Javascript 正则表达式中的捕获组
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capturing group in regex
提问by Madhur Ahuja
I am exploring capturing groups in Regex and I am getting confused about lack of documentation on it. For ex, can anyone tell me difference between two regex:
我正在探索 Regex 中的捕获组，但我对缺乏有关它的文档感到困惑。例如，谁能告诉我两个正则表达式之间的区别：
/(?:madhur)?/
and
和
/(madhur)?/
As per me, ? in second suggests matching madhurzero or once in the string.
在我看来，? in second 建议madhur在字符串中匹配零或一次。
How is the first different from second ?
第一个和第二个有什么不同？
回答by alex
The first one won't store the capturing group, e.g. $1will be empty. The ?:prefix makes it a non capturing group. This is usually done for better performance and un-cluttering of back references.
第一个不会存储捕获组，例如$1将是空的。该?:前缀使其成为一个非捕获组。这样做通常是为了更好的性能和整洁的反向引用。
In the second example, the characters in the capturing group will be stored in the backreference $1.
在第二个示例中，捕获组中的字符将存储在反向引用中$1。
回答by brymck
Here's the most obvious example:
这是最明显的例子：
"madhur".replace(/(madhur)?/, " ahuja"); // returns "madhur ahuja"
"madhur".replace(/(?:madhur)?/, " ahuja"); // returns " ahuja"
Backreferences are stored in order such that the first match can be recalled with $1, the second with $2, etc. If you capture a match (i.e. (...)instead of (?:...)), you can use these, and if you don't then there's nothing special. As another example, consider the following:
反向引用按顺序存储，这样第一个匹配可以用 调用$1，第二$2个匹配用等等。如果您捕获匹配（即(...)代替(?:...)），您可以使用这些，如果您不这样做，则没有什么特别的。作为另一个示例，请考虑以下内容：
/(mad)hur/.exec("madhur"); // returns an array ["madhur", "mad"]
/(?:mad)hur/.exec("madhur"); // returns an array ["madhur"]
回答by AndreKR
It doesn't affect the matching at all.
完全不影响匹配。
It tells the regex engine
它告诉正则表达式引擎
- not to store the group contents for use (as $1, $2, ...) by the
replace()method - not to return it in the return array of the
exec()method and - not to count it as a backreference (\1, \2, etc.)
- 不存储组内容的使用（如$ 1，$ 2，...）由
replace()方法 - 不要在方法的返回数组中返回它，
exec()并且 - 不要将其视为反向引用（\1、\2 等）
