PHP preg_replace - 在结果中包含模式/完整主题

Question

I've a problem that I couldn't solve myself: replace...

locale("Sendx", "Send")
locale("System", "System")

should become:

locale("Sendx", "Subsub")
locale("System", "Newsys")

I tried a simple replace:

$mysearchword = "System"; #changes in a loop
$myreplaceword = "Newsys"; #also changes in the loop
$oneline = str_replace($mysearchword, $myreplaceword, $oneline);

but the result looks like

locale("Sendx", "53ND")
locale("Newsys", "Newsys") #problem with the doubled word

of course the System was replaced both times. So I decided to use preg_replace

$pattern = '/locale\\(["|\']([^"\']*)["|\'], ["|\']([^"\']*)["|\']\\)/';
$replacement = '${1}, Newsys';
$subject = 'locale("System", "System")';
echo preg_replace($pattern, $replacement, $subject, -1 );

But now nearly anything is missing, because only the words in the brackets are returned and I have no idea how to include the pattern or return a replaced $subject. The $pattern changes, so I couldn't write "locale(..." into the $replacement / I somehow must return a replaced pattern...

System, Newsys # No idea how to combine $replacement with $pattern...

Could you please help me to get the right result?

Answer 1

I guess you are looking for something like this:

$input  = 'locale("Sendx", "Send")';
$output = preg_replace('/, "(.*?)"/', ', "Subsub"', $input);
echo $output;
echo "
";

$input  = 'locale("System", "System")';
$output = preg_replace('/, "(.*?)"/', ', "Newsys"', $input);
echo $output;

Output:

locale("Sendx", "Subsub")
locale("System", "Newsys")

The pattern /, "(.*?)"/ searches for the word between double quotes " after the comma and replaces it with , "NEW_WORD"

Using this pattern, you can easily replace them in a loop:

$input = array(
  'locale("Sendx", "Send")' => 'Subsub',  
  'locale("System", "System")' => 'Newsys'
);

foreach($input as $string => $replacement) {
    $output = preg_replace('/, "(.*?)"/', ', "' . $replacement . '"', $string);
    echo $output. PHP_EOL;
}