Question

我有两个SQL查询，其中首先使用内部联接根据条件进行匹配，而另一个不使用。最终，我想要每个查询创建的列之间的区别。我怎样才能做到这一点？

我已经尝试像在一些类似的帖子中那样合并和加入查询，但这是行不通的。我想知道问题是否在每个查询中的联接周围。

查询1：

SELECT A.date, COUNT(DISTINCT A.id)
FROM A
INNER JOIN B
ON A.id = B.id AND A.date = B.date
AND B.col1 = 'value1'
LEFT JOIN C on C.key = A.key
WHERE A.col1 = 'value2'
AND C.category = 'cat1'
GROUP BY 1
ORDER BY 1 DESC

查询2：

SELECT A.date, COUNT(DISTINCT A.id)
FROM A
LEFT JOIN C on C.key = A.key
WHERE A.col1 = 'value2'
AND C.category = 'cat1'
GROUP BY 1
ORDER BY 1 DESC

Answer 1

SELECT A.date, COUNT(DISTINCT A.id)
FROM A
INNER JOIN B
ON A.id = B.id AND A.date = B.date
AND B.col1 = 'value1'
LEFT JOIN C on C.key = A.key
WHERE A.col1 = 'value2'
AND C.category = 'cat1'
GROUP BY 1
ORDER BY 1 DESC

UNION 

SELECT A.date, COUNT(DISTINCT A.id)
FROM A
LEFT JOIN C on C.key = A.key
WHERE A.col1 = 'value2'
AND C.category = 'cat1'
GROUP BY 1
ORDER BY 1 DESC

Answer 2

一种简单的方法是使用JOIN列对两个查询进行date：

SELECT x.date, x.cnt, y.cnt, y.cnt - x.cnt
FROM 
(
    SELECT A.date, COUNT(DISTINCT A.id) AS cnt
    FROM A
    INNER JOIN B ON A.id = B.id AND A.date = B.date AND B.col1 = 'value1'
    LEFT JOIN C on C.key = A.key
    WHERE A.col1 = 'value2' AND C.category = 'cat1'
    GROUP BY 1
) AS x 
INNER JOIN (
    SELECT A.date, COUNT(DISTINCT A.id) AS cnt
    FROM A
    LEFT JOIN C on C.key = A.key
    WHERE A.col1 = 'value2' AND C.category = 'cat1'
    GROUP BY 1
) AS y ON x.date = y.date
ORDER BY 1 DESC

您可能要根据数据布局调整联接类型：

LEFT JOIN，如果所有日期都在第一个子查询中可用，但在第二个子查询中可能缺少
RIGHT JOIN（如果情况相反）
FULL OUTER JOIN，如果您希望两端都有可用的日期

如果您选择以上任一选项，则当其中一项为COALESCE时，需要使用NULL来防止减法返回NULL。

Answer 3

您的c的左联接实际上变成了内部联接，因为它用在NULL子句中的WHERE排除表达式中。因此，您可以直接内部联接c和左联接b。然后，您可以在一个count()中使用一个个案，以仅计算来自b的行被联接的实例。从另一个count()中减去该值，然后计算所有出现的次数以获得差值。

SELECT a.date,
       count(DISTINCT a.id)
       -
       count(DISTINCT CASE
                        WHEN b.id IS NOT NULL THEN
                          a.id
                      END)
       FROM a
            INNER JOIN c
                       ON c.key = a.key
                          AND c.category = 'cat1'
            LEFT JOIN b
                      ON a.id = b.id
                         AND a.date = b.date
                         AND b.col1 = 'value1'
       WHERE a.col1 = 'value2'
       GROUP BY 1
       ORDER BY 1 DESC;

将两个SQL查询合并到一个表中

3 个答案: