Javascript数组。先进的

Javascript数组。先进的,javascript,arrays,sorting,hasownproperty,Javascript,Arrays,Sorting,Hasownproperty,我的php脚本中有JSON字符串,如下所示: var r.co = { "A20018425":[ {"balance":"1390.31"}, // 1 {"balance":"1304.11"}, // 2 {"balance":"1188.11"},

我的php脚本中有JSON字符串,如下所示:

var r.co = {
                "A20018425":[
                                {"balance":"1390.31"},      // 1
                                {"balance":"1304.11"},      // 2
                                {"balance":"1188.11"},      // 3
                                {"balance":"1421.71"}       // 4
                            ],

                "A25005922":[
                                {"balance":"1000"},         // 1
                                {"balance":"1000.86"},      // 2
                                {"balance":"986.32"},       // 3
                                {"balance":"988.96"},       // 4
                                {"balance":"980.26"},       // 5
                                {"balance":"980.16"}        // 6 MAX 
                            ],

                "A25005923":[
                                {"balance":"1001"},         // 1
                                {"balance":"1000.16"},      // 2
                            ]           
            }
我不知道有多少AXXXXXXX元素以及它包含多少元素。 要获得我所拥有的元素,请使用以下代码:

var accounts = [];
for(var key in r.co) {
   if(r.co.hasOwnProperty(key)) {
      accounts.push(key);
   }
}
现在我知道我的A元素长度了

var accounts_length = accounts.length; // 3 for example
现在我需要知道A中元素的最大长度:

var accounts_elements_length = [];

    for (var c = 0; c < accounts.length; c++) {   

          accounts_elements_length.push(r.co[accounts[c]].length);
    }

    var accounts_elements_length_max = accounts_elements_length.max() // 6 For example
谢谢

代码:

var outputData = [];
for (var i = 0; i < 6; i++) { // filter should be - i < accounts_elements_length_max
    var temp = {
        'count': i + 1
    };

    for (var j = 0; j < accounts.length; j++) {
        if (r[accounts[j]][i]) temp[accounts[j]] = r[accounts[j]][i].balance;
    }

    outputData.push(temp);
}

工作

只是结合了您的算法:

var outputData=[];
用于(r.co中的var键){
if(r.co.hasOwnProperty(钥匙)){
var账户长度=r.co[key]。长度;
对于(var c=0;c

虽然完全可以做您想做的事情,但如果可能的话,我建议更改您的数据结构:您现在拥有的数据不适合进行有用/高效的处理。我无法更改输入数据。您到底想做什么?
outputData
的数据结构与
r.co
var outputData = [];
for (var i = 0; i < 6; i++) { // filter should be - i < accounts_elements_length_max
    var temp = {
        'count': i + 1
    };

    for (var j = 0; j < accounts.length; j++) {
        if (r[accounts[j]][i]) temp[accounts[j]] = r[accounts[j]][i].balance;
    }

    outputData.push(temp);
}
[{
    "count": 1,
    "A20018425": "1390.31",
    "A25005922": "1000",
    "A25005923": "1001"},
{
    "count": 2,
    "A20018425": "1304.11",
    "A25005922": "1000.86",
    "A25005923": "1000.16"},
{
    "count": 3,
    "A20018425": "1188.11",
    "A25005922": "986.32"},
{
    "count": 4,
    "A20018425": "1421.71",
    "A25005922": "988.96"},
{
    "count": 5,
    "A25005922": "980.26"},
{
    "count": 6,
    "A25005922": "980.16"}]​