Php 获取laravel 5和mysql中groupby对象的第一个和最后一个值

我有一张叫做收入的桌子

+----+-------------+---------------------+--------+--------+--------+-----------+---------------------+---------------------+
| id | employee_id | date                | gross  | income | credit | comission | created_at          | updated_at          |
+----+-------------+---------------------+--------+--------+--------+-----------+---------------------+---------------------+
|  1 |           1 | 2016-03-30 19:21:09 | 100.00 |  29.00 |  11.00 |     60.00 | 2016-03-31 19:21:46 | 2016-03-31 19:21:46 |
|  2 |           1 | 2016-03-31 19:24:44 | 110.00 |  43.00 |   1.00 |     60.00 | 2016-03-31 19:24:56 | 2016-03-31 19:24:56 |
|  3 |           2 | 2016-03-31 21:44:09 |  77.00 |  30.80 |   0.00 |     60.00 | 2016-03-31 21:44:19 | 2016-03-31 21:44:19 |
+----+-------------+---------------------+--------+--------+--------+-----------+---------------------+---------------------+

我要做的是查询这些数据并按

employee\u id

对其进行分组，然后我要它的开始日期和结束日期。我得到的是

$records = Income::whereBetween('date', [$start, $end])
                ->groupBy('employee_id')
                ->selectRaw('store_incomes.* , sum(gross) as total_gross , sum(income) as total_income, sum(credit) as total_credit')
                ->get();

这始终返回该员工的第一个日期。例如，

employee_id 1

始终返回日期

2016-03-30 19:21:09

。我想要的是为每个

员工id

获取第一个和最后一个日期，因此

员工id 1

的开始日期为

2016-30-30

，结束日期为

2016-03-31

。有没有一种方法可以做到这一点而不需要混乱的手动代码

对于当前代码，我的输出是：

[
{"id":"1","employee_id":"1","date":"2016-03-30 19:21:09","gross":"100.00","income":"29.00","credit":"11.00","comission":"60.00","created_at":"2016-03-31 19:21:46","updated_at":"2016-03-31 19:21:46","total_gross":"210.00","total_income":"72.00","total_credit":"12.00"},
{"id":"3","employee_id":"2","date":"2016-03-31 21:44:09","gross":"77.00","income":"30.80","credit":"0.00","comission":"60.00","created_at":"2016-03-31 21:44:19","updated_at":"2016-03-31 21:44:19","total_gross":"77.00","total_income":"30.80","total_credit":"0.00"}
]

我的预期输出包括另外两个值（

start\u date

和

end\u date

）：

---

您需要的原始MySQL查询大致如下：

SELECT MIN(date) AS start_date, MAX(date) AS end_date, store_incomes.*,
    SUM(gross) AS total_gross , SUM(income) AS total_income, SUM(credit) AS total_credit
FROM income
GROUP BY employee_id

要在Laravel中执行此操作，只需在原始查询中添加

MIN（）

和

MAX（）

：

$records = DB::Income
               ->selectRaw('store_incomes.* , min(date) as start_date, max(date) as end_date, sum(gross) as total_gross , sum(income) as total_income, sum(credit) as total_credit')
               ->groupBy('employee_id')
               ->get();

---

向我们展示您的预期输出。感谢您的评论。我已编辑以包含所需的输出。只是想知道我是否可以在selectRaw中实现这一点？您是只想要日期还是想要其他信息？如果有两个日期，您的聚合可能没有意义。因为我可能会说3个员工id，然后说我想显示从2月到4月的记录。当我查询时，我希望我的结果按每个员工分组，每个员工在开始日期和结束日期之间的收入。但是，并非所有员工都在同一日期开始，并在同一日期结束。因此，我必须显示每个员工的开始和结束日期。

在两者之间（'date'，[$start，$end]）的目的是什么

那么，如果您为每个员工取最小和最大日期？

$records = DB::Income
               ->selectRaw('store_incomes.* , min(date) as start_date, max(date) as end_date, sum(gross) as total_gross , sum(income) as total_income, sum(credit) as total_credit')
               ->groupBy('employee_id')
               ->get();