php - 一次查询中的多个数学值

Question

我对一个查询有疑问。

SELECT
        CASE WHEN b.value IS NULL 
        THEN sum(d.value)
        ELSE sum(d.value) + sum(b.value)
        END AS sumaWydatkow

        FROM data d
        LEFT JOIN data_bills b 
        ON  d.id = b.data_id
        WHERE   d.type = 0

        UNION

        SELECT

        CASE WHEN b.value IS NULL 
        THEN sum(d.value)
        ELSE sum(d.value) + sum(b.value)
        END AS sumaPrzychodow

        FROM data d
        LEFT JOIN data_bills b 
        ON  d.id = b.data_id
        WHERE   d.type = 1

我想对两列求和，或者如果其中一列（data_bills）为空，则只对一个表求和，列类型有两个版本，0 和 1。

但现在是一个 clumn 有两个记录。我想要两列，每列一列。我在问是否可以通过一个查询来完成？

谢谢

Answer 1

你为什么做这个：

CASE WHEN b.value IS NULL 
    THEN sum(d.value)
    ELSE sum(d.value) + sum(b.value)
    END

? 这将基于是否在某个任意行的 b.value 上包含 sum(b.value)。您可能的意思是：（sum(d.value) + coalesce(sum(b.value),0) 说CASE WHEN sum(b.value) IS NULL...是等价的。）

我会这样做（感谢小提琴，kordirko）：

SELECT
    sum( if(d.type=0, coalesce(d.value,0)+coalesce(b.value,0), null) ) type0,
    sum( if(d.type=1, coalesce(d.value,0)+coalesce(b.value,0), null) ) type1
FROM data d
LEFT JOIN data_bills b
ON  d.id = b.data_id
WHERE d.type in (0,1);

Answer 2

试试这个查询：

SELECT
      ( SELECT
        CASE WHEN b.value IS NULL 
        THEN sum(d.value)
        ELSE sum(d.value) + sum(b.value)
        END AS sumaWydatkow

        FROM data d
        LEFT JOIN data_bills b 
        ON  d.id = b.data_id
        WHERE   d.type = 0
       ) type0,

      ( SELECT
        CASE WHEN b.value IS NULL 
        THEN sum(d.value)
        ELSE sum(d.value) + sum(b.value)
        END AS sumaPrzychodow

        FROM data d
        LEFT JOIN data_bills b 
        ON  d.id = b.data_id
        WHERE   d.type = 1
      ) type1

演示 --> http://www.sqlfiddle.com/#!2/6ec52/2

Answer 3

SELECT   d.type, SUM(d.value) + SUM(b.value)
FROM     data d LEFT JOIN data_bills b ON b.data_id = d.id
WHERE    d.type IN (0,1)
GROUP BY d.type

在sqlfiddle上查看（再次感谢 @kordirko）。