PHP preg_replace string path given

Question

I have issue with preg_replace function in PHP. I cant figure out a pattern and a replacement.

I have this two strings and some code:

    $dirname1 = 'hdadas/dasdad/dasd/period_1min';
    $dirname2 = 'hdadas/dasdad/dasd/period_1min/abcdrfg.php';

    $pieces1 = explode('/', $dirname1);
    $pieces2 = explode('/', $dirname2);

    $dirname1 = end($pieces1);  // output will be period_1min
    $dirname2 = end($pieces2); // output will be abcdrfg.php

    $output = preg_replace($pattern, $replacement, $dirname1); // or (..,..,$dirname2

    echo $output; // i need 1min(without period_) or abcdrfg (without .php)

UPD:

   function Cat($dirname)
   {
      $name = explode('/', $dirname);

      $pattern = ???;
      $replacement = ???;

      return preg_replace($pattern, $replacement, $dirname1);
    }

    print(Cat('hdadas/dasdad/dasd/period_1min'))); // output need 1min only
    print(Cat('hdadas/dasdad/dasd/period_1min/abcdrfg.php'))); // output  need abcdrfg only

1. The first 8 lines is the same as: $dirname1 = basename($dirname1); 2. Which pattern does this follow when you want period_ for the first and abcdrfg for the second ? — Rizier123
– Rizier123, Commented Mar 18, 2015 at 12:36
i cant figure out what to write in $pattern and $replacement — Maris
– Maris, Commented Mar 18, 2015 at 12:37

Rizier123 · Accepted Answer · 2015-03-18 12:49:14Z

2

This should work for you:

(Here I use basename() to get only the last part of the path, then I use pathinfo() to get the last part without the extension. After this I just use preg_replace() to replace everything before the underscore with an empty string)

<?php

    $dirname1 = "hdadas/dasdad/dasd/period_1min";
    $dirname2 = "hdadas/dasdad/dasd/period_1min/abcdrfg.php";

    $dirname1 = pathinfo(basename($dirname1))["filename"];
    $dirname2 = pathinfo(basename($dirname2))["filename"];

    echo $output = preg_replace("/(.*)_/", "", $dirname1); 


?>

output:

1min
abcdrfg

answered Mar 18, 2015 at 12:49

Rizier123

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6 Comments

Rizier123 Over a year ago

@Undertaker echo phpversion(); = ? <5.4 ?

Maris Over a year ago

Yes 5.3.3 that why :)

Rizier123 Over a year ago

@Undertaker Maybe you want to upgrade your php version, we live in 2015, so you can use array dereferencing

Maris Over a year ago

I cant project run on < 5.4

Rizier123 Over a year ago

@Undertaker Then you can change these lines from: $dirname1 = pathinfo(basename($dirname1))["filename"]; to this: $dirname1 = pathinfo(basename($dirname1)); $dirname1 = $dirname1["filename"]

|

Toto · Accepted Answer · 2015-03-18 13:10:48Z

1

How about this regex /^.+_|\.[^.]+$/:

$dirname1 = 'hdadas/dasdad/dasd/period_1min';
$dirname2 = 'hdadas/dasdad/dasd/period_1min/abcdrfg.php';

$pieces1 = explode('/', $dirname1);
$pieces2 = explode('/', $dirname2);

$dirname1 = end($pieces1);  // output will be period_1min
$dirname2 = end($pieces2); // output will be abcdrfg.php

$output = preg_replace('/^.+_|\.[^.]+$/', '', $dirname1); // or (..,..,$dirname2
echo $output,"\n"; // i need 1min(without period_) or abcdrfg (without .php)

$output = preg_replace('/^.+_|\.[^.]+$/', '', $dirname2); // or (..,..,$dirname2
echo $output,"\n"; // i need 1min(without period_) or abcdrfg (without .php)

Output:

1min
abcdrfg

edited Mar 18, 2015 at 13:10

answered Mar 18, 2015 at 12:52

Toto

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7 Comments

Maris Over a year ago

Thank you, but it outputs abcdrfg.php (with .php) on version 5.3.3

Toto Over a year ago

@Undertaker: Sorry, typo I forget the + quantifier in the second example. See my edit, it works now.

Maris Over a year ago

How the regex look like if it will be period_week_sunday?

Toto Over a year ago

@Undertaker: what do you want to keep?

Toto Over a year ago

@Yes, you're right, change the the regex to: ^[^_]+_|\.[^.]+$

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