Using preg_replace with a function

[BradleySmith]

Hey, I am trying to write a sort of replace function where {user:1} is converted into a set of information from a database using preg_replace. So I have created the following preg_replace:

 

$patterns = array();
$patterns[1] = '/\{user:([0-9]+)\}/';

$replacements = array();
$replacements[1] = UserCredit('$1');

ksort($patterns);
ksort($replacements);

$content_show = preg_replace($patterns, $replacements, $content);

echo $content_show;

 

and then the function UserCredit which is just above that on the page:

 

function UserCredit($user_id) {
require 'connect.php';
	$user_lookup = mysql_query("SELECT * FROM prop_users WHERE ID = '$user_id'");
	while($row = mysql_fetch_array($user_lookup)) {
		$user_name = $row['full_name'];
		$user_file = $row['credit_file'];
		$user_url = $row['link_url'];
		$user_url_alt = $row['link_alt'];
	}
	if(mysql_num_rows($user_lookup)==0){
	  die("An error has occured, Sorry for any Inconvenience Caused! User ID=".$user_id);
	}
require 'close.php';
$output = "<li class=\"clearfix\"><div class=\"credimg\"><img src=\"".$user_file."\" width=\"75px\" height=\"75px\" alt=\"".$user_name."\" /></div><p><strong><a href=\"".$user_url."\" title=\"".$user_url_alt."\">".$user_name."</a></strong><br /><em>Writing / Code</em></p></li>";
return $output;
}

 

I have tested the function and it works when I echo UserCredit('1');

 

And the preg_replace works at simply extracting the value for the user.

 

The only problem is that I can't figure a way for the preg_replace to carry the value of $1 into the function it just literally carries "$1"

 

Is there something obvious that I am missing or am I going around this the wrong way?

[BlueSkyIS]

single quotes means literal string

 

$replacements[1] = UserCredit($1);

[BradleySmith]

Parse error: syntax error, unexpected T_LNUMBER, expecting T_VARIABLE or '$'

 

It's calling to the variable $1 which is invalid, and I don't want to call the variable, I want to use the preg_replace value.

[AbraCadaver]

single quotes means literal string

 

$replacements[1] = UserCredit($1);

 

Actually, in this case the $1 is a regex backreference so it would be OK.  Here is the problem:

 

When you do this, usercredit() is immediately called and sent the literal string '$1', so it will always be the same:

 

$replacements[1] = UserCredit('$1');

 

What you want to do is have literal PHP code assigned and then use the e modifier in your pattern to actually have preg_replace() execute the code with the substituted backreference:

 

$patterns[1] = '/\{user:([0-9]+)\}/e';

$replacements[1] = 'UserCredit("$1")';

 

You can also look at preg_replace_callback().

[BradleySmith]

Thank You so much! It works.