	echo $y;


Any idea what is going inside? I am confused.  

Function inside a function.?

I would like to run a Zend Framework action to generate some files, from command line. Is this possible and how much change would I need to make to my existing Web project that is using ZF?

Thanks!

Running a Zend Framework action from command line

I like how this works in Zend Framework. I can know which environment I&#39;m currently using by checking APPLICATION_ENV constant in my controller. 

    &lt;VirtualHost *:80&gt;
    	#ServerName 
    	#DocumentRoot
        
            SetEnv APPLICATION_ENV &quot;development&quot;
    
    	# Directory
    &lt;/VirtualHost&gt;

But unfortunately I can&#39;t use ZF in my current project. How can I check this environment variable in my PHP code?

Set Application_ENV via virtual host config and read this in PHP

I have started to learn Zend Framework with the Book &quot;Zend Framework in Action&quot; in German.

Right there where it starts to get interesting, my PHP Unit Test throws this Error:
&quot;Zend_Db_Adapter_Exception: SQLSTATE[HY000] [2002] No such file or directory&quot;

I can&#39;t find any hints through Google searches. I did everything like it is in the book.
Can anyone give me a hint as to where to search for the fault?

Is this a common beginner mistake?

Starting with Zend Tutorial - Zend_DB_Adapter throws Exception: &quot;SQLSTATE[HY000] [2002] No such file or directory&quot;

I have two `*.sql` files that I use when creating a new web site database.  The first file creates all the tables.  The second file populates some default records.  I would like to execute these files from PHP.  I also use the Zend_Framework, if that will help accomplish this.

**Additional Info**

  1. I don&#39;t have console access
  1. I&#39;m trying to automate site generation from within our application.

**SOLUTION**

Using `shell_exec()`...

    $command = &#39;mysql&#39;
            . &#39; --host=&#39; . $vals[&#39;db_host&#39;]
            . &#39; --user=&#39; . $vals[&#39;db_user&#39;]
            . &#39; --password=&#39; . $vals[&#39;db_pass&#39;]
            . &#39; --database=&#39; . $vals[&#39;db_name&#39;]
            . &#39; --execute=&quot;SOURCE &#39; . $script_path
    ;
    $output1 = shell_exec($command . &#39;/site_db.sql&quot;&#39;);
    $output2 = shell_exec($command . &#39;/site_structure.sql&quot;&#39;);

...I never did get useful output, but followed [some suggestions][1] on [another thread][2] and finally got it all working.  I switch to the `--option=value` format for the commands and used `--execute=&quot;SOURCE ...&quot;` instead of `&lt;` to execute the file.

Also, I never got a good explanation of the difference between `shell_exec()` and `exec()`.


  [1]: https://stackoverflow.com/questions/814586/calling-mysql-exe-using-php-exec-doesnt-work/814722#814722
  [2]: https://stackoverflow.com/questions/814586/calling-mysql-exe-using-php-exec-doesnt-work

Running MySQL *.sql files in PHP

I have the following code:


     $final = array();
        foreach ($words as $word) {
            $query = &quot;SELECT Something&quot;;
            $result = $this-&gt;_db-&gt;fetchAll($query, &quot;%&quot;.$word.&quot;%&quot;);
            foreach ($result as $row)
            {
                $id = $row[&#39;page_id&#39;];
                if (!empty($final[$id][0]))
                {
                    $final[$id][0] = $final[$id][0]+3;
                }
                else
                {
                    $final[$id][0] = 3;
                    $final[$id][&#39;link&#39;] = &quot;/&quot;.$row[&#39;permalink&#39;];
                    $final[$id][&#39;title&#39;] = $row[&#39;title&#39;];
                }
            } 
        }

The code SEEMS to work fine, but I get this warning:

    Warning: Cannot use a scalar value as an array in line X, Y, Z (the line with: $final[$id][0] = 3, and the next 2).

Can anyone tell me how to fix this?

