Remove a string from the beginning of a string

Question

I have a string that looks like this:

$str = \"bla_string_bla_bla_bla\";

How can I remove the first bla_; but only if it\'s fo

Answer 1

<?php
$str = 'bla_string_bla_bla_bla';
echo preg_replace('/bla_/', '', $str, 1); 
?>

Answer 2

Plain form, without regex:

$prefix = 'bla_';
$str = 'bla_string_bla_bla_bla';

if (substr($str, 0, strlen($prefix)) == $prefix) {
    $str = substr($str, strlen($prefix));
} 

Takes: 0.0369 ms (0.000,036,954 seconds)

And with:

$prefix = 'bla_';
$str = 'bla_string_bla_bla_bla';
$str = preg_replace('/^' . preg_quote($prefix, '/') . '/', '', $str);

Takes: 0.1749 ms (0.000,174,999 seconds) the 1st run (compiling), and 0.0510 ms (0.000,051,021 seconds) after.

Profiled on my server, obviously.

Answer 3

This will remove first match wherever it is found i.e., start or middle or end.

$str = substr($str, 0, strpos($str, $prefix)).substr($str, strpos($str, $prefix)+strlen($prefix));

Answer 4

I think substr_replace does what you want, where you can limit your replace to part of your string: http://nl3.php.net/manual/en/function.substr-replace.php (This will enable you to only look at the beginning of the string.)

You could use the count parameter of str_replace ( http://nl3.php.net/manual/en/function.str-replace.php ), this will allow you to limit the number of replacements, starting from the left, but it will not enforce it to be at the beginning.

Answer 5

Here.

$array = explode("_", $string);
if($array[0] == "bla") array_shift($array);
$string = implode("_", $array);