php string matching with wildcard *?

Question

I want to give the possibility to match string with wildcard *.

Example

$mystring = 'dir/folder1/file';
$pattern = 'dir/*/file';

stringMatchWithWildcard($mystring,$pattern);  //> Returns true

Example 2:

$mystring = 'string bl#abla;y';
$pattern = 'string*y'; 

stringMatchWithWildcard($mystring,$pattern);  //> Returns true

I thought something like:

function stringMatch($source,$pattern) {
    $pattern = preg_quote($pattern,'/');        
    $pattern = str_replace( '\*' , '.*?', $pattern);   //> This is the important replace
    return (bool)preg_match( '/^' . $pattern . '$/i' , $source );
}

Basically replacing * to .*? (considering in *nix environment * matches empty string) ©vbence

Any improvments/suggests?

// Added return (bool) because preg_match returns int

mario · Accepted Answer · 2011-05-28 18:30:20Z

62

There is no need for preg_match here. PHP has a wildcard comparison function, specifically made for such cases:

fnmatch()

And fnmatch('dir/*/file', 'dir/folder1/file') would likely already work for you. But beware that the * wildcard would likewise add further slashes, like preg_match would.

answered May 28, 2011 at 18:30

mario

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2 Comments

dynamic Over a year ago

hi mario. fnmatches has some other feature like []. I just need * as "special char"

Pinoniq Over a year ago

fnmatch is not supported on non-POSIX systems. (I see that php5.3+ also supports windows now). So this is not always the best way

Dogbert · Accepted Answer · 2011-05-28 17:12:01Z

4

You should just use .* instead.

$pattern = str_replace( '*' , '.*', $pattern);   //> This is the important replace

Edit: Also your ^ and $ were in the wrong order.

<?php

function stringMatchWithWildcard($source,$pattern) {
    $pattern = preg_quote($pattern,'/');        
    $pattern = str_replace( '\*' , '.*', $pattern);   
    return preg_match( '/^' . $pattern . '$/i' , $source );
}

$mystring = 'dir/folder1/file';
$pattern = 'dir/*/file';

echo stringMatchWithWildcard($mystring,$pattern); 



$mystring = 'string bl#abla;y';
$pattern = 'string*y'; 

echo stringMatchWithWildcard($mystring,$pattern);

Working demo: http://www.ideone.com/mGqp2

edited May 28, 2011 at 17:12

answered May 28, 2011 at 17:05

Dogbert

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6 Comments

dynamic Over a year ago

I don't know if I want to give the possibility to match this 'dir//file' with 'dir/*/file'

vbence Over a year ago

.* is greedy, it will eat all the rest of your string, no just the minimal number of characters to continue with the patters. So if you have anything after .* the result will be always false.

Dogbert Over a year ago

I thought that was exactly what you wanted to do with .+? from your description.

Dogbert Over a year ago

@vbence preg_match('/.*a/', 'aa') ?

vbence Over a year ago

Ok it seems match will give the same results.

|

vbence · Accepted Answer · 2011-05-28 17:07:50Z

3

.+?

Causes non-greedy matching for all characters. This is NOT equal to "*" becuase it will not match the empty string.

The following pattern will match the empty string too:

.*?

so...

stringMatchWithWildcard ("hello", "hel*lo"); // will give true

answered May 28, 2011 at 17:07

vbence

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7 Comments

dynamic Over a year ago

do you think I should allow for empty string? Consider dir too: 'dir//file' matching 'dir/*/file' ?

vbence Over a year ago

If you want it to work analogous to how OS shells work, then you have to allow for the empty string. Just try echo > hello followed by dir hello* in windows or ls hello* in *nix. In this context * matches the empty string.

vbence Over a year ago

It would be polite to give a comment with your -1.

dynamic Over a year ago

ah nice! thanks for the *nix comparsion! (also i didnt -1 your answer, no reason for me to do that)

vbence Over a year ago

@yes123 Yep, I know, not accusing. :)

|

netcoder · Accepted Answer · 2011-05-28 17:12:43Z

2

You're mixing up ending ($) and beginning (^). This:

preg_match( '/$' . $pattern . '^/i' , $source );

Should be:

preg_match( '/^' . $pattern . '$/i' , $source );

answered May 28, 2011 at 17:12

netcoder

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Comments

Spudley · Accepted Answer · 2011-05-28 17:11:30Z

0

The one problem you'll have is that the call to preg_quote() will escape the asterisk character. Given that, your str_replace() will replace the *, but not the escape character in front of it.

Therefore you should change the str_replace('*' ..) with str_replace('\*'..)

edited May 28, 2011 at 17:11

answered May 28, 2011 at 17:05

Spudley

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1 Comment

dynamic Over a year ago

i missed that from the documetnatnion my bad. fixed

Tim13 · Accepted Answer · 2019-02-25 08:52:58Z

0

$pattern = str_replace( '\*' , '.+?', $pattern);   // at least one character

answered Feb 25, 2019 at 8:52

Tim13

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1 Comment

Nico Haase Over a year ago

Please add some explanation to your code - have a look at the other answers to this question for some inspiration

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