javascript Regex（正则表达式），替换javascript中的第二次出现
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Regex (regular expressions), replace the second occurence in javascript
提问by MintDeparture
This is an example of the string that's being worked with:
这是正在使用的字符串的示例：
xxxxxx[xxxxxx][7][xxxxxx][9][xxxxxx]
I'm having a little trouble matching the second occurrence of a match, I want to return the 2nd square brackets with a number inside. I have some regex finding the first square backets with numbers in a string:
我在匹配第二次匹配时遇到了一些麻烦，我想返回带有数字的第二个方括号。我有一些正则表达式在字符串中找到第一个带有数字的方形 backets：
\[+[0-9]+\]
This returns [7], however I want to return [9].
这将返回 [7]，但是我想返回 [9]。
I'm using Javascript's replace function, the following regex matches the second occurrence (the [9]) in regex testeing apps, however it isn't replaced correctly in the Javascript replace function:
我正在使用 Javascript 的替换功能，以下正则表达式与正则表达式测试应用程序中的第二次出现（[9]）相匹配，但是它在 Javascript 替换功能中没有被正确替换：
(?:.*?(\[+[0-9]+\])){2}
My question is how do I use the above regex to replace the [9] in Javasctipt or is there another regex that matches the second occurrence of a number in square brackets.
我的问题是如何使用上述正则表达式替换 Javasctipt 中的 [9] 或者是否有另一个正则表达式匹配方括号中数字的第二次出现。
Cheers!
干杯!
采纳答案by moinudin
If xxxis just any string, and not necessarily a number, then this might be what you want:
如果xxx只是任何字符串，而不一定是数字，那么这可能就是您想要的：
(\[[0-9]+\]\[.*?\])\[([0-9]+)\]
This looks for the second number in []. Replace it with $1[<replacement>]. Play with it on rubular.
这将在 中查找第二个数字[]。将其替换为$1[<replacement>]. 在 rubular 上玩它。
Your regular expression fails to work as intended because groups followed by +only end up holding the last[xxx].
您的正则表达式无法按预期工作，因为后面的组+只保留最后一个[xxx].
回答by Tim Pietzcker
Try
尝试
result = subject.replace(/(\[\d\]\[[^\]]+\])\[\d\]/, "[replace]");
As a commented regex:
作为评论的正则表达式：
( # capture the following in backref 1:
\[\d\] # first occurrence of [digit]
\[ # [
[^\]]+ # any sequence of characters except ]
\] # ]
) # end of capturing group
\[\d\] # match the second occurence of [digit]
If the number of [xxx]groups between the first and second [digit]group is variable, then use
如果[xxx]第一[digit]组和第二组之间的组数是可变的，则使用
result = subject.replace(/(\[\d\](?:\[[^\]]+\])*?)\[\d\]/, "[replace]");
By surrounding the part that matches the [xxx]groups with (non-capturing) parentheses and the lazy quantifier *?I'm asking the regex engine to match as few of those groups as possible, but as many as necessary so the next group is a [digit]group.
通过[xxx]用（非捕获）括号和惰性量词包围匹配组的部分，*?我要求正则表达式引擎匹配尽可能少的组，但尽可能多，所以下一组是一个[digit]组。
回答by Steven de Salas
console.log( "xxxxxx[xxxxxx][7][xxxxxx][9][xxxxxx]".replace(
/^(.*\[[0-9]+\].*)(\[[0-9]+\])(.*)$/,
'[15]')); // replace with matches before () and after () your match ()
returns:
返回：
// xxxxxx[xxxxxx][7][xxxxxx][15][xxxxxx]
It will match where [n] is preceeded by 1 set of brackets with numbers inside.
它将匹配 [n] 前面是 1 组括号内的数字。
