如何合并两个具有不同 WHERE 和特殊条件的 select

Question

我有table这样的东西：

date|status|value

日期是日期，状态为 1 为待处理，2 为已确认价值是订单的价值

我想得到 3 列：

date|#status pending|#status pending+confirmed

数据示例：

+------------+-----------------+-----------------+
| date       | status          | value           |
+------------+-----------------+-----------------+
| 2015-11-17 |              1  |               89|
| 2015-11-16 |              1  |               6 |
| 2015-11-16 |              2  |              16 |
| 2015-11-16 |              2  |              26 |
| 2015-11-15 |              2  |              26 |
| 2015-11-14 |              2  |              24 |
+------------+-----------------+-----------------+

我想要的例子：

+------------+-----------------+-----------------+
| date       | confirmed       |confirmed+pending|
+------------+-----------------+-----------------+
| 2015-11-17 |              0  |               1 |
| 2015-11-16 |              2  |               3 |
| 2015-11-15 |              1  |              1 |
| 2015-11-14 |              1  |              1 |
+------------+-----------------+-----------------+

我正在尝试做：

    SELECT array1.DATE
        ,array1.confirmed
        ,array2.total
    FROM (
        SELECT DATE (DATE) AS DATE
            ,count(value) AS confirmed
        FROM Orders
        WHERE STATUS = '2'
        GROUP BY DATE (DATE) DESC limit 5
        ) AS array1
    INNER JOIN (
        SELECT DATE (DATE) AS DATE
            ,count(value) AS total
        FROM Orders
        GROUP BY DATE (DATE) DESC limit 5
        ) AS array2

但我每个日期得到 4 个结果，重复确认价值和不同的总交易。

如果我尝试分开，我可以得到两个正确的信息：

将仅列出最近 5 天的已确认订单总数：

SELECT array1.DATE
    ,array1.confirmed
    ,array2.total
FROM (
    SELECT DATE (DATE) AS DATE
        ,count(valor) AS confirmed
    FROM Orders
    WHERE STATUS = '2'
    GROUP BY DATE (DATE) DESC limit 5;
    )

将列出最近 5 天的所有订单总和：

SELECT DATE (DATE) AS DATE
    ,count(valor) AS total
FROM Orders
GROUP BY DATE (DATE) DESC limit 5

我观察到至少一个大问题：

有时我们会有一天有很多未确认订单和零确认订单，所以内部连接可能会失败。

Answer 1

您的 INNER JOIN 中缺少 ON 子句。或者，由于在您的情况下您加入的列在两侧相同，您可以使用 USING:

SELECT array1.DATE
    ,array1.confirmed
    ,array2.total
FROM (
    SELECT DATE (DATE) AS DATE
        ,count(value) AS confirmed
    FROM Orders
    WHERE STATUS = '2'
    GROUP BY DATE (DATE) DESC limit 5
    ) AS array1
INNER JOIN (
    SELECT DATE (DATE) AS DATE
        ,count(value) AS total
    FROM Orders
    GROUP BY DATE (DATE) DESC limit 5
    ) AS array2
USING (DATE)

Answer 2

你可以使用CASE WHEN，为了得到预期的输出，你已经给了。

SELECT `date`,
    (SUM(CASE WHEN `status`=1 THEN 1 ELSE 0 END)) AS Confirmed,
    (SUM(CASE WHEN `status`=1 OR `status`=2 THEN 1 ELSE 0 END)) AS Confirmed_Pending
    FROM
    table_name
    GROUP BY DATE(`date`) DESC

希望对您有所帮助。

Answer 3

一种更简单的方法是使用 case 表达式来评估状态是否是您想要计算的状态，并将 count 函数应用于该表达式：

SELECT   DATE (`date`) AS `date`,
         COUNT(CASE status WHEN 2 THEN 1 END) AS `confirmed`,
         COUNT(CASE WHEN status IN (1, 2) THEN 1 END) AS `pending and confirmed`,
FROM     orders
GROUP BY DATE (`date`) DESC

如何合并两个具有不同 WHERE 和特殊条件的 select

How merge two select with different WHERE and special conditions

mysql

sql

inner-join