Error trying to pass regex match to function - php

Question

I'm getting Syntax error, unexpected T_LNUMBER, expecting T_VARIABLE or '$'
This is the code i'm using
function wpse44503_filter_content( $content ) {
$regex = '#src=("|\')'.
'(/images/(19|20)(0-9){2}/(0|1)(0-9)/[^.]+\.(jpg|png|gif|bmp|jpeg))'.
'("|\')#';
$replace = 'src="'.get_site_url( $2 ).'"';
$output = preg_replace( $regex, $replace, $content );
return $output;
}
This is the line where i'm getting that error $replace = 'src="'.get_site_url( $2 ).'"';
Can anyone help me to fix it?
Thanks

Answer 1

You can't have '$2' as a variable name. It must start with a letter or underscore.
http://php.net/manual/en/language.variables.basics.php
Variable names follow the same rules as other labels in PHP. A valid variable name starts with a letter or underscore, followed by any number of letters, numbers, or underscores. As a regular expression, it would be expressed thus: '[a-zA-Z_\x7f-\xff][a-zA-Z0-9_\x7f-\xff]*'
Edit Above was my original answer and is the correct answer to the simple "syntax error" question. More in-depth answer below...
You are trying to use $2 to represent "the second capture group", but you haven't done anything at that point to match your regex. Even if $2 was a valid PHP variable name, it still wouldn't be set at that point in your script. Because of this, you can determine that you are using preg_replace improperly and that it may not suit your actual needs.
Note that the preg_replace documentation doesn't support using $n as a separate variable outside of the replacement operation. In other words, 'foo' . $1 . 'bar' is not a valid replacement string, but 'foo$1bar' is.
Depending on the complexity of get_site_url, you have 2 options:
If get_site_url is simply adding a root directory or server name, you could change your replacement string to src="/myotherlocation$2". This will effectively replace "/image/..." with "/myotherlocation/image/..." in the img src. This will not work if get_site_url is doing something more complex.
If get_site_url is complex, you should use preg_replace_callback per other answers. Give the documentation a read and post a new question (or I guess update this question?) if you have trouble with the implementation.

Answer 2

What you're trying to do (ie replacing the matched string with the result of a function call) can't be done using preg_replace, you'll need to use preg_replace_callback instead to get a function called for every match.
A short example of preg_replace_callback;
$get_site_url = // Returns replacement
function($row) {
return '!'.$row[1].'!'; // row[1] is first "backref"
};
$str = 'olle';
$regex = '/(ll)/'; // String to match
$output = preg_replace_callback( // Match, calling get_site_url for replacement
$regex,
$get_site_url,
$str);
var_dump($output); // output "o!ll!e"

Answer 3

PHP variable names cant begin with a number.

Answer 4

$2 is not a valid PHP variable. If you meant the second group in the regex then you want to put \2 in a string. However, since you're passing it to a function then you'll need to use preg_replace_callback() instead and substitute appropriately in the callback.

Answer 5

if PHP variable begins with number use following:
when I was getting the following as the result set from thrid party API
Code Works
$stockInfo->original->data[0]->close_yesterday
Code Failed
$stockInfo->original->data[0]->52_week_low
Solution
$stockInfo->original->data[0]->{'52_week_high'}