为什么捕获组会导致双重匹配正则表达式 [英] Why capturing group results in double matches regex
问题描述
考虑以下两个脚本:
第一个:" ".match(/(\s)/)
和
2nd:" ".match(/\s/)
结果
第一名:[" "," "]
2nd:[" "]
我不了解这种行为.据我所知，捕获组/同形异义的目的是要有一部分匹配项稍后在正则表达式中再次引用.但是很明显，这还不是全部.还是这种行为特定于 match 和 split 方法.
I don't understand this behavior. As far as i knew purpose of capturing group/paranthesis was to have a section of match to be refered again later within regex. But clearly that's not all. Or is this behavior specific to match and split methods.
推荐答案
第一个脚本:第一个结果是整个模式，第二个结果是捕获组
First script: The first result is the whole pattern, the second is the capturing group
第二个脚本:唯一的结果就是整个模式.
Second script: the only result is the whole pattern.
捕获组不仅要在模式中稍后引用，它们还会显示在结果中.
Capturing groups are not only to refer later in the pattern, they are displayed in results too.
当您将捕获组与split一起使用时，捕获组将返回结果，并且由于应该使用分隔符对字符串进行切片，因此通常以
" "作为输入获得["", " ", ""]作为结果字符串和/(\s)/作为模式.
When you use a capturing group with split, the capturing group is returned with results and since the separator is supposed to slice the string, it is normal that you obtain ["", " ", ""] as result with " " as input string and /(\s)/ as pattern.
当您写" ".match(/(\s)/)时，返回的结果是第一个匹配项.此结果是唯一的，并且包含:
When you write " ".match(/(\s)/) the result returned is the first match. This result is unique and contains:
- 整个比赛
- 捕获组
- 比赛索引
- 输入字符串
当您编写" ".match(/(\s)/g)时，返回的结果是所有匹配项:
When you write " ".match(/(\s)/g) the result returned is all the matches:
- 全场比赛1
- 全场比赛2
- 等
(在当前情况下，您只有一场比赛)
(in the present case you have only one match)
此行为是正常的. match方法具有两种不同的行为(带有或不带有/g).它是两种功能合而为一.为了在没有g修饰符的PHP(或其他语言)中进行比较，您具有两个不同的功能:preg_match和preg_match_all
This behaviour is normal. The match method as two different behaviours (with or without /g). It is a kind of two functions in one. For comparison in PHP (or other languages) which doesn't have the g modifier, you have two different functions: preg_match and preg_match_all
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