String replace multiple values

Question

I have a string that looks like this:

Bla bla %yada yada% bla bla %yada yada%

Is there a way to replace only the first two "%" (or the last two) so I can get the next output:

Bla bla <a href='link1'>yada yada</a> bla bla %yada yada%

and also, if necessary, the last two "%" so it outputs:

Bla bla <a href='link1'>yada yada</a> bla bla <a href='link2'>yada yada</a>

I can't figure out how to make the distinction between the first two and the last two so, if I want, I can be able to replace either the first or the last two marks "%" with a link. I'm using php. Thanks in advance

Regards

With your link identifier being %, it can cause a lot of errors a lot of different ways. You can get creative with the strpos function for it though. — Jon
– Jon, Commented Apr 24, 2013 at 8:04
u have to find first occurence then replace this and then find second occurnce then replace this — Mayur Kukadiya
– Mayur Kukadiya, Commented Apr 24, 2013 at 8:04
@Jon I don't necessarily have to use "%". I can mark it any way I want. — Kapn0batai
– Kapn0batai, Commented Apr 24, 2013 at 8:06
Then you should come up with something that you are going to stick to that is more unique and you can create a good regular expression from. BB Code was a good example of how to format something in that fashion and allow for expression replacement. ^^ — Jon
– Jon, Commented Apr 24, 2013 at 8:08

HamZa · Accepted Answer · 2013-04-24 08:24:29Z

3

Using regex (PHP 5.3+ required) :

$string = 'Bla bla %yada yada% bla bla %yada yada%';
echo preg_replace('/%([^%]*)%/', '<a href="http://example.com">$1</a>', $string, 1) . '<br>'; // to replace the first instance.
// ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ //
$links = array('http://example.com', 'http://stackoverflow.com', 'http://google.com');
$index = 0;
echo preg_replace_callback('/%([^%]*)%/', function($m) use($links, &$index){
    $m[1] = '<a href="'.$links[$index].'">'.$m[1].'</a>';
    $index++;
    // reset the index if it exceeds (N links - 1)
    if($index >= count($links)){
        $index = 0;
    }
    return $m[1];
}, $string).'<br>'; // to replace according to your array
// ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ //
// To test with a string that contains more %% than the links
$string2 = 'Bla bla %yada yada% bla bla %yada yada% wuuut dsfsf %yada yada% sjnfsf %yada yada% jnsfds';
$links = array('http://example.com', 'http://stackoverflow.com', 'http://google.com');
$index = 0;
echo preg_replace_callback('/%([^%]*)%/', function($m) use($links, &$index){
    $m[1] = '<a href="'.$links[$index].'">'.$m[1].'</a>';
    $index++;
    // reset the index if it exceeds (N links - 1)
    if($index >= count($links)){
        $index = 0;
    }
    return $m[1];
}, $string2).'<br>'; // to replace according to your array

Online demo.

edited Apr 24, 2013 at 8:24

answered Apr 24, 2013 at 8:06

HamZa

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5 Comments

Jules Colle Over a year ago

Works, but on some (older) servers you will get the T_FUNCTION error. To fix this you will need to name your function an then call it like this: preg_replace_callback('/%([^%]*)%/','my_function',$string). Example: codepad.org/zUMXvbEN

HamZa Over a year ago

@Jules I know, this is assuming you're using PHP 5.3+

Kapn0batai Over a year ago

In this manner wouldn't I have to know the text between the %% ? Because that is my problem... I can't mess with the actual text since there are many strings in many different languages. All I have to do is make it linkable.

HamZa Over a year ago

@Kapn0batai You wouldn't have to know what's between the %% but you will have unwanted results if it contains a % !

HamZa Over a year ago

@Kapn0batai If you're not sure whether there is a % between %%, then you maybe should take the BB code approach like Jon said. With that said, I have a little (close) example here

Community · Accepted Answer · 2023-11-17 19:52:31Z

0

Try this
Support for PHP 4 and PHP 5

Solution:

$string ='Bla bla %yada yada% bla bla %yada yada%';

// Count no of %
$count = substr_count($string,'%');

// Valid string pattern
if ( 0 == ($count % 2) ) {

    $urlString = $string;
    
    // Iterate for each pair of % to make it link
    for ( $i=1; $i <= $count/2 ; $i++  ) {
    
        $urlString = preg_replace('/%/', "<a href='link$i'>", $urlString, 1);
        $urlString = preg_replace('/%/', "</a>", $urlString, 1);
    }
}
// Invalid string pattern
else {
    echo "Invalid string pattern";
}

// Display generated link 
echo $urlString;

Working of preg_replace function

Remove first two %

    $str ='Bla bla %yada yada% bla bla %yada yada%';
    $newStr = preg_replace('/%/', '', $str, 2);
    
    echo $newStr;
    
    // Output => Bla bla yada yada bla bla %yada yada%

Remove last two %

    $str ='Bla bla %yada yada% bla bla %yada yada%';
    $newStr = preg_replace('/%/', '', strrev($str), 2);
    $newStr =  strrev($newStr);
    
    echo $newStr;
    
    // Output => Bla bla %yada yada% bla bla yada yada

Remove all %

    $str ='Bla bla %yada yada% bla bla %yada yada%';
    $newStr = preg_replace('/%/', '', $str);
    
    echo $newStr;
    
    // Output => Bla bla yada yada bla bla yada yada

Reference
https://www.php.net/preg%5Freplace

edited Nov 17, 2023 at 19:52

CommunityBot

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answered Apr 24, 2013 at 8:03

Navnath Godse

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2 Comments

Jon Over a year ago

There's no pattern, and it doesn't allow for the opening or closing of the a tag.

Jon Over a year ago

If only a link, should be a comment.

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