排序并将空值添加到数组如果需要

-3

我有一个数组，我要计数和排序。我怎样才能做到这一点？我的尝试会导致无限循环。排序并将空值添加到数组如果需要

我想这个数组

[ 
{activityId: 201609011024550000, Employee_employeeId: 4075, projectId: 300000000000000200, Date_date: "2017-12-05", timestamp: "2017-12-05 15:58:25", …} 
{activityId: 201609011024550000, Employee_employeeId: 4075, projectId: 300000000000000200, Date_date: "2017-12-06", timestamp: "2017-12-06 10:13:12", …} 
{activityId: 201609011024550000, Employee_employeeId: 4075, projectId: 300000000000000200, Date_date: "2017-12-06", timestamp: "2017-12-07 09:38:12", …} 
{activityId: 201609011024550000, Employee_employeeId: 4075, projectId: 300000000000000200, Date_date: "2017-12-08", timestamp: "2017-12-08 09:35:25", …} 
{activityId: 201609011024550000, Employee_employeeId: 4075, projectId: 300000000000000200, Date_date: "2017-12-11", timestamp: "2017-12-11 11:25:31", …} 
{activityId: 201609011024550000, Employee_employeeId: 4075, projectId: 300000000000000200, Date_date: "2017-12-12", timestamp: "2017-12-12 15:05:17", …} 
{activityId: 201609011024550000, Employee_employeeId: 4075, projectId: 300000000000000200, Date_date: "2017-12-13", timestamp: "2017-12-13 11:08:40", …} 
{activityId: 201609011024550000, Employee_employeeId: 100, projectId: 300000000000000200, Date_date: "2017-12-15", timestamp: "2017-12-15 11:51:33", …} 
{activityId: 201609011024550000, Employee_employeeId: 4075, projectId: 300000000000000200, Date_date: "2017-12-18", timestamp: "2017-12-18 11:04:56", …} 
{activityId: 201609011024550000, Employee_employeeId: 4075, projectId: 300000000000000200, Date_date: "2017-12-15", timestamp: "2017-12-18 12:04:56", …} 
]

看起来像基于此阵/排序上Date_date

（Date_date： “2017年12月1日” 不存在，所以值是0， Date_date：“2017-12-06”存在两次，因此值为2）

[0,0,0,0...,1,2,0,1...]

好的，如果新建数组已创建。没有jquery。

编辑：对不起，这里是循环不停止

for(var i = 0; i < a.length; i++){ 
    for(var j = 0; j <= b.length; j++){ 
     if(parseInt(a[i].Date_date.substring(8))-1==j){ 
      var num = b[j]; 
      b[j] = num+1; 
     } 
     else{ 
      var num = b[j]; 
      b[j] = num+0; 
     } 
    } 
}

来源

2017-12-18 Softy

请更新这个职位与你已经尝试过的代码， – 31piy

添加您尝试过的代码，而不是直接要求的解决方案。另外，添加日期范围。 –

看来你的代码有问题。但是，除非我们有[可以重现问题的代码或信息]（// stackoverflow.com/help/mcve），否则我们无能为力。否则，我们只是盲目猜测。 – Cerbrus

首先排序阵列上日期，然后创建一个对象，查找基于Date_date。然后从排序后的数组中获取第一个和最后一个日期，并为该月的第一天和该月的最后一天创建日期对象。然后使用while循环，通过检查对象查找添加记录直到最后一个日期。

var data = [{activityId: 201609011024550000, Employee_employeeId: 4075, projectId: 300000000000000200, Date_date: "2017-12-05", timestamp: "2017-12-05 15:58:25"},{activityId: 201609011024550000, Employee_employeeId: 4075, projectId: 300000000000000200, Date_date: "2017-12-06", timestamp: "2017-12-06 10:13:12"},{activityId: 201609011024550000, Employee_employeeId: 4075, projectId: 300000000000000200, Date_date: "2017-12-06", timestamp: "2017-12-07 09:38:12"},{activityId: 201609011024550000, Employee_employeeId: 4075, projectId: 300000000000000200, Date_date: "2017-12-08", timestamp: "2017-12-08 09:35:25"},{activityId: 201609011024550000, Employee_employeeId: 4075, projectId: 300000000000000200, Date_date: "2017-12-11", timestamp: "2017-12-11 11:25:31"},{activityId: 201609011024550000, Employee_employeeId: 4075, projectId: 300000000000000200, Date_date: "2017-12-12", timestamp: "2017-12-12 15:05:17"},{activityId: 201609011024550000, Employee_employeeId: 4075, projectId: 300000000000000200, Date_date: "2017-12-13", timestamp: "2017-12-13 11:08:40"},{activityId: 201609011024550000, Employee_employeeId: 100, projectId: 300000000000000200, Date_date: "2017-12-15", timestamp: "2017-12-15 11:51:33"},{activityId: 201609011024550000, Employee_employeeId: 4075, projectId: 300000000000000200, Date_date: "2017-12-18", timestamp: "2017-12-18 11:04:56"},{activityId: 201609011024550000, Employee_employeeId: 4075, projectId: 300000000000000200, Date_date: "2017-12-15", timestamp: "2017-12-18 12:04:56"},]; 
 

 
var dateCount = data 
 
    .sort((a,b) => new Date(a.Date_date) - new Date(b.Date_date)) 
 
