PHP正则表达式 - $ 1和\\ 1之间有什么区别？

Question

The two following codes give the same result:

$test = "this is a test";
echo preg_replace('/a (test)/','\1',$test); //this is test

$test = "this is a test";
echo preg_replace('/a (test)/',"$1",$test); //this is test

But two the following codes give different result:

$test = "this is a test";
echo preg_replace('/a (test)/',md5('\1'),$test); //this is b5101eb6e685091719013d1a75756a53

$test = "this is a test";
echo preg_replace('/a (test)/',md5("$1"),$test); //this is 06d3730efc83058f497d3d44f2f364e3

Does this mean \\1 and $1 is different?

Answer 1

Neither of those does what you want. In your code, the md5 function is called on the literal string '\\1' or "$1" (the difference between backslash and dollar sign being the difference in the checkums), and then preg_replace is passed the result of that md5 call, which never had a chance to consider the string it's trying to match.

What you want in this case is to use preg_replace_callback :

echo preg_replace_callback('/a (test)/', function($m) { return md5($m[1]); }, $test);

The argument passed to the callback function is an array containing the captured subexpressions, so the equivalent of \\1 or $1 is $m[1] ; \\2 or $2 would be $m[2] , etc.

Answer 2

Well that's because you aren't passing \\1 or $1 to the preg_replace call any more, are you??

\\1 and $1 are synonymous in regex replacement strings.