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正则表达式抓取方括号之间的字符串

[英]Regex to grab strings between square brackets

我有以下字符串: pass[1][2011-08-21][total_passes]

如何将方括号之间的项目提取到数组中? 我试过了

match(/\\[(.*?)\\]/);

 var s = 'pass[1][2011-08-21][total_passes]'; var result = s.match(/\\[(.*?)\\]/); console.log(result);

但这只会返回[1]

不知道该怎么做.. 提前致谢。

你快到了,你只需要一个全局匹配(注意/g标志):

match(/\[(.*?)\]/g);

示例: http : //jsfiddle.net/kobi/Rbdj4/

如果你想要只捕获组的东西(来自MDN ):

var s = "pass[1][2011-08-21][total_passes]";
var matches = [];

var pattern = /\[(.*?)\]/g;
var match;
while ((match = pattern.exec(s)) != null)
{
  matches.push(match[1]);
}

示例: http : //jsfiddle.net/kobi/6a7XN/

另一种选择(我通常更喜欢)是滥用替换回调:

var matches = [];
s.replace(/\[(.*?)\]/g, function(g0,g1){matches.push(g1);})

示例: http : //jsfiddle.net/kobi/6CEzP/

var s = 'pass[1][2011-08-21][total_passes]';

r = s.match(/\[([^\]]*)\]/g);

r ; //# =>  [ '[1]', '[2011-08-21]', '[total_passes]' ]

example proving the edge case of unbalanced [];

var s = 'pass[1]]][2011-08-21][total_passes]';

r = s.match(/\[([^\]]*)\]/g);

r; //# =>  [ '[1]', '[2011-08-21]', '[total_passes]' ]

将全局标志添加到您的正则表达式,并迭代返回的数组。

 match(/\[(.*?)\]/g)
'pass[1][2011-08-21][total_passes]'.match(/\[.+?\]/g); // ["[1]","[2011-08-21]","[total_passes]"]

解释

\[       # match the opening [
           Note: \ before [ tells that do NOT consider as a grouping symbol.
   .+?   # Accept one or more character but NOT greedy
\]       # match the closing ] and again do NOT consider as a grouping symbol
/g       # do NOT stop after the first match. Do it for the whole input string.

您可以使用正则表达式的其他组合https://regex101.com/r/IYDkNi/1

我不确定您是否可以将其直接放入数组中。 但是下面的代码应该可以找到所有出现的情况然后处理它们:

var string = "pass[1][2011-08-21][total_passes]";
var regex = /\[([^\]]*)\]/g;

while (match = regex.exec(string)) {
   alert(match[1]);
}

请注意:我真的认为您需要这里的字符类 [^\\]]。 否则,在我的测试中,表达式将匹配孔字符串,因为 ] 也与 .* 匹配。

[C#]

        string str1 = " pass[1][2011-08-21][total_passes]";
        string matching = @"\[(.*?)\]";
        Regex reg = new Regex(matching);
        MatchCollection matches = reg.Matches(str1);

您可以将 foreach 用于匹配的字符串。

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