PHP - cannot use a scalar as an array warning

I have a `res/layout/main.xml` including these elements and others:

    &lt;some.package.MyCustomView android:id=&quot;@+id/foo&quot; (some other params) /&gt;
    &lt;TextView android:id=&quot;@+id/boring&quot; (some other params) /&gt;

In my Activity&#39;s onCreate, I do this:

    setContentView(R.layout.main);
    TextView boring = (TextView) findViewById(R.id.boring);
    // ...find other elements...
    MyCustomView foo = (MyCustomView) findViewById(R.id.foo);
    if (foo == null) { Log.d(TAG, &quot;epic fail&quot;); }

The other elements are found successfully, but `foo` comes back null. MyCustomView has a constructor `MyCustomView(Context c, AttributeSet a)` and a `Log.d(...)` at the end of that constructor appears successfully in logcat just before the &quot;epic fail&quot;.

Why is `foo` null?

findViewById() returns null for custom component in layout XML, not for other components

I have two images that I want to display on a page as figures. Each eats up little less than half of the space available so there&#39;s not much room for any other stuff on that page, but I know there is enough space for both of the figures. I tried to place the figures with [ht] and [hb], both [h] and both [ht] but still I can&#39;t get those two images on the same page but instead at least few paragraphs between them. 

How do I force those two figures to stay on the same page?

How to force two figures to stay on the same page in LaTeX?

I would like to add an image in my template for my ruby on rails project where i currenly have the code `&lt;img src=&quot;../../../public/images/rss.jpg&quot; alt=&quot;rss feed&quot; /&gt;` in a the layout `stores.html.erb` file however this doesn&#39;t seem to load as it looks like its missing a route which i&#39;m not sure what its supposed to be. 

Any ideas please?

Add image to layout in ruby on rails

I don&#39;t know if it&#39;s a common problem, but I can&#39;t find the solution in web so far.
I would like to have two divs wrapped inside another div, however these two divs inside have to be align the same level (for example: left one takes 20%width of the wrappedDiv, right one take another 80%). To achieve this purpose, I used the following example CSS. However, now the wrap DIV didn&#39;t wrap those divs all. The wrap Div has a small height than those two divs contained inside. How could I make sure that the wrap Div has the largest height as the contained divs? Thanks!!!
     
    &lt;!DOCTYPE HTML PUBLIC &quot;-//W3C//DTD HTML 4.01 Transitional//EN&quot;
            &quot;http://www.w3.org/TR/html4/loose.dtd&quot;&gt;
    &lt;html lang=&quot;en&quot;&gt;
    &lt;head&gt;
    	&lt;meta http-equiv=&quot;content-type&quot; content=&quot;text/html; charset=iso-8859-1&quot;&gt;
    	&lt;title&gt;liquid test&lt;/title&gt;
    	&lt;style type=&quot;text/css&quot; media=&quot;screen&quot;&gt;
    		body
    		{
    			margin: 0;
    			padding: 0;
    			height:100%;
    		}
    		#nav
    		{
    			float: left;
    			width: 25%;
    			height: 150px;
    			background-color: #999;
    			margin-bottom: 10px;
    		}
    		
    		#content
    		{
    			float: left;
    			margin-left: 1%;
    			width: 65%;
    			height: 150px;
    			background-color: #999;
    			margin-bottom: 10px;
    		}		
    		#wrap
    		{
    		  background-color:#DDD;
    		  height:100%;
    		}
    		
    &lt;/style&gt;
    &lt;/head&gt;
    &lt;body&gt;
    &lt;div id=&quot;wrap&quot;&gt;
        &lt;h1&gt;wrap1 &lt;/h1&gt;
        &lt;div id=&quot;nav&quot;&gt;&lt;/div&gt;
        &lt;div id=&quot;content&quot;&gt;&lt;a href=&quot;index.htm&quot;&gt;&amp;lt; Back to article&lt;/a&gt;&lt;/div&gt;
    &lt;/div&gt;
    &lt;/body&gt;
    &lt;/html&gt;

how to make a div to wrap two float divs inside?

How can I prevent automatic line breaks in a column of table (not a single cell)?

Content Type	Original Author	Original Content on Stackoverflow
Question	Andrew	View Question on Stackoverflow
Solution 1 - Php	Josh Lindsey	View Answer on Stackoverflow
Solution 2 - Php	Md Moin Uddin	View Answer on Stackoverflow

How to switch layout files in Zend Framework?

Php Problem Overview

Php Solutions

Solution 1 - Php

Solution 2 - Php

Is the CLR a virtual machine?

Maintaining order in MySQL "IN" query

Attributions