Question

我有两个SQL SELECT COUNT语句：

SELECT COUNT(*) FROM table1 t1 
INNER JOIN table2 t2 ON t2.id = t1.rowid
WHERE t1.flag1 = false AND t2.flag2 = true;

SELECT COUNT(*) FROM table1 t1 
INNER JOIN table2 t2 ON t2.id = t1.rowid
WHERE t1.flag1 = true AND t2.flag2 = false;

可以看出，这两个陈述的唯一区别是翻转条件。

但我想要做的是将两个语句合并为一个，以便输出成为一个包含两列的表，第一列包含第一个语句的结果，第二列包含第二个语句的结果。类似的东西：

   count1   |   count 2
-------------------------
    3456    |     9864

我顺便使用PostgreSQL。有人能告诉我怎么做吗？

非常感谢

Answer 1

这应该为你做。我不熟悉PostgreSql，但我认为它会起作用。

SELECT 
    SUM(CASE WHEN t1.Flag1 = false AND t2.flag2 = true THEN 1 ELSE 0 END) Count1,
    SUM(CASE WHEN t1.Flag1 = true AND t2.flag2 = false THEN 1 ELSE 0 END) Count2
FROM 
    table1 t1 
    INNER JOIN table2 t2 ON t2.id = t1.rowid

Answer 2

如果你真的需要这种方式（使用两个sql查询并将它们组合起来），那么：

select * from
 (SELECT COUNT(*) FROM table1 t1 INNER JOIN table2 t2 ON t2.id = t1.rowid WHERE t1.flag1 = false AND t2.flag2 = true) a,
 (SELECT COUNT(*) FROM table1 t1 INNER JOIN table2 t2 ON t2.id = t1.rowid WHERE t1.flag1 = true AND t2.flag2 = false) b

根据您的SQL，这将是更好的解决方案：

select
    sum (case when not t1.flag1 and t2.flag2 then 1 else 0 end) as count1,
    sum (case when t1.flag1 and not t2.flag2 then 1 else 0 end) as count2
FROM 
    table1 t1 
    INNER JOIN table2 t2 ON t2.id = t1.rowid

你也可以将boolean类型转换为整数并缩短sql（true :: int = 1，false :: int = 0）：

 select 
    sum((flag1::int<flag2::int)::int) count1,
    sum((flag1::int>flag2::int)::int) count2    
 from
    table1 t1 
    join table2 t2 ON t2.id = t1.rowid

因为真的＆gt;假和假＆lt;是的（至少在PostgreSQL中）你可以写：

 select 
    sum((flag1 < flag2)::int) count1,
    sum((flag1 > flag2)::int) count2    
 from
    table1 t1 
    join table2 t2 ON t2.id = t1.rowid

Answer 3

Select * from
(
    SELECT COUNT(*) FROM table1 t1 
    INNER JOIN table2 t2 ON t2.id = t1.rowid
    WHERE t1.flag1 = false AND t2.flag2 = true) tab1,
(
    SELECT COUNT(*) FROM table1 t1 
    INNER JOIN table2 t2 ON t2.id = t1.rowid
    WHERE t1.flag1 = true AND t2.flag2 = false) tab2

将两个SQL select合并为一个

3 个答案: