javascript 正则表达式抓取方括号之间的字符串
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Regex to grab strings between square brackets
提问by cabaret
I have the following string: pass[1][2011-08-21][total_passes]
我有以下字符串： pass[1][2011-08-21][total_passes]
How would I extract the items between the square brackets into an array? I tried
如何将方括号之间的项目提取到数组中？我试过
match(/\[(.*?)\]/);
match(/\[(.*?)\]/);
var s = 'pass[1][2011-08-21][total_passes]';
var result = s.match(/\[(.*?)\]/);
console.log(result);but this only returns [1].
但这只会返回[1].
Not sure how to do this.. Thanks in advance.
不知道如何做到这一点.. 提前致谢。
回答by Kobi
You are almost there, you just need a global match(note the /gflag):
你快到了，你只需要一个全局匹配（注意/g标志）：
match(/\[(.*?)\]/g);
Example: http://jsfiddle.net/kobi/Rbdj4/
示例：http: //jsfiddle.net/kobi/Rbdj4/
If you want something that only captures the group (from MDN):
如果您想要仅捕获组的内容（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]);
}
Example: http://jsfiddle.net/kobi/6a7XN/
示例：http: //jsfiddle.net/kobi/6a7XN/
Another option (which I usually prefer), is abusing the replace callback:
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);})
Example: http://jsfiddle.net/kobi/6CEzP/
回答by James Kyburz
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]' ]
回答by amal
add the global flag to your regex , and iterate the array returned .
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)
回答by Chris
I'm not sure if you can get this directly into an array. But the following code should work to find all occurences and then process them:
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]);
}
Please note: i really think you need the character class [^\]] here. Otherwise in my test the expression would match the hole string because ] is also matches by .*.
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。
回答by Suriya
[C#]
[C＃]
string str1 = " pass[1][2011-08-21][total_passes]";
string matching = @"\[(.*?)\]";
Regex reg = new Regex(matching);
MatchCollection matches = reg.Matches(str1);
you can use foreach for matched strings.
您可以将 foreach 用于匹配的字符串。
