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php - 对数组中存在的休假时间求和

问题描述

我想计算每个员工的休假时间。

我有以下表格：

使用 (id, empid, check_in, check_out, date) 列的出勤率

离开(id,empid,reason,time_long from_date, to_date) 列

具有 (id,name,....) 列的员工

这是我的查询：

select `emp`.*, `p`.*, `a`.*, `l`.`id` as `leaveId`, `l`.`time_long` as `leaveLong`, `l`.`from_date` as `leaveFrom`, `l`.`to_date` as `leaveTo` from `employee` as `emp` inner join `attendance` as `a` on `emp`.`id` = `a`.`empid`  left join `leave` as `l` on `emp`.`id` = `l`.`empid` where `a`.`date` between 2019-03-01 and 2019-03-31 order by `emp`.`id` asc)

查询返回以下记录。

[
   {
      "log_id": 1310,
      "name": "ahmad",
      "empid": 3,
      "check_in": "11:56",
      "check_out": "17:25",
      "date": "2019-03-23",
      "time_long": "5:28",
      "leaveId": 5,
      "leaveLong": 16,
      "leaveFrom": "2019-03-15",
      "leaveTo": "2019-03-17"
    },
    {
      "log_id": 1311,
      "name": "ahmad",
      "empid": 3,
      "check_in": "07:14",
      "check_out": "17:24",
      "date": "2019-03-24",
      "time_long": "10:9",
      "leaveId": 5,
      "leaveLong": 16,
      "leaveFrom": "2019-03-15",
      "leaveTo": "2019-03-17"
    },{
      "log_id": 1312,
      "name": "ahmad",
      "empid": 3,
      "check_in": "06:58",
      "check_out": "17:21",
      "date": "2019-03-25",
      "time_long": "10:23"
      "leaveId": 5,
      "leaveLong": 16,
      "leaveFrom": "2019-03-15",
      "leaveTo": "2019-03-17"
    },{
      "log_id": 1313,
      "name": "ahmad",
      "empid": 3,
      "check_in": "07:58",
      "check_out": "17:21",
      "date": "2019-03-26",
      "time_long": "9:23"
      "leaveId": 15,
      "leaveLong": 8.0,
      "leaveFrom": "2019-03-28",
      "leaveTo": "2019-03-29"
    },
    {
      "log_id": 1314,
      "name": "ahmad",
      "empid": 3,
      "check_in": "07:58",
      "check_out": "17:21",
      "date": "2019-03-26",
      "time_long": "9:23"
      "leaveId": 15,
      "leaveLong": 8.0,
      "leaveFrom": "2019-03-28",
      "leaveTo": "2019-03-29"
    },
    {
      "log_id": 1315,
      "name": "ahmad",
      "empid": 3,
      "check_in": "08:00",
      "check_out": "16:00",
      "date": "2019-03-27",
      "time_long": "8:00"
      "leaveId": 15,
      "leaveLong": 8.0,
      "leaveFrom": "2019-03-28",
      "leaveTo": "2019-03-29"
    }
    { ... }
  ]

所以我期望这个输出有以下结果：

ID 3 员工的假期 = 24 小时

标签： phparrayslaravelarraylist

解决方案

Javascript

编辑：更改代码以适应最新要求。

什么概念？我们用来reduce在一个对象中累积结果。基本上，我们取出一个对象empid并从中取出对象内部的一个键。如果键已经存在，我们获取键的现有值并将当前值添加leaveLong到它，或者如果键不存在，我们开始我们的值，0但仍然添加当前值leaveLong并创建一个键值对。

(a[c.empid] || 0)可以读作：如果定义，则使用给定键的值，否则使用 0 作为值。

因此，当reduce遍历我们的整个对象数组时，我们有一个对象，它将所有empid的 's 作为键，并将它们各自的 leaveLong 值作为总和值。

编辑：在此之前，我们必须filter. 我们只需找到第一次出现的 leaveId 并过滤掉所有其余的。

var arr = [{"log_id":1310,"name":"ahmad","fname":"Mohammad","photo":"images/user_profile//1550473469.jpg","title":"Doctor","description":null,"empid":3,"check_in":"11:56","check_out":"17:25","date":"2019-03-23","time_long":"5:28","leaveId":5,"leaveLong":16,"leaveFrom":"2019-03-15","leaveTo":"2019-03-17"},{"log_id":1311,"name":"ahmad","fname":"Mohammad","photo":"images/user_profile//1550473469.jpg","title":"Doctor","description":null,"empid":3,"check_in":"07:14","check_out":"17:24","date":"2019-03-24","time_long":"10:9","leaveId":5,"leaveLong":16,"leaveFrom":"2019-03-15","leaveTo":"2019-03-17"},{"log_id":1312,"name":"ahmad","fname":"Mohammad","photo":"images/user_profile//1550473469.jpg","title":"Doctor","description":null,"empid":3,"check_in":"06:58","check_out":"17:21","date":"2019-03-25","time_long":"10:23","leaveId":5,"leaveLong":16,"leaveFrom":"2019-03-15","leaveTo":"2019-03-17"},{"log_id":1313,"name":"ahmad","fname":"Mohammad","photo":"images/user_profile//1550473469.jpg","title":"Doctor","description":null,"empid":3,"check_in":"07:58","check_out":"17:21","date":"2019-03-26","time_long":"9:23","leaveId":15,"leaveLong":8.0,"leaveFrom":"2019-03-28","leaveTo":"2019-03-29"},{"log_id":1314,"name":"ahmad","fname":"Mohammad","photo":"images/user_profile//1550473469.jpg","title":"Doctor","description":null,"empid":3,"check_in":"07:58","check_out":"17:21","date":"2019-03-26","time_long":"9:23","leaveId":5,"leaveLong":8.0,"leaveFrom":"2019-03-28","leaveTo":"2019-03-29"},{"log_id":1315,"name":"ahmad","fname":"Mohammad","photo":"images/user_profile//1550473469.jpg","title":"Doctor","description":null,"empid":3,"check_in":"08:00","check_out":"16:00","date":"2019-03-27","time_long":"8:00","leaveId":5,"leaveLong":8.0,"leaveFrom":"2019-03-28","leaveTo":"2019-03-29"},{"log_id":1316,"name":"Neda Mohammad","fname":"Gada Mohammad","photo":"images/user_profile//1550473758.jpg","title":"Pharmacist","description":null,"empid":8,"check_in":"07:36","check_out":"17:57","date":"2019-03-25","time_long":"10:20","leaveId":null,"leaveLong":null,"leaveFrom":null,"leaveTo":null,},{"log_id":1317,"name":"Neda Mohammad","fname":"Gada Mohammad","photo":"images/user_profile//1550473758.jpg","title":"Pharmacist","description":null,"empid":8,"check_in":"08:00","check_out":"16:00","date":"2019-03-26","time_long":"8:00","leaveId":null,"leaveLong":null,"leaveFrom":null,"leaveTo":null,},{"log_id":1318,"name":"Neda Mohammad","fname":"Gada Mohammad","photo":"images/user_profile//1550473758.jpg","title":"Pharmacist","description":null,"empid":8,"check_in":"08:00","check_out":"16:00","date":"2019-03-27","time_long":"8:00","leaveId":null,"leaveLong":null,"leaveFrom":null,"leaveTo":null,}];

