Question

我有以下表结构：

表___BillableDatas：

|--------|---------------|------------|----------|--------------|------------|
| BIL_Id | BIL_BookingId | BIL_Date   | BIL_Rate | BIL_Quantity | BIL_Status |
|--------|---------------|------------|----------|--------------|------------|
| 1      | 21            | 2017-12-23 | 10.00    | 2            | charged    |
| 2      | 21            | 2017-12-23 | 105.00   | 1            | charged    |
| 3      | 21            | 2017-12-24 | 105.00   | 1            | charged    |
| 4      | 21            | 2017-12-25 | 105.00   | 1            | notcharged |
| 5      | 21            | 2017-12-26 | 105.00   | 1            | notcharged |
| 6      | 21            | 2017-12-26 | 30.00    | 2            | charged |
|--------|---------------|------------|----------|--------------|------------|

我希望array显示每天的项目总数，具体取决于状态charged和notcharged。

所需的输出应如下：

Array
(
    [2017-12-23] => Array
        (
            [charged] => 125.00
            [notcharged] => 0.00
        )
    [2017-12-24] => Array
        (
            [charged] => 105.00
            [notcharged] => 0.00
        )
    [2017-12-25] => Array
        (
            [charged] => 0.00
            [notcharged] => 105.00
        )
    [2017-12-26] => Array
        (
            [charged] => 30.00
            [notcharged] => 105.00
        )
)

我试过的查询：

select a.BIL_BookingId,a.BIL_Date,a.BIL_Rate*a.BIL_Quantity as total, a. BIL_Status
from ___BillableDatas as a

Answer 1

有条件地总结，检查分组：

select a.BIL_BookingId
      ,a.BIL_Date
      ,CASE WHEN BIL_Status = 'charged' THEN a.BIL_Rate*a.BIL_Quantity ELSE 0 END as charged
      ,CASE WHEN BIL_Status = 'notcharged' THEN a.BIL_Rate*a.BIL_Quantity ELSE 0 END as notcharged
  from ___BillableDatas as a
  group by a.BIL_BookingId, a.BIL_Date

Answer 2

你应该使用sum和group by（如果null，则为null时为0）

select BIL_Date, 
  ifnull(sum(case when BIL_Status ='charged' then BIL_Rate else 0 end), 0) as charged,
  ifnull(sum(case when BIL_Status ='notcharged' then BIL_Rate else 0 end), 0) as notcharged
from ___BillableDatas
group by BIL_Date

Answer 3

以下查询将每个日期返回一行：

php7.0-xml

由于select BIL_Date, sum((BIL_Status = 'charged') * BIL_Rate * BIL_Quantity) as charged, sum((BIL_Status = 'notcharged') * BIL_Rate * BIL_Quantity) as notcharged from ___BillableDatas where BIL_BookingId = 21 group by BIL_Date不在结果数组中，我假设您只需要在WHERE子句中定义的特定BookingId的数据。

请注意，像BookingId这样的表达式将返回(BIL_Status = 'charged')表示TRUE，或1表示MySQL中的FALSE。这样就可以避免长CASE语句。也不需要0或COALESCE()，因为IFNULL()始终至少有一行具有非NULL值。

在PHP中执行查询后，您可以在获取数据的同时在一个循环中构建嵌套数组：

SUM()

查询表并根据其状态求和数据

3 个答案: