Question

status_name=Array("a","b","c","b","e","f");
status_id=Array( 1, 2, 3, 4, 5, 6);

如何组合这两个数组并构建多维数组预期的多维数组就像这样

[["a", 1],["b", 2],["c", 3],["d", 4],["e", 5],["f", 6]]

帮助我如何使用上面两个数组值并构建我期望的多维数组

Answer 1

由于你包含了jQuery，你可以使用jQuery.map与Linus的回答类似：

var result      = [],
    status_name = ["a","b","c","b","e","f"],
    status_id   = [1, 2, 3, 4, 5, 6];

result = $.map(status_name, function (el, idx) {
    return [[el, status_id[idx]]];
});

看看你的变量名，我猜你是来自一种语言（比如PHP）。如果是这种情况，请确保记住使用var关键字声明局部变量，否则您将污染全局范围，并且您将在IE中遇到一些可怕的错误。

Answer 2

JavaScript没有buitin方法，但您可以自己轻松编写：

function zip(arrayA, arrayB) {
    var length = Math.min(arrayA.length, arrayB.length);
    var result = [];
    for (var n = 0; n < length; n++) {
        result.push([arrayA[n], arrayB[n]]);
    }
    return result;
}

选择名称zip是因为执行此类操作的函数通常在其他语言中称为zip。

Answer 3

var combined = [], length = Math.min(status_name.length, status_id.length);
for(var i = 0; i < length; i++) {
    combined.push([status_name[i], status_id[i]]);
}

您也可以使用Array.prototype.map，但并非所有浏览器都支持：

var combined = status_name.map(function(name, index) { return [name, status_id[index]] });

Answer 4

我尝试了自己，并提出了这个解决方案，这可能对某人有所帮助

  status_name=Array("a","b","c","b","e","f");
    status_id=Array( 1, 2, 3, 4, 5, 6);

脚本：

            Values=[];
            for (i = 0; i < status_name.length; ++i)
            {
                Values[i] =Array(status_name[i], status_id[i]);
            }

Answer 5

试

function array_combine (keys, values) {
    // Creates an array by using the elements of the first parameter as keys and the elements of the second as the corresponding values  
    // 
    // version: 1102.614
    // discuss at: http://phpjs.org/functions/array_combine
    // +   original by: Kevin van Zonneveld (http://kevin.vanzonneveld.net)
    // +   improved by: Brett Zamir (http://brett-zamir.me)
    // *     example 1: array_combine([0,1,2], ['kevin','van','zonneveld']);
    // *     returns 1: {0: 'kevin', 1: 'van', 2: 'zonneveld'}
    var new_array = {},
        keycount = keys && keys.length,
        i = 0;

    // input sanitation
    if (typeof keys !== 'object' || typeof values !== 'object' || // Only accept arrays or array-like objects
    typeof keycount !== 'number' || typeof values.length !== 'number' || !keycount) { // Require arrays to have a count
        return false;
    }

    // number of elements does not match
    if (keycount != values.length) {
        return false;
    }

    for (i = 0; i < keycount; i++) {
        new_array[keys[i]] = values[i];
    }

    return new_array;

参考
- arr combine
- array combine

Answer 6

使用jQuery.map

var status_name = ["a","b","c","b","e","f"],
    status_id = [1,2,3,4,5,6],
    r = [];

r = $.map(status_name, function(n, i) {
    return [[n, status_id[i]]]; 
});

请注意return [[n, status_id[i]]]和return [n, status_id[i]]之间的区别。使用前者将产生2d数组，而使用后者将产生1d数组。

在javascript中将两个数组合并为单个多维数组

6 个答案: