由网友(而你无动于衷)分享简介：我有对象的数组。数组中的每个对象都有一个ID和一个项属性，它是包含其他对象的数组。我需要能够找到在一个阵列按id的元素。下面是我迄今所做的一个样本，但递归函数总是返回undefined。I have an array of objects. Every object in the array has an id an...

我有对象的数组。数组中的每个对象都有一个ID和一个项属性，它是包含其他对象的数组。我需要能够找到在一个阵列按id的元素。下面是我迄今所做的一个样本，但递归函数总是返回undefined。

I have an array of objects. Every object in the array has an id and an item property that is an array containing other object. I need to be able to find an element in an array by id. Here is a sample of what I have done so far, but the recursive function is always returning undefined.

我怎么能退出的功能和回报的项目时，我已经调用的函数递归几次？

How can I quit the function and return the item when I have called the function recursively several times?

   $(function () {
    var treeDataSource = [{
        id: 1,
        Name: "Test1",
        items: [{
            id: 2,
            Name: "Test2",
            items: [{
                id: 3,
                Name: "Test3"
            }]
        }]
    }];
    var getSubMenuItem = function (subMenuItems, id) {
        if (subMenuItems && subMenuItems.length > 0) {
            for (var i = 0; i < subMenuItems.length; i++) {
                var item;
                if (subMenuItems[i].Id == id) {
                    item = subMenuItems[i];
                    return item;
                };
                getSubMenuItem(subMenuItems[i].items, id);
            };
        };
    };
    var searchedItem = getSubMenuItem(treeDataSource, 3);
    alert(searchedItem.id);
});

的jsfiddle

推荐答案

您应该替换

  getSubMenuItem(subMenuItems[i].items, id);

与

  var found = getSubMenuItem(subMenuItems[i].items, id);
  if (found) return found;

为了返回元素的时候才发现。

in order to return the element when it is found.

和小心的属性名，JavaScript是大小写敏感的，所以你也必须更换

And be careful with the name of the properties, javascript is case sensitive, so you must also replace

  if (subMenuItems[i].Id == id) {

与

  if (subMenuItems[i].id == id) {

示范

决赛（清洗）code：

Final (cleaned) code :

var getSubMenuItem = function (subMenuItems, id) {
    if (subMenuItems) {
        for (var i = 0; i < subMenuItems.length; i++) {
            if (subMenuItems[i].id == id) {
                return subMenuItems[i];
            }
            var found = getSubMenuItem(subMenuItems[i].items, id);
            if (found) return found;
        }
    }
};