Cannot print variable name in PHP file

[gdebojyoti.mail96]

I want to save the following line in a file in PHP:

<?php $value=15; ?>

For this, I am using the following code:

$myFileHandle=fopen($myurl,'w');   // $myurl = URL of the file
$myData="<?php $value=".$value."; ?>"; // $value = 15
fwrite($myFileHandle,$myData);
fclose($myFileHandle);

But when I open the file, I find the following statement:

<?php 15=15; ?>

What am I doing wrong?

By the way, I am using WAMP server.

 

[essellar]

When you use double quotes as a string delimiter, variables within the string are evaluated. With single quotes, you'd get a literal "$value" in $myData instead of the evaluated version. In double quotes, you need to escape the sigil:

$myData="<?php \$value=".$value."; ?>"; // $value = 15

 

[gdebojyoti.mail96]

Thank you.

 

[misson]

Note that if you're using double-quotes, you don't need to use the concatenation operator when combining data from a variable:

$myData="<?php \$value=$value; ?>"; // $value = 15

Alternatively, if you use single quotes, you don't need to escape the dollar sign:

$myData='<?php $value=' . (int) $value. '; ?>'; // $value = 15

All in all, you'd do well to read up on PHP string syntax.

Also, be wary of injection attacks. If the data comes from user input, filter it appropriately.