let res = arr.filter((v,i) => arr.findIndex(o => o.leaveId == v.leaveId) == i)  
             .reduce((a,c) => {a[c.empid] = (a[c.empid] || 0) + c.leaveLong; return a},{})

console.log(res)

如果您需要一个数组作为答案，请将对象放入Object.entries.

var arr = [{"log_id":1310,"name":"ahmad","fname":"Mohammad","photo":"images/user_profile//1550473469.jpg","title":"Doctor","description":null,"empid":3,"check_in":"11:56","check_out":"17:25","date":"2019-03-23","time_long":"5:28","leaveId":5,"leaveLong":16,"leaveFrom":"2019-03-15","leaveTo":"2019-03-17"},{"log_id":1311,"name":"ahmad","fname":"Mohammad","photo":"images/user_profile//1550473469.jpg","title":"Doctor","description":null,"empid":3,"check_in":"07:14","check_out":"17:24","date":"2019-03-24","time_long":"10:9","leaveId":5,"leaveLong":16,"leaveFrom":"2019-03-15","leaveTo":"2019-03-17"},{"log_id":1312,"name":"ahmad","fname":"Mohammad","photo":"images/user_profile//1550473469.jpg","title":"Doctor","description":null,"empid":3,"check_in":"06:58","check_out":"17:21","date":"2019-03-25","time_long":"10:23","leaveId":5,"leaveLong":16,"leaveFrom":"2019-03-15","leaveTo":"2019-03-17"},{"log_id":1313,"name":"ahmad","fname":"Mohammad","photo":"images/user_profile//1550473469.jpg","title":"Doctor","description":null,"empid":3,"check_in":"07:58","check_out":"17:21","date":"2019-03-26","time_long":"9:23","leaveId":15,"leaveLong":8.0,"leaveFrom":"2019-03-28","leaveTo":"2019-03-29"},{"log_id":1314,"name":"ahmad","fname":"Mohammad","photo":"images/user_profile//1550473469.jpg","title":"Doctor","description":null,"empid":3,"check_in":"07:58","check_out":"17:21","date":"2019-03-26","time_long":"9:23","leaveId":5,"leaveLong":8.0,"leaveFrom":"2019-03-28","leaveTo":"2019-03-29"},{"log_id":1315,"name":"ahmad","fname":"Mohammad","photo":"images/user_profile//1550473469.jpg","title":"Doctor","description":null,"empid":3,"check_in":"08:00","check_out":"16:00","date":"2019-03-27","time_long":"8:00","leaveId":5,"leaveLong":8.0,"leaveFrom":"2019-03-28","leaveTo":"2019-03-29"},{"log_id":1316,"name":"Neda Mohammad","fname":"Gada Mohammad","photo":"images/user_profile//1550473758.jpg","title":"Pharmacist","description":null,"empid":8,"check_in":"07:36","check_out":"17:57","date":"2019-03-25","time_long":"10:20","leaveId":null,"leaveLong":null,"leaveFrom":null,"leaveTo":null,},{"log_id":1317,"name":"Neda Mohammad","fname":"Gada Mohammad","photo":"images/user_profile//1550473758.jpg","title":"Pharmacist","description":null,"empid":8,"check_in":"08:00","check_out":"16:00","date":"2019-03-26","time_long":"8:00","leaveId":null,"leaveLong":null,"leaveFrom":null,"leaveTo":null,},{"log_id":1318,"name":"Neda Mohammad","fname":"Gada Mohammad","photo":"images/user_profile//1550473758.jpg","title":"Pharmacist","description":null,"empid":8,"check_in":"08:00","check_out":"16:00","date":"2019-03-27","time_long":"8:00","leaveId":null,"leaveLong":null,"leaveFrom":null,"leaveTo":null,}];

let res = Object.entries(
              arr.filter((v,i) => arr.findIndex(o => o.leaveId == v.leaveId) == i)
                 .reduce((a,c) => {a[c.empid] = (a[c.empid] || 0) + c.leaveLong; return a},{}))

console.log(res)

在这两种情况下，您都可以相应地替换(a,c)和(a, { empid, leaveLong})调整功能。但这只是个人喜好。

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