jQuery的：为什么不jQuery.inArray（）工作？ [英] jQuery: why doesn't jQuery.inArray() work?

问题描述

由于图所示，我只是做inArray阵列上，找一个节点。和$previously_selected_node和指数37 $形状的项目是同一对象....所以...为什么不工作？

As shown in the image, I just do inArray on an array, looking for a node. and $previously_selected_node and the item at index 37 in $shapes are the same object.... so... why isn't it working?

编辑：
我找到了另一种方式来搜索后aswerers之一postedd他的回答：

I found another way to search after one of the aswerers postedd his answer:

var result = -1;
jQuery.each(shapes, function(key, value){
    if (value.id == shape.id){
        result = key;
    }
});
return result;

显然，我的问题的一部分，我不能在一个循环的中间返回。 （我回来的瞬间找到匹配，这是造成一些问题。）

apparently, part of my problem is that I can't return in the middle of a loop. (I was returning the instant a match was found, which was causing some issues.)

推荐答案

您的对象不是一个数组。结果
$。inArray 只能用长度和一组上的阵列状物体的工作命名属性 0 到长度 - 1 的

Your object is not an array.
$.inArray only work on array-like objects with a length and a set of properties named 0 through length - 1.

您需要手动搜索你的非数组。结果
例如，你可以在遍历实际存在的所有属性中使用在 / 循环，看看是否有任何它们符合你的对象：

You need to search your non-array manually.
For example, you could use a for / in loop to loop through all properties that actually exist and see if any of them match your object:

for (var key in $shapes) {
    if ($shapes[key] === yourObject) {
        //Match!
    }
}

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