CODEWITHSUNDEEP
phppreg-replace
abe1432181
abe1432181

Reputation: 1374

php preg_replace strings with brackets

I am trying to replace some text if it isn't preceded with a dot (.), but my function seems to result include an unexpected slash ...

<?php

$test="function insertXYZ(e){e.insertXYZ()};";
$search="insertXYZ(";
$replace="zzzz(";

$search=str_replace("(", '\(', $search);
//echo $search."\n\n";
$pattern = '\b' . $search;
$replaceException="999";

$test=preg_replace('/' . "\.".$pattern . '/', "$replaceException", $test);
$test=preg_replace('/' . $pattern . '/', $replace, $test);
$test=preg_replace('/' . $replaceException . '/', ".$search", $test);

echo $test;
?>

A sample of this code can be found at http://sandbox.onlinephpfunctions.com/code/24cab4eece20d22a11dd887da44d63e393b51aa9

which outputs...

function zzzz(e){e.insertXYZ\()};
                            ^

but I want the output to read...

function zzzz(e){e.insertXYZ()};
                            ^

Where am I going wrong?

Upvotes: 0

Views: 574

Answers (1)

Lucan
Lucan

Reputation: 3645

It's because in this line, you're adding the backslash and then never remove it

$search=str_replace("(", '\(', $search);

You could change your code to the following so that you don't need to manually escape this bracket.

// $search=str_replace("(", '\(', $search); <-- Removed
$pattern = '\b' . preg_quote($search);

However, you could get the same output with much less code

$testString = "function insertXYZ(e){e.insertXYZ()};";
$regex = "/[^.](insertXYZ)/s";
echo(preg_replace($regex, " zzzz", $testString));

[^.] - Character group looking where there's NOT a dot character
(insertXYZ) - matches 'insertXYZ' exactly (case sensitive)
/s - Single line modifier so there's no need to escape the dot character

Upvotes: 2

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