Regular Expression matching syntax

Question

I need to replace every instance of any text in square brackets with something else, with each square bracket block being treated separately. For example, start with:

[quote author=joe link=topic=765.msg4476#msg4476 date=1330380346] This is the quoted text [/quote] This is the new post

being turned into:

** This is the quoted text ** This is the new post

I tried using the following:

preg_replace('/\[.*\]/', '**', $msgtext);

What I get is:

** This is the new post

It seems to be matching from the first '[' character to the last ']' character in the entire string, even if there are a bunch of separate blocks of square brackets in the larger text. How do I change my regex to replace each block between the square brackets individually? Obviously my .* in the regex is matching everything including right brackets until the last one, but I want it to stop at the first right bracket it encounters, and then repeat that logic throughout the entire string.

Answer 1

You need to use a non-greedy match, either by using the /U flag to make the whole pattern greedy:

preg_replace('/\[.*\]/U', '**', $msgtext);

or by using .*? ("zero or more, and preferably as few as possible") instead of .* ("zero or more, and preferably as many as possible"):

preg_replace('/\[.*?\]/', '**', $msgtext);

Alternatively, you can use [^\\]] ("any character except ] ") instead of . ("any character except newline"):

preg_replace('/\[[^\]]*\]/', '**', $msgtext);

Answer 2

默认情况下，PHP的regexp会进行贪心匹配，您需要将其设置为不贪心（例如，使用u开关）

preg_replace('/\[.*\]/U', '**', $msgtext);

Answer 3

这对我有用：

preg_replace('/\[[^\]]*\]/', '**', $msgtext);