php - 正则表达式用大括号和引号替换字符串

Question

I see a lot of questions here regarding regex in general , but the problem is that they are usually (like mine) quite localized, and difficult to deduct if one is not a regex expert ..

My string involves characters like quotes and braces , which are quite known for making regex more difficult.

I would like to know the expression strings (search, replace) I need in order to perform this task.

in other words , in :

 ereg_replace (string pattern, string replacement, string subject) 

I will need the string pattern and string replacement expressions.

my string is

array('val' => 'something', 'label' => 'someword'),

I need to change the last part :

'label' => 'someword'),

to

'label' => __('someword','anotherstring')),

I will be using php for that , but I would also like to test it ( and also use in other cases) with Notepad ++ . ( I do not know if it actually changes something regarding the search and replace strings ).

Note that the string someword can also be SomeWord or SOMEWORD or even Some word or Some_Word on cases, meaning it can contain spaces, underscores or actually almost any character from within...)

Edit I : forgot to mantion that the __() part is of course wordpress textdomain function for translations . eg __('string','texdomain')

Edit II :

I am sorry if I come off too exigent or demanding in the comments , I really do try to UNDERSTAND and not only copy-paste a solution that might not work for me in other cases .. :-)

Edit III :

By the help of THIS tool , I understood that my basic misunderstanding is the possibility to use VARIABLES inside regex . The $1 is actually all I needed for better understanding .

the (incredibly simple) pattern that will work also in notepad++

Pattern: 'label' => ('.*')

Replace: 'label' => __(\1,'textdomain')

(In notepad++ it is called Tag Region (not var) and it is marked as \\1

Answer 1

If you will always be looking for the label key, you should be able to do something like this:

$pattern = "/array\((.*), 'label' => '(.*)'/U";
$added_string = 'anotherstring';
$replacement = 'array($1, ' . "'label' => __('" . '$2' . "','$added_string'";
$final_string = preg_replace($pattern, $replacement, $original_string);

Answer 2

For given inputs and outputs in question pattern: 'label' => ('.*') is enough to match strings and perform replacement. This pattern matches following part from a string: 'label' => any character between ' . Part of pattern in braces will group any character between ' that can be later accessed with $1 . Eg:

$str = "array('val' => 'something', 'label' => 'some testing string_with\$specialchars\/'),";
$str = preg_replace('/\'label\' => (\'.*\')/', '\'label\' => __($1, \'some other string\')', $str);
echo $str;
//Outputs:
//   array('val' => 'something', 'label' => __('some testing string_with$specialchars\/', 'some other string')),

Answer 3

<?php
$strings = array('some word', 'some Word', 'SOMEword', 'SOmE_Word', 'sOmE_ WOrd');
$pattern = '/([a-z]+)([^a-z]*)([a-z]+)/i';
foreach($strings as $v){
 echo preg_replace($pattern, 'otherword', $v)."<br>";
}
?> 

output:

otherword
otherword
otherword
otherword
otherword

EDIT:

$pattern = "/('label'\s=>\s')(([a-z]+)([^a-z]*)([a-z]+))('\),)/i";
$otherword = 'otherword';
$replacement = "'label' => __('$2','$otherword')),";
echo preg_replace($pattern, $replacement, "'label' => 'someword'),");

output:

'label' => __('someword','otherword')),

DEMO