javascript - 为什么 string.split() 结果包含 undefined？

Question

我想在 %\d+ 或 \n 上拆分字符串。我能够在这两个中的任何一个上成功拆分，但不能同时在两个上拆分:

> msg = 'foo %1 bar \n baz %2'

> msg.split(/(%\d+)/)
["foo ", "%1", " bar 
 baz ", "%2", ""]

> msg.split(/(\n)/)
["foo %1 bar ", "
", " baz %2"]

> msg.split(/(\n)|(%\d)/)
["foo ", undefined, "%1", " bar ", "
", undefined, " baz ", undefined, "%2", ""]

在最后一种情况下，为什么 undefined 在结果数组中，我应该做什么？

更新:我忘记说明我需要分隔符。我想要的结果是:

["foo ", "%1", " bar ", "\n", " baz ", "%2"]

Answer 1

引用MDN doc对于 String.prototype.split:

If separator is a regular expression that contains capturing parentheses, then each time separator is matched, the results (including any undefined results) of the capturing parentheses are spliced into the output array.

重点是任何捕获组都会被拼接——即使是未命中目标的组。您示例中的第一个 undefined 是 \n 匹配的“无”(当 %\d 匹配时发生拆分)，第二个是 for %\d(当 \n 匹配时)...你看到图片了。

要解决这个问题，您可以去掉捕获组(因为交替运算符的优先级最低):

msg.split(/\n|%\d/); // ["foo ", " bar ", " baz ", ""]

如果您确实也需要分离部分，请只使用一个捕获组:

msg.split(/(\n|%\d)/); 
// ["foo ", "%1", " bar ", "\n", " baz ", "%2", ""]