将阵列中存在的休假时间总和

时间：2019-07-04 07:47:57

标签： php arrays laravel arraylist

我想计算每个员工的休假时间。

我有以下表格：

出勤，带有（id，Empid，check_in，check_out，date）列

离开带有（id，empid，原因，time_long from_date，to_date）列

员工，其中包含（id，name，....）列

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"
    }
    { ... }
  ]

所以我期望从此输出获得以下结果：

ID 3 = 24小时的员工的空缺

3 个答案:

答案 0 :(得分：2)

JavaScript

编辑：更改代码以符合最新要求。

概念是什么？我们使用reduce将结果累积到一个对象中。基本上，我们采用对象的empid并在其中创建键。如果键已经存在，我们将使用键的现有值并向其中添加当前的leaveLong；如果键不存在，那么我们将从0开始我们的值，但是仍然添加当前的{ {1}}，然后创建一个键值对。

leaveLong可以理解为：使用给定键的值，如果定义了 else ，则使用0作为值。

因此，当(a[c.empid] || 0)遍历整个对象数组时，我们有一个对象将所有reduce都作为键，并将它们各自的LeaveLong值作为和值。

编辑：在此之前，我们必须empid。我们只需找到第一个离开的id并过滤掉其余的所有东西。

filter

如果您需要一个数组作为答案，请将对象放入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)。

Object.entries

在两种情况下，您都可以将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)替换为(a,c)并相应地调整功能。但这只是个人喜好。

答案 1 :(得分：2)

在这种情况下，您可以使用foreach将数据分组为Empid，

$temp = [];
foreach ($arr as $key => $value) {
    // fetching all data as per empid and leave id 
    $temp[$value['empid']][$value['leaveId']][] = $value['leaveLong'];
}
$result = [];
foreach ($temp as $key => $value) {
    foreach ($value as $key1 => $value1) {
        // fetching max value for empid and leave id
        $result[$key][$key1] = max($value1);        
    }   
}
// summing per emp id
$result = array_map("array_sum", $result);
print_r($result);

Demo。

输出 

Array
(
    [3] => 24
    [8] => 0
)

答案 2 :(得分：0)

对于javascript

假设我们将数据存储在empLeave 

// Assuming we stored the data in a variable empLeave

// fetch the data in a object format
let empLeaveArray = JSON.parse(empLeave)

// declare empty object
let result = {} 

// loop through every object in the data i.e every leave sanctioned
empLeaveArray.forEach((ele) => {
// if the employee has already taken a leave add the leave to it
if(ele.empid in result)
   result[ele.empid] += ele.leavelong
// if employee has not taken the leave already add the employee to the object
else
   result[ele.empid] = ele.leavelong
})

for (let key in result) {
  console.log("employee " + key + " has taken " + result[key] + "leaves");
}
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