Question

我有2个对象，我想将它合并为一个对象数组，但我需要先使用JavaScript或AngularJS进行比较。

A = [
     {date: "2013-07-31", start_time:"2013-07-31 17:30:00+10", finish_time:"2013-07-31 20:30:00+10"},
     {date: "2013-08-03", start_time:"2013-08-03 17:00:00+10", finish_time:"2013-08-03 20:00:00+10"},
     {date: "2013-09-03", start_time:"2013-09-03 17:00:00+10", finish_time:"2013-09-03 20:00:00+10"}
    ]

B = [
     {date: "2013-07-31", start_time:"2013-07-31 17:37:49+10", finish_time:"2013-07-31 20:32:04+10"},
     {date: "2013-08-03", start_time:"2013-08-03 16:57:34+10", finish_time:"2013-08-03 20:00:57+10"}
    ]

Expected output
C = [
     {date: "2013-07-31", start_time:"late", finish_time:"on time"},
     {date: "2013-08-03", start_time:"on time", finish_time:"on time"},
    ]

如果两个对象数组具有相同的日期，我将首先进行比较，然后我将比较同一日期的开始，然后如果B上的start_time值超过A start_time，那么它将更改为字符串，它是＆＃ 34;晚期＆＃34 ;.同样对于finish_time，如果B上的值低于A，那么字符串将是＃34;太早＆＃34;。

Answer 1

对我来说，这只是一个JavaScript问题，而不是一个真正的AngularJS问题。

我认为这样的事情是你正在寻找的：

const A = [
     {date: "2013-07-31", start_time:"2013-07-31 17:30:00+10", finish_time:"2013-07-31 20:30:00+10"},
     {date: "2013-08-03", start_time:"2013-08-03 17:00:00+10", finish_time:"2013-08-03 20:00:00+10"},
     {date: "2013-08-03", start_time:"2013-08-03 17:00:00+10", finish_time:"2013-08-03 20:00:00+10"}
    ];

const B = [
     {date: "2013-07-31", start_time:"2013-07-31 17:37:49+10", finish_time:"2013-07-31 20:32:04+10"},
     {date: "2013-08-03", start_time:"2013-08-03 16:57:34+10", finish_time:"2013-08-03 20:00:57+10"},
     {date: "2013-08-03", start_time:"2013-08-03 16:57:34+10", finish_time:"2013-08-03 19:00:57+10"}
    ];


let C = [];
const maxLength = Math.min(A.length, B.length);

for (let i = 0; i < maxLength; i += 1) {
  const startTimeResult = B[i].start_time > A[i].start_time ? 'late' : 'on time';
  const finishTimeResult = B[i].finish_time > A[i].finish_time ? 'on time' : 'too early';
  C[i] = { date: A[i].date, start_time: startTimeResult, finish_time: finishTimeResult };
  console.log(C[i]);
}

https://codepen.io/joseph4tw/pen/mXGeoO?editors=1012

Answer 2

根据我的理解，基本上你想要针对B中存在的值搜索数组A内的值，并计算开始时间是早期，晚期还是准时。

https://stackoverflow.com/users/1148564/joseph-serido的解决方案将完美运行，但需要O（n ^ 2）时间。在Javascript中，您可以通过键名访问对象。如果将A和B作为对象，则可以利用此行为并使代码仅使用一个循环运行。

var A_Times = {}; //Make an Object
//Use the date as Key
A_Obj["2013-07-31"] = {start_time:"2013-07-31 17:30:00+10", finish_time:"2013-07-31 20:30:00+10"};
A_Obj["2013-08-03"] ={ start_time:"2013-08-03 17:00:00+10", finish_time:"2013-08-03 20:00:00+10"};

var B_Times = {};
B_Times["2013-07-31"] = {start_time:"2013-07-31 17:37:49+10", finish_time:"2013-07-31 20:32:04+10"};
B_Times["2013-08-03"] ={start_time:"2013-08-03 16:57:34+10", finish_time:"2013-08-03 20:00:57+10"};

var A_Days = Object.keys(A_Times); //Get all the days in A

for(var i=0; i<A_Times_Days.length; i++){
    var day = A_Times_Days[i];
    console.log(day); //Log the Key here
    var A_Data = A_Times[day];
    var B_Data = B_Times[day];
    console.log(A_Data);
    console.log(B_Data);
    //Compute punctuality here based on A_Data and B_Data
}

但是，如果您的用例只涉及数天的数据，那么这种增加的复杂性可能不值得。由你来决定权衡。希望它有所帮助。

Answer 3

嗨，这就是我想出来的，我不知道该说些什么。我认为代码不言自明：p

var A = [
     {date: "2013-07-31", start_time:"2013-07-31 17:30:00+10", finish_time:"2013-07-31 20:30:00+10"},
     {date: "2013-08-03", start_time:"2013-08-03 17:00:00+10", finish_time:"2013-08-03 20:00:00+10"}
    ];

var B = [
     {date: "2013-07-31", start_time:"2013-07-31 17:37:49+10", finish_time:"2013-07-31 20:32:04+10"},
     {date: "2013-08-03", start_time:"2013-08-03 16:57:34+10", finish_time:"2013-08-03 20:00:57+10"}
     ];

function getResult()
{
    var results = [];
    for(var i = 0; i < A.length; i++)
    {
        var objA = A[i];

        for(var j = 0; j < B.length; j++)
        {
            var objB = B[j];

            if(objB.date === objA.date)
            {
                var o = {};
                o.date = objA.date;

                //if start_time of A is less than start_time of B
                if(Date.parse(objA.start_time) < Date.parse(objB.start_time))
                    o.start_time = "late";
                else
                    o.start_time = "on time";

                //if end_time of A is less than end_time of B
                if(Date.parse(objA.finish_time) < Date.parse(objB.finish_time))
                    o.finish_time = "too early";
                else
                    o.finish_time = "on time";

                results.push(o);
            }
        }
    }

    if(results.length !== 0)
        return results;

    return null;
}

P.S。它只输出A日期等于B日期的对象

如何使用javascript / angularjs比较两个对象数组？

3 个答案: