jQuery.each：重新排序数组

发布于 2024-11-25 02:32:34 字数 429 浏览 2 评论 0原文

我最近一直在摆弄数组和数组排序，并偶然发现了一些奇怪的东西。

考虑这种情况：

arr = {};
arr[1] = "one";
arr[2] = "two";
arr[105] = "three";
arr[4] = "four";

$.each(arr, function (key, val) {
    $(body).html(key + " => " + val);
});

现在，我们应该希望得到以下结果：

1 => one
2 => two
105 => three
4 => four

对吗？不幸的是没有。我收到数字排序，导致索引 105 成为序列中的最后一项。有人知道我如何克服这个问题吗？非常感谢您的指导，谢谢。

原文

I've been fiddling around with arrays and array ordering recently and stumbled across something peculiar.

Take this situation:

arr = {};
arr[1] = "one";
arr[2] = "two";
arr[105] = "three";
arr[4] = "four";

$.each(arr, function (key, val) {
    $(body).html(key + " => " + val);
});

Now, we should hope for the following results:

1 => one
2 => two
105 => three
4 => four

Right? Unfortunately not. I am receiving a numeric sorting which results in index 105 being the last item in the sequence. Anybody have an idea of how I can overcome this problem? Words of guidance are very much appreciated, thank you.

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檐上三寸雪 2024-12-02 02:32:34

那不是一个数组。它是一个对象。因此，没有保证订单。

为了保证某种序列，您可以在数组中定义序列，然后迭代该数组，从对象中选择每个数组值的索引。

arr = {};
arr[1] = "one";
arr[2] = "two";
arr[105] = "three";
arr[4] = "four";

var order = [1,2,105,4];

$.each(order, function(i,val) {
    console.log( val + '=>' + arr[ val ] );
});

That's not an Array. It's an Object. And as such, there is no guaranteed order.

To guarantee some sort of sequence, you could define the sequence in an Array, then iterate that Array, selecting the index of each array value from the object.

arr = {};
arr[1] = "one";
arr[2] = "two";
arr[105] = "three";
arr[4] = "four";

var order = [1,2,105,4];

$.each(order, function(i,val) {
    console.log( val + '=>' + arr[ val ] );
});

回复收藏 0 原文

失去的东西太少 2024-12-02 02:32:34

也许你可以尝试这个：

arr =
[   ['1', 'one']
,   ['2', 'two']
,   ['105', 'three']
,   ['4', 'four']
];

$.each(arr, function(key, values)
{
    $(body).html(values[0] + " => " + values[1]);
});

Maybe you could try this instead:

arr =
[   ['1', 'one']
,   ['2', 'two']
,   ['105', 'three']
,   ['4', 'four']
];

$.each(arr, function(key, values)
{
    $(body).html(values[0] + " => " + values[1]);
});

回复收藏 0 原文

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