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Parsing PHP variable within highlight_string syntax display

Posted on 2009-04-10
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Last Modified: 2012-05-06
Hiya
The code below shows a variable $form_code in the value for $mycode.

$mycode is echoed as syntax for copy and paste and this script is to aventually become a form builder where the output is a dynamic PHP and HTML form. Something on another level to what I've done but I think I can do it.

So far I've only been able to get an error or the actual variable $form_code being displayed on the output. Is it possible to parse $form_code so that the output would end up being...

<form></form>
                         <?php
                        echo "PHP will be displayed";
                  ?><h1>and HTML will be displayed</h1>
//This file is to test how I create some HTML with PHP mixed in then output it for use as normal script
 
	$a = "1";
	$b = "2";
	
	if($a == "1")
	{
		$form_code = "<form>";
	}
	
	if($b == "2")
	{
		$form_code .= "</form>";
	}
	
	//This can be used to make the final display of code on the page
	$mycode='$form_code <?php 
				echo "PHP will be displayed"; 
			?><h1>and HTML will be displayed</h1>'; 
	echo highlight_string($mycode,true);

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Question by:Ryan Bayne
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6 Comments
 
LVL 19

Expert Comment

by:Michael701
ID: 24116634
look at using <pre> and </pre> around the php code you wish to display. Mush like um THIS page.

Michael
0
 
LVL 9

Expert Comment

by:Mahdii7
ID: 24119013
Single Quotes.

You can't include a PHP variable in single quotes and have it output the contents of the variable. Either append it with concatenation ($var . 'some stuff') or use double quotes ("$var some stuff")

See snippet:
<?php
//This file is to test how I create some HTML with PHP mixed in then output it for use as normal script
 
        $a = "1";
        $b = "2";
        
        if($a == "1")
        {
                $form_code = "<form>";
        }
        
        if($b == "2")
        {
                $form_code .= "</form>";
        }
        
        //This can be used to make the final display of code on the page
        $mycode = $form_code . '<?php 
                                echo "PHP will be displayed"; 
                        ?><h1>and HTML will be displayed</h1>'; 
        echo highlight_string($mycode,true);
		
		?>

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LVL 9

Expert Comment

by:Mahdii7
ID: 24119020
Also, if you didn't want the $form_code to be escaped, you would need to output it separately:
<?php
//This file is to test how I create some HTML with PHP mixed in then output it for use as normal script
 
        $a = "1";
        $b = "2";
        
        if($a == "1")
        {
                $form_code = "<form>";
        }
        
        if($b == "2")
        {
                $form_code .= "</form>";
        }
        
        //This can be used to make the final display of code on the page
        $mycode = '<?php 
                                echo "PHP will be displayed"; 
                        ?><h1>and HTML will be displayed</h1>'; 
        echo $form_code . highlight_string($mycode,true);
		
		?>

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LVL 2

Author Comment

by:Ryan Bayne
ID: 24121828
Good ok that all works for me but I plan to get slighly more complicated and have script thats very mixed with html  and php.

What about this example? The PHP variable in the 2nd if statement shows as blank but I'e not been able to make it be display as PHP.

This is all for a form builder by the way. Using values stored by a wizard type form creatore my script will go through a series of if statements building the form as it goes. It will be a heavy mix of HTML and PHP.

My code here is slowly working towards putting the whole form into a single variable then displaying it for copying. I'm wondering if it would be better to echo each part of the code as the script runs. Outputting each line with echo I just think it would be a slower.

What do you think?
<?php 
        $a = "1";
        $b = "2";
        
        if($a == "1")
        {
                $form_code = "<h1>HTML Only example</h1>";
        }
        
        if($b == "2")
        {
                $form_code .= '<?php echo $php_example;?>';
        }
        
        //This can be used to make the final display of code on the page
        $mycode = $form_code; 
        echo highlight_string($mycode,true);            
?>

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LVL 9

Accepted Solution

by:
Mahdii7 earned 1200 total points
ID: 24123769
I get your problem. Unfortunately what you probably need is the exact opposite of "eval()" which doesn't exist in PHP. The best approach may be a class that handles data, to clean up syntax structure. Note the " (double) and ' (single) quotes though - they are still needed!
<?php 
class PHPOut {
	var $file_name;
	var $file_resource;
	
	function PHPOut(){
		$this->file_name = md5(rand()*time());
		if(!$this->file_resource = fopen($this->file_name, "w")){
		 die("<strong>There was an error creating the file. Check the permissions of this script! Script needs '644', directory needs '777'</strong>"); 
		}
	}
	
	function add($data){
		$data .= "\n";
		fwrite($this->file_resource, $data);
	}
	
	function output(){
		highlight_file($this->file_name);
		fclose($this->file_resource);
		unlink($this->file_name);
	}
}
 
 
$something = "Hi mom!";
 
$form = new PHPOut;
$form->add("<strong>");
$form->add('<?php $some_undefined_variable ?>');
$form->add("</strong>");
$form->add($something);
 
$output = $form->output();
?>

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0
 
LVL 2

Author Comment

by:Ryan Bayne
ID: 24124017
Ah see now I have not worked a great deal with files but this may be the project to learn it on. I'll end up going with the 2 approaches probably, better lessons to be learned that way.

Thanks for your help and that bit of code it will give me somewhere to start thinking on the file approach.
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