JavaScript通过NodeList进行迭代 [英] JavaScript iterate through NodeList

问题描述

我正在寻找一种遍历NodeList的最佳方法(不包括长度).我正在使用:

I am looking for a best way to iterate through NodeList excluding length. I am using:

var foo =  document.querySelectorAll('[id^=foo_id]')
console.log(foo)

返回的NodeList具有所有必需的元素以及最后一个长度为length的条目.

Returned NodeList has all the required elements + the last entry of length:.

  0: input#foo_id_...
  1: input#foo_id_...
  ..................
  n: input#foo_id_...
  length: n+1

我想知道哪种最有效的方法来迭代该NodeList.我可以利用列表长度等，但是想知道是否还有更优雅"的方式.

I wonder what the most efficient way to iterate through that NodeList. I can utilize list length etc, but would like to know if there is a more "elegant" way.

推荐答案

最简单的方法是for循环:

for (let i = 0; i < foo.length; i++) {
  // Do stuff
}

这是最好的解决方案，如此处所述，使用数组方法或将NodeList转换为数组-如果需要，可以使用其他变量，但是for循环只需要在NodeList上循环即可. (感谢异端猴子向我指出了这一点.)

This is the best solution, as pointed out here it's bad practice to use array methods or convert a NodeList to an array - use a different variable if you need to, but a for loop is all you need to loop over a NodeList. (Thanks Heretic Monkey for pointing this out to me).

如果要使用forEach，map等数组方法，则最好先转换为数组-扩展非常简单:

If you want to use array methods like forEach, map, etc., it's best convert to an array first - this is incredibly simple with spreading:

[...foo].forEach(e /* Do stuff */);

如果要专门使用map，请使用Array.from作为第二个参数是要应用于map的回调.

If you want to specifically use map, use Array.from as the second argument is the callback to be applied to map.

Array.from(foo, e => /* .map callback */);

在较旧的环境中:

Array.prototype.slice.call(foo).forEach(e => /* Do stuff */);

_{(我知道您可以在NodeList上使用数组方法，但是如果坚持使用一种数据类型会更容易.)}

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