A loosely-typed indexOf()?

Question

Let's suppose I have the array:

[1, 2, 3, 4]

Now let's suppose I want to know the first index of a value that ==s (not ===s) a given value, like "3". For this case I would want to return 2, but Array.indexOf returns -1.

:(

Answer 1

Update 2021-04-13: ES6 introduced findIndex() and arrow functions, so you can now do...

> [1,2,3,4].findIndex(v => v == '3')
2

[Original answer]

嗯，有一个天真的实现......

function looseIndex(needle, haystack) {
  for (var i = 0, len = haystack.length; i < len; i++) {
    if (haystack[i] == needle) return i;
  }
  return -1;
}

例如

> [0,1,2].indexOf('1')
-1  // Native implementation doesn't match
> looseIndex('1', [0, 1, 2])
1   // Our implementation detects `'1' == 1`
> looseIndex('1', [0, true, 2])
1   // ... and also that `true == 1`

（这种方法的一个好处是它适用于可能不支持完整 Array API 的NodeLists 和arguments等对象。）

如果您的数组已排序，您可以进行二分搜索以获得更好的性能。有关示例，请参阅Underscore.js 的 sortedIndex 代码

Answer 2

indexOf does strict equals, so do a conversion on either of the sides, You can try this way:

Either:

arr.map(String).indexOf("3");

or

var str = "3";
arr.indexOf(+str);

Fiddle