Question

为什么：

/(\[#([0-9]{8})\])/g.exec("[#12345678] [#87654321] [#56233001] [#36381069] [#23416459] [#56435355]")

返回

[“[＃12345678]”，“[＃12345678]”，“12345678”]

我希望它能匹配所有这些数字，但它看起来太贪心了。

[＃12345678] [＃87654321] [＃56233001] [＃36381069] [＃23416459] [＃56435355] 12345678 87654321 56233001 36381069 23416459 56435355

Answer 1

这就是.exec()的工作方式。要获得多个结果，请循环运行。

var re = /(\[#([0-9]{8})\])/g,
    str = "[#12345678] [#87654321] [#56233001] [#36381069] [#23416459] [#56435355]",
    match;

while (match = re.exec(str)) {
    console.log(match);
}

此外，外部捕获组似乎无关紧要。你可能应该摆脱它。

/\[#([0-9]{8})\]/g,

结果：

[
    "[#12345678]",
    "12345678"
],
[
    "[#87654321]",
    "87654321"
],
[
    "[#56233001]",
    "56233001"
],
[
    "[#36381069]",
    "36381069"
],
[
    "[#23416459]",
    "23416459"
],
[
    "[#56435355]",
    "56435355"
]

Answer 2

regex.exec返回正则表达式中的组（括号中包含的内容）。

您正在寻找的功能是您在字符串match上调用的功能。

string.match(regex)会返回所有匹配项。

"[#12345678] [#87654321] [#56233001] [#36381069] [#23416459] [#56435355]".match(/(\[#([0-9]{8})\])/g)
// yields: ["[#12345678]", "[#87654321]", "[#56233001]", "[#36381069]", "[#23416459]", "[#56435355]"]

编辑：

如果您只想要没有方括号和＃的数字，只需将正则表达式更改为/\d{8}/g

"[#12345678] [#87654321] [#56233001] [#36381069] [#23416459] [#56435355]".match(/[0-9]{8}/g)
// yields: ["12345678", "87654321", "56233001", "36381069", "23416459", "56435355"]

Answer 3

您可以使用字符串的replace方法收集所有匹配项：

var s = "[#12345678] [#87654321] [#56233001] [#36381069] [#23416459] [#56435355]";
var re = /\[#([0-9]{8})\]/g;
var l = [];
s.replace(re, function($0, $1) {l.push($1)});
// l == ["12345678", "87654321", "56233001", "36381069", "23416459", "56435355"]

Answer 4

试试这个：

    var re = /\[#(\d{8})\]/g;
    var sourcestring = "[#12345678] [#87654321] [#56233001] [#36381069] [#23416459] [#56435355]";
    var results = [];
    var i = 0;
    var matches;
    while (matches = re.exec(sourcestring)) {
        results[i] = matches;
        alert(results[i][1]);
        i++;
    }

Answer 5

在ES2020中，添加了一个新功能matchAll由女巫完成。

只有匹配第一个匹配的全局标志的正则表达式

5 个答案: