PHP Regular replaces the first bit with 0. is ignored

$subject = '0300020F005176C5A11A730004F955A2A20CD8-367C-EC5B-9D8C-2CAA0B8FE45D';echo preg_replace('/^(\w{8})\w{2}/i','$1'.'08',$subject);// 输出结果：85176C5A11A730004F955A2A20CD8-367C-EC5B-9D8C-2CAA0B8FE45D// 期望结果：0300020F085176C5A11A730004F955A2A20CD8-367C-EC5B-9D8C-2CAA0B8FE45D// 但是：preg_replace('regex','$1'.'__',$subject);  这样是没有问题的

Reply content:

$subject = '0300020F005176C5A11A730004F955A2A20CD8-367C-EC5B-9D8C-2CAA0B8FE45D';echo preg_replace('/^(\w{8})\w{2}/i','$1'.'08',$subject);// 输出结果：85176C5A11A730004F955A2A20CD8-367C-EC5B-9D8C-2CAA0B8FE45D// 期望结果：0300020F085176C5A11A730004F955A2A20CD8-367C-EC5B-9D8C-2CAA0B8FE45D// 但是：preg_replace('regex','$1'.'__',$subject);  这样是没有问题的

Quote from PHP manual preg_replace function to a back-reference description: Preg_replace

When working in replacement mode and the back reference is followed by a second number (for example, adding a text number immediately after a matching pattern), you cannot use syntax such as \1 to describe the back reference. For example, \11 will make preg_replace () do not understand what you want is a \1 back to the reference followed by a text 1, or a \11 after the reference is not followed by anything. In this case the solution is to use \${1}1. This creates a separate, forward-referencing, 1-independent source.

Your replacement part is "$108", so you need to replace it with ${1} to avoid confusion.

echo preg_replace_callback('/^(\w{8})(\w{2})/i',create_function(        // single quotes are essential here,        // or alternative escape all $ as \$        '$matches',        'return $matches[1].\'08\';'    ),$subject);

这样,callback处理

