Javascript 如何匹配但排除正则表达式模式？

How can I match on, but exclude a regex pattern?

提问by Adam

I have this URL:

我有这个网址：

http://example.com/createSend/step4_1.aspx?cID=876XYZ964D293CF&snap=true&jlkj=kjhkjh&

http://example.com/createSend/step4_1.aspx?cID=876XYZ964D293CF&snap=true&jlkj=kjhkjh&

And this regex pattern:

这个正则表达式模式：

cID=[^&]*

Which produces this result:

产生这个结果：

cID=87B6XYZ964D293CF

How do I REMOVE the "cID="?

如何删除“cID=”？

Thanks

谢谢

回答by Kerry Jones

You can either use lookbehind (not in Javascript):

您可以使用lookbehind（不在Javascript中）：

(?<=cID=)[^&]*

Or you can use grouping and grab the first group:

或者您可以使用分组并获取第一组：

cID=([^&]*)

回答by polygenelubricants

Generally speaking, to accomplish something like this, you have at least 3 options:

一般来说，要完成这样的事情，您至少有 3 个选择：

• Use lookarounds, so you can match precisely what you want to capture
  • No lookbehind in Javascript, unfortunately
• Use capturing group to capture specific strings
  • Near universally supported in all flavors
• If all else fails, you can always just take a substringof the match
  • Works well if the length of the prefix/suffix to chop is a known constant

• 使用环视，以便您可以精确匹配要捕获的内容
  • 不幸的是，Javascript 中没有回溯
• 使用捕获组捕获特定字符串
  • 几乎所有口味都支持
• 如果所有其他方法都失败了，您可以随时参加substring比赛
  • 如果要切割的前缀/后缀的长度是已知常量，则效果很好

References

参考

• w3schools - jsref- substring
• regular-expressions.info/Capturing groupsand Lookarounds
  • Flavor comparison

• w3schools - jsref-substring
• 正则表达式.info/捕获组和环视
  • 口味对比

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Examples

例子

Given this test string:

鉴于此测试字符串：

i have 35 dogs, 16 cats and 10 elephants

These are the matches of some regex patterns:

这些是一些正则表达式模式的匹配：

• \d+ cats-> 16 cats(see on rubular.com)
• \d+(?= cats)-> 16(see on rubular.com)
• (\d+) cats-> 16 cats(see on rubular.com)
  • Group 1 captures 16

• \d+ cats-> 16 cats（见rubular.com）
• \d+(?= cats)-> 16（见rubular.com）
• (\d+) cats-> 16 cats（见rubular.com）
  • 第 1 组捕获 16

You can also do multiple captures, for example:

您还可以进行多次捕获，例如：

• (\d+) (cats|dogs)yields 2 match results (see on rubular.com)
  • Result 1: 35 dogs
    • Group 1 captures 35
    • Group 2 captures dogs
  • Result 2: 16 cats
    • Group 1 captures 16
    • Group 2 captures cats

• (\d+) (cats|dogs)产生 2 个匹配结果（参见 rubular.com）
  • 结果 1： 35 dogs
    • 第 1 组捕获 35
    • 第 2 组捕获 dogs
  • 结果 2： 16 cats
    • 第 1 组捕获 16
    • 第 2 组捕获 cats

回答by gnarf

With JavaScript, you'll want to use a capture group(put the part you want to capture inside ()) in your regular expression

使用 JavaScript，您需要在正则表达式中使用捕获组（将要捕获的部分放在里面()）

var url = 'http://example.com/createSend/step4_1.aspx?cID=876XYZ964D293CF&snap=true&jlkj=kjhkjh&';

var match = url.match(/cID=([^&]*)/);
// ["cID=876XYZ964D293CF", "876XYZ964D293CF"]

// match[0] is the whole pattern
// match[1] is the first capture group - ([^&]*)
// match will be 'false' if the match failed entirely

回答by cherouvim

By using capturing groups:

通过使用捕获组：

cID=([^&]*)

and then get $1:

然后得到 1 美元：

87B6XYZ964D293CF

回答by David A Moss

Here's the Javascript code:

这是Javascript代码：

 var str = "http://example.com/createSend/step4_1.aspx?cID=876XYZ964D293CF&snap=true&jlkj=kjhkjh&";
    var myReg = new RegExp("cID=([^&]*)", "i");
    var myMatch = myReg.exec(str);
    alert(myMatch[1]);

回答by tylik

There is a special syntax in javascript which allows you to exclude unwanted match from the result. The syntax is "?:" In your case the solution would be the following

javascript 中有一种特殊的语法，它允许您从结果中排除不需要的匹配。语法是“？：”在您的情况下，解决方案如下

'http://example.com/createSend/step4_1.aspx?cID=876XYZ964D293CF&snap=true&jlkj=kjhkjh&'.match(/(?:cID=+)([^&]*)/)[1];