仅在指定字符的第一个实例上拆分字符串 [英] split string only on first instance of specified character

问题描述

在我的代码中，我根据 _ 拆分了一个字符串，然后抓住数组中的第二项。

In my code I split a string based on _ and grab the second item in the array.

var element = $(this).attr('class');
var field = element.split('_')[1];

取得 good_luck 并为我提供运气。效果很棒！

Takes good_luck and provides me with luck. Works great!

但是，现在我的课程看起来像 good_luck_buddy 。如何让我的javascript忽略第二个 _ 并给我 luck_buddy ？

But, now I have a class that looks like good_luck_buddy. How do I get my javascript to ignore the second _ and give me luck_buddy?

我在ac＃stackoverflow回答中找到了这个 var field = element.split（new char [] {'_'}，2）; 但是它不起作用。我在jsFiddle试了一下...

I found this var field = element.split(new char [] {'_'}, 2); in a c# stackoverflow answer but it doesn't work. I tried it over at jsFiddle...

推荐答案

使用捕获括号：

"good_luck_buddy".split(/_(.+)/)[1]
"luck_buddy"

它们定义为

如果分隔符包含捕获括号，匹配的结果是数组中返回的
。

If separator contains capturing parentheses, matched results are returned in the array.

所以在这种情况下我们要分割为 _。+ （即拆分分隔符是以 _ 开头的子字符串）但让结果包含我们分隔符的一部分（即所有在 _ 之后）。

So in this case we want to split at _.+ (i.e. split separator being a sub string starting with _) but also let the result contain some part of our separator (i.e. everything after _).

在此示例中，我们的分隔符（匹配 _（。+））是 _luck_buddy 和捕获的组（在分隔符内）是 lucky_buddy 。如果没有捕获括号， luck_buddy （匹配。+ ）将不会包含在结果数组中，因为它是使用简单 split 的情况，结果中不包含分隔符。

In this example our separator (matching _(.+)) is _luck_buddy and the captured group (within the separator) is lucky_buddy. Without the capturing parenthesis the luck_buddy (matching .+) would've not been included in the result array as it is the case with simple split that separators are not included in the result.

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