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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