mysql - 你能把2个完全不同的查询结果分成1个结果吗

Question

我正在尝试将数字结果从 2 个完全不同的查询中分离出来。

最终结果应该是查询 1 除以查询 2

查询 1 =

SELECT COUNT(DISTINCT(table1.ID)) AS count_1
FROM table1 
    INNER JOIN op 
    INNER JOIN Org 
    ON table1.EID = op.id 
        AND Op.OrgID = Org.ID
WHERE table1.TitleID = 123
    AND op.BrandID = 1 
    AND op.Start <= NOW() AND op.End >= NOW();

查询 2 =

SELECT COUNT(DISTINCT user.id) AS count_2
FROM table1 INNER JOIN user INNER JOIN ur
    ON table1.EID = user.id AND ur.userID = user.id
WHERE 
    user.BrandID = 1 
    AND table1.TitleID = 123
    AND ur.role = 0
    AND user.Inactive = 0;

Answer 1

当然！您可以使用子选择来实现这一点，尽管它会非常冗长！

SELECT
(
    SELECT COUNT(DISTINCT(table1.ID)) AS count_1
    FROM table1 
        INNER JOIN op 
        INNER JOIN Org 
        ON table1.EID = op.id 
            AND Op.OrgID = Org.ID
    WHERE table1.TitleID = 123
        AND op.BrandID = 1 
        AND op.Start <= NOW() AND op.End >= NOW()
) / (
    SELECT COUNT(DISTINCT user.id) AS count_2
    FROM table1 INNER JOIN user INNER JOIN ur
        ON table1.EID = user.id AND ur.userID = user.id
    WHERE 
        user.BrandID = 1 
        AND table1.TitleID = 123
        AND ur.role = 0
        AND user.Inactive = 0
);

格式但是它对你来说感觉最不难看。

Answer 2

使用这样的子查询：

SELECT Q1.count_1 / Q2.Count_2 
FROM 
( ... Query1 ...) AS Q1
JOIN
( ... Query2 ...) AS Q2
ON 1=1

将 Query1 和 Query2 替换为您的代码。

Answer 3

像这样：

SELECT count_1 / count_2
FROM (SELECT COUNT(*) count_1 FROM foo) f
JOIN  (SELECT COUNT(*) count_2 FROM bar) b ON 1=1;

http://sqlfiddle.com/#!2/c215e/1