如何从嵌套数组中获取项目的索引

Question

如何从嵌套数组中获取项目的索引但不知道嵌套的深度，例如 arr=[a,[b,[c],[d],e],f,[g],h]

Answer 1

For getting the full path, you might use an iterative and recursive approach for the nested array.

The result is an array with all indices.

 function findPath(array, value) { var path = []; array.some(function iter(p) { return function (a, i) { if (a === value) { path = p.concat(i); return true; } if (Array.isArray(a)) { return a.some(iter(p.concat(i))); }; }; }([])); return path; } var array = ['a', ['b', ['c'], ['d'], 'e'], 'f', ['g'], 'h']; console.log(findPath(array, 'a')); console.log(findPath(array, 'b')); console.log(findPath(array, 'c')); console.log(findPath(array, 'd')); console.log(findPath(array, 'e')); console.log(findPath(array, 'f')); console.log(findPath(array, 'g')); console.log(findPath(array, 'h'));

 .as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 2

If I understand your question correctly you are trying to get the index of for example value g .

I created a little script that iterates trough the array, and possibly nested arrays, and if found the correct value returns the index .

Here is a screenshot of the array with the keys that the function getIndexOfValue returns.

 var arr = [ "a", [ "b", [ "c" ], [ "d" ], "e" ], "f", [ "g" ], "h" ] function getIndexOfValue(haystack, needle) { for(i in haystack) { if(haystack[i] instanceof Array) { result = getIndexOfValue(haystack[i], needle); if(result) { return result; } } else if(haystack[i] == needle) { return i; } } return false; } var indexOfA = getIndexOfValue(arr, "a"); var indexOfB = getIndexOfValue(arr, "b"); var indexOfC = getIndexOfValue(arr, "c"); var indexOfD = getIndexOfValue(arr, "d"); var indexOfE = getIndexOfValue(arr, "e"); var indexOfF = getIndexOfValue(arr, "f"); var indexOfG = getIndexOfValue(arr, "g"); var indexOfH = getIndexOfValue(arr, "h"); console.log(indexOfA); //0 console.log(indexOfB); //0 console.log(indexOfC); //0 console.log(indexOfD); //0 console.log(indexOfE); //3 console.log(indexOfF); //2 console.log(indexOfG); //0 console.log(indexOfH); //4

It might still contain bugs, I created this quickly in a few minutes.