删除字符串末尾方括号之间的文本

Question

I need a regex to remove last expression between brackets (also with brackets)

source: input[something][something2] target: input[something]

I've tried this, but it removes all two:

"input[something][something2]".replace(/\[.*?\]/g, '');

Answer 1

Note that \\[.*?\\]$ won't work as it will match the first [ (because a regex engine processes the string from left to right), and then will match all the rest of the string up to the ] at its end. So, it will match [something][something2] in input[something][something2] .

You may specify the end of string anchor and use [^\\][]* (matching zero or more chars other than [ and ] ) instead of .*? :

\[[^\][]*]$

See the JS demo:

 console.log( "input[something][something2]".replace(/\\[[^\\][]*]$/, '') ); 

Details :

• \\[ - a literal [
• [^\\][]* - zero or more chars other than [ and ]
• ] - a literal ]
• $ - end of string

Another way is to use .* at the start of the pattern to grab the whole line, capture it, and the let it backtrack to get the last [...] :

 console.log( "input[something][something2]".replace(/^(.*)\\[.*]$/, '$1') ); 

Here, $1 is the backreference to the value captured with (.*) subpattern. However, it will work a bit differently, since it will return all up to the last [ in the string, and then all after that [ including the bracket will get removed.

Answer 2

不要使用g修饰符，而应使用$锚点：

"input[something][something2]".replace(/\[[^\]]*\]$/, '');

Answer 3

try this code

 var str = "Hello, this is Mike (example)"; alert(str.replace(/\\s*\\(.*?\\)\\s*/g, ''));