如何用正则表达式模式替换正则表达式模式？

Question

I'm feeling very silly for not being able to figure this one out, but I just cant figure it. I need to replace a regex string with a pattern, I found two examples of doing it, but they honestly left me more confused than ever.

This is my current attempt:

$string = '26 14:46:54",,"2011-08-26 14:47:02",8,0,"BUSY","DOCUMENTATION","1314370014.18","","","61399130249","7466455647","from-internal","""Oh Snap"" <61399130249>","SIP/1037-00000014","SIP/CL-00000015","Dial","SIP/CL/61436523277,45","2011-08-26 14:47:06","2011-08-26 14:47:15","2011-08-26 ';
$pattern = '["SIP/CL/\d*?,\d*?",]';
$replacement = '"SIP/CL/\1|\2",';
$string = preg_replace($pattern, $replacement, $string);
print($string);

But that just replaces the \\1 and \\2 with blanks. So obviously I'm not getting the entire concept.

What I'm wanting at the end is to change:

this: "SIP/CL/61436523277,45"
to: "SIP/CL/61436523277|45"

That comma in a poorly formatted CSV throws off some of my other scripts.

Answer 1

你缺少大括号，模式应如下所示：

$pattern = '["SIP/CL/(\d*),(\d*)",]';

Answer 2

I think you need this:

$pattern = '["SIP/CL/(\d+),(\d+)",]';

The parens - () - capture the match inside, allowing you to reference it in your replacement pattern.

You could also simplify the replacement pattern by expanding the captures in the match pattern:

$pattern = '[("SIP/CL/\d+),(\d+",)]';   
$replacement = '\1|\2';

Answer 3

The backreferences \\1 and \\2 in your $replacement are meant to refer to captured groups in $pattern . Captured groups are saved by placing brackets around them () .

Try (I changed your regex delimiters [] to ! pending an answer from comments below):

$pattern = '!"SIP/CL/(\d*?),(\d*?)",!';
$replacement = '"SIP/CL/\1|\2",';

Also note that since you have a trailing comma on your $pattern , if your $string ended in "SIP/CL/61436523277,45" that would not be converted. I'd recommend removing that trailing comma.

Also your current regex will convert "SIP/CL/," to "SIP/CL/|" . If that is not your intent, change the * after the \\d (ie 0 or more matches) to a + (one or more matches).

Answer 4

I did this

$string = '26 14:46:54",,"2011-08-26 14:47:02",8,0,"BUSY","DOCUMENTATION","1314370014.18","","","61399130249","7466455647","from-internal","""Oh Snap"" <61399130249>","SIP/1037-00000014","SIP/CL-00000015","Dial","SIP/CL/61436523277,45","2011-08-26 14:47:06","2011-08-26 14:47:15","2011-08-26 ';
$pattern = '/SIP\/CL\/([0-9]+),([0-9]+)/';
$replacement = 'SIP/CL/\1|\2';
$string = preg_replace($pattern, $replacement, $string);
print($string);