Question

我想生成JSON数组，如下所示。

完全详细的数组;

MyArray = [{ prj : "P1", days : "8", ot : "2" }, { prj : "P2", days : "8", ot : "2" }, { prj : "P2", days : "8", ot : "2" }, { prj : "P1", days : "8", ot : "2" }{prj : "P3", days : "8", ot : "2" }, { prj : "P2", days : "8", ot : "2" }, { prj : "P3", days : "8", ot : "2" }];

但我想按如下方式压缩它;

MyArray = [{ prj : "P1", days : "16", ot : "4" }, { prj : "P2", days : "24", ot : "6" }, { prj : "P3", days : "16", ot : "4" }];

如果存在project_number，它应该得到天数和OT;

See the example

Answer 1

试试这个，这符合你的需要。

var MyArray = [];

//This will be used to construct your JSON object
function ConstructJson(prj, days, ot) {

    var xObj = {};
    xObj.prj = prj;
    xObj.days = days;
    xObj.ot = ot;

    MyArray[MyArray.length] = xObj;

}

 //This will minimise your json object.
function MinimiseJson() {

    var xTempArray = [];
    var xTempObj = {};
    var xPrj = '';

    for (var i = 0; i < MyArray.length; i++) {

        if (xPrj == '' && MyArray[i]) {

            xPrj = MyArray[i].prj;

            xTempObj.prj = xPrj;
            xTempObj.days = MyArray[i].days;
            xTempObj.ot = MyArray[i].ot;

            for (var j = i + 1; j < MyArray.length; j++) {

                if (MyArray[j] && MyArray[j].prj == xPrj) {

                    xTempObj.days += MyArray[j].days;
                    xTempObj.ot += MyArray[j].ot;

                    MyArray[j] = null;
                }

            }

            xPrj = '';
            xTempArray[xTempArray.length] = xTempObj;
            xTempObj = {};
        }
    }

    return xTempArray;

}

LIVE - DEMO

Answer 2

一个非常简单的解决方案就是遍历所有日子，并在您经历时跟踪值的总和。我按如下方式分叉你的jsfiddle：http://jsfiddle.net/BM9VE/，这是JS：

$(function () {
    var proj, score, MyArray = [], tally = {};
    for (var i = 1; i <= 7; i++) {
        proj = $('#p' + i).val();
        score = tally[proj] || {prj: proj, days: 0, ot: 0}; // default val if we haven't seen this day yet
        score.days += parseInt($('#d' + i).val());
        score.ot += parseInt($('#ot' + i).val());
        tally[proj] = score;
    }
    for (var day in tally) {
        // your example has days and ot as strings, this code keeps them as numbers; feel free to convert them back into strings if you need to, here
        MyArray.push(tally[day]);
    }
    $('body').append('<p>MyArray: ' + JSON.stringify(MyArray) + '</p>');
});

jquery使用天数和OT后跟project_no创建数组

2 个答案:

LIVE - DEMO