删除字符串末尾方括号之间的文本 [英] Remove text between square brackets at the end of string

问题描述

我需要一个正则表达式来删除括号之间的最后一个表达式(也有括号)

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

来源:输入[东西][东西2]目标:输入[东西]

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

我试过这个，但它删除了所有两个:

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

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

推荐答案

请注意 \[.*?\]$ 将不起作用，因为它将匹配 first [(因为正则表达式引擎从左到右处理字符串)，然后将匹配字符串的所有其余部分，直到其末尾的 ] .因此，它将匹配 input[something][something2] 中的 [something][something2].

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].

您可以指定字符串锚点的结尾并使用[^\][]*(匹配除[ 和] 之外的零个或多个字符)code>) 而不是 .*?:

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

\[[^\][]*]$

查看 JS 演示:

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

详情:

• \[ - 文字 [
• [^\][]* - 除了 [ 和 ]
之外的零个或多个字符
• ] - 文字 ]
• $ - 字符串结束

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

另一种方法是在模式的开头使用 .* 来抓取整行，捕获它，然后让它回溯以获取最后一个 [...]:

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')
);

这里，$1 是对使用 (.*) 子模式捕获的值的反向引用.但是，它的工作方式会有所不同，因为它将返回字符串中最后一个 [ 之前的所有内容，然后所有之后的 [ 包括括号将被删除.

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.

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