Preg_Replace: Add Number After Backreference

preg_replace: add number after backreference

The solution is to wrap the backreference in ${}.

$out = preg_replace( '/([aeiou])/', '${1}8',  $in);

which will output a8bcde8fghi8j

See the manual on this special case with backreferences.

RegEx backreference followed by numbers

I found my answer here

Turns out all I had to do was to wrap the back-reference in curly brackets

Solution

Sublime: Find What: ([a-zA-Z]*)2013 Replace With: ${1}2014

PHP: preg_replace('~([a-zA-Z]*)2013~', '${1}2014', $file_contents);

Named backreferences with preg_replace

They exist:

http://www.php.net/manual/en/regexp.reference.back-references.php

With preg_replace_callback:

function my_replace($matches) {
    return '/user/profile/' . $matches['id'];
}
$newandimproved = preg_replace_callback('#^user/(?P<id>[^/]+)$#Di', 'my_replace', $string);

Or even quicker

$newandimproved = preg_replace('#^user/([^/]+)$#Di', '/user/profile/$1', $string);

Avoid backreference replacement in php's preg_replace

Escape the dollar character before using a character translation (strtr):

$repValue = strtr('$0.0 $00.00 $000.000 $1.1 $11.11 $111.111', ['$'=>'\$']);

For more complicated cases (with dollars and escaped dollars) you can do this kind of substitution (totally waterproof this time):

$str = strtr($str, ['%'=>'%%', '$'=>'$%', '\\'=>'\\%']);
$repValue = strtr($repValue, ['%'=>'%%', '$'=>'$%', '\\'=>'\\%']);
$pattern = '/{{' . strtr($field, ['%'=>'%%', '$'=>'$%', '\\'=>'\\%']) . '}}/';
$str = preg_replace($pattern, $repValue, $str );
echo strtr($str, ['%%'=>'%', '$%'=>'$', '\\%'=>'\\']);

Note: if $field contains only a literal string (not a subpattern), you don't need to use preg_replace. You can use str_replace instead and in this case you don't have to substitute anything.

Capture Group Not Captured in Preg_replace Replacement

In fact, the issue is the concatenation of the capture groups with time(). Use this version instead:

$f = 'IMG_1474.PNG';
$output = preg_replace('/(.*)([.][^.]+$)/', '${1}' . time() . '${2}', $f);
echo $output;

This prints:

IMG_14741604148316.PNG
        ^^^^^^^^^^  second since epoch (1970-01-01)

To understand why this is happening, we have to consider that variable substitution occurs before the preg_replace function call does. So your concatenated replacement term:

'$1' . time() . '$2'

actually first becomes:

'$1' . '1604148623' . '$2'

which is:

'$11604148623' . '$2'

The "first" capture group here is probably being interpreted as $11 (or $116, etc.), and is not defined. This is why you are currently only getting the second capture group, but the first one is empty. The solution I suggested uses ${1} which preserves the capture group reference so it is available during the preg_replace call as you intend.

PHP preg_replace add tag to digit in string

You would want to use a backreference, like this:

$formula = preg_replace('/(\d+)/','<sub>\1</sub>',$formula);

I added the // as delimiters to make it a proper pattern. I also added a + so that it would place all connecting digits in the same <sub> tag.

Using preg_replace to back reference array key and replace with a value

You need to use the /e eval flag, or if you can spare a few lines preg_replace_callback.

  $string = preg_replace(
     '|http://mysite.com/script.php\?fruit=([a-zA-Z0-9_-]*)|e',
     ' "http://mysite.com/" . $fruitArray["$1"] ',
     $string
  );

Notice how the whole URL concatenation expression is enclosed in single quotes. It will be interpreted as PHP expression later, the spaces will vanish and the static URL string will be concatenated with whatever is in the fruitArray.

PHP preg_replace Backreferencing

\w is only [A-Za-z0-9_] so you need to handle other characters, spaces etc. Either add all the types you need or maybe match everything up to ' %>'.

You can use preg_replace and then get the file size afterwards or use preg_replace_callback.

A couple of examples:

$oldStr = "<% size_../files/file with spaces.rar %>";

$size1 = preg_replace_callback("/<% size_([\w\.\/\s]+) %>/", 
    function ($matches) {
        return filesize($matches[1]);
    }, 
    $oldStr);

$newStr = preg_replace("/<% size_(.+) %>/", "$1", $oldStr);
if (is_file($newStr)) {
    $size2 = filesize($newStr);
} else {
    //not a file
}

PHP: preg_replace replace abbreviation using backreferences

The idea is to find all single letter + dot \b[a-z]\. followed by an other single letter + dot.

you can use:

$txt = preg_replace('~\b[a-z]\.(?=[a-z]\.)~i', '$0\,', $txt);

where (?=..) is a lookahead that performs only a check (followed by)

Second solution:

$txt = preg_replace('~(?<=\b[a-z]\.)(?=[a-z]\.)~i', '\,', $txt);

Instead of a backreference to the whole match, I use a lookbehind (?<=..). Nothing is captured and the regex engine is at the good offset to add \,

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