    .reduce((r, o) => { 
 
     r[o.Date_date] = r[o.Date_date] || []; 
 
     r[o.Date_date].push(o) 
 
     return r; 
 
    }, {}); 
 

 
//Get the first date from the sorted array 
 
var date1 = new Date(data[0].Date_date); 
 
var firstDay = new Date(date1.getFullYear(), date1.getMonth(), 2); 
 

 
//Get the last date from the sorted array 
 
var date2 = new Date(data[data.length - 1].Date_date); 
 
var lastDay = new Date(date2.getFullYear(), date2.getMonth() + 1, 1); 
 

 
var result = []; 
 
while(firstDay <= lastDay){ 
 
    var count = (dateCount[firstDay.toISOString().split('T')[0]] || []).length; 
 
    result.push(count); 
 
    firstDay.setDate(firstDay.getDate() + 1); 
 
} 
 

 
console.log(result);

来源

2017-12-18 14:15:53

非常感谢！ – Softy

哈桑，看起来00天正在取代第31天。阵列长度为31，但第5天在[5]，应该在pos [4]。我怎样才能解决这个问题？我会提出一个新问题吗？ – Softy

更新了解决方案，请现在检查。 –

首先我试过的代码，你需要得到每个日期出现在一个阵列的次数。所以我使用data.forEach迭代了数组中的每个对象。

请注意，我正在跟踪名为counter的对象中的日期计数。

然后，我使用data.map将阵列中的每个对象映射到新阵列中的项目。因此，对于data阵列中的每个项目，我都会返回观察次数。

如果你不熟悉Array#map和Array#forEach，你可以阅读更多。

var data = [ 
 
{activityId: 201609011024550000, Employee_employeeId: 4075, projectId: 300000000000000200, Date_date: "2017-12-05", timestamp: "2017-12-05 15:58:25"}, 
 
{activityId: 201609011024550000, Employee_employeeId: 4075, projectId: 300000000000000200, Date_date: "2017-12-06", timestamp: "2017-12-06 10:13:12"}, 
 
{activityId: 201609011024550000, Employee_employeeId: 4075, projectId: 300000000000000200, Date_date: "2017-12-06", timestamp: "2017-12-07 09:38:12"}, 
 
{activityId: 201609011024550000, Employee_employeeId: 4075, projectId: 300000000000000200, Date_date: "2017-12-08", timestamp: "2017-12-08 09:35:25"}, 
 
{activityId: 201609011024550000, Employee_employeeId: 4075, projectId: 300000000000000200, Date_date: "2017-12-11", timestamp: "2017-12-11 11:25:31"}, 
 
{activityId: 201609011024550000, Employee_employeeId: 4075, projectId: 300000000000000200, Date_date: "2017-12-12", timestamp: "2017-12-12 15:05:17"}, 
 
{activityId: 201609011024550000, Employee_employeeId: 4075, projectId: 300000000000000200, Date_date: "2017-12-13", timestamp: "2017-12-13 11:08:40"}, 
 
{activityId: 201609011024550000, Employee_employeeId: 100, projectId: 300000000000000200, Date_date: "2017-12-15", timestamp: "2017-12-15 11:51:33"}, 
 
{activityId: 201609011024550000, Employee_employeeId: 4075, projectId: 300000000000000200, Date_date: "2017-12-18", timestamp: "2017-12-18 11:04:56"}, 
 
{activityId: 201609011024550000, Employee_employeeId: 4075, projectId: 300000000000000200, Date_date: "2017-12-15", timestamp: "2017-12-18 12:04:56"} 
 
]; 
 

 
var counter = {}; 
 

 
data.forEach(function(item) { 
 
    var date = item.Date_date; 
 
    
 
    if(!!counter[date]) { 
 
    counter[date]++; 
 
    } else { 
 
    counter[date] = 1; 
 
    } 
 
}); 
 

 
console.log("Date-wise counts: ", counter); 
 

 
var result = data.map(function(item) { 
 
    return counter[item.Date_date]; 
 
}); 
 

 
console.log("Result:", result);

来源

2017-12-18 14:08:13 Nisarg

嗨Nisarg和谢谢你的答案。这看起来在正确的道路上，我仍然需要添加失踪日期aswell（“2017-12-01”：0，）。我是否创建一个新的循环并使用计数器创建一个新的数组？ – Softy

@Softy是的，类似的东西。我错过了这个问题中的那一部分。你想让我展示一下吗？ – Nisarg

没问题，我会用哈桑提供的答案。您的回答仍将由我维持（我的声望在15以下，因此投票将在稍后开始） – Softy

排序并将空值添加到数组如果需要

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