Question

我创建了一个函数abcd_insert（），它将数据插入到表abcd中，该表有8列。函数内部的代码类似于下面的代码：

BEGIN  
 INSERT INTO abcd

 VALUES 
      (
    x                 ,
    y                 ,
    select sum(count) from (select count(*) from a where a1 = x and a2 = y and a3 = 1 union all select count(*) from b where b1 = x and b2 = y and b3 = 1 ) as n1,
    select sum(count) from (select count(*) from a where a1 = x and a2 = y and a2 = 2 union all select count(*) from b where b1 = x and b2 = y and b3 = 2 ) as n2 ,
    select sum(count) from (select count(*) from a where a1 = x and a2 = y and a2 = 3 union all select count(*) from b where b1 = x and b2 = y and b3 = 3 ) as n3 ,
    select sum(count) from (select count(*) from a where a1 = x and a2 = y and a2 = 4 union all select count(*) from b where b1 = x and b2 = y and b3 = 4 ) as n4 ,
    select sum(count) from (select count(*) from a where a1 = x and a2 = y and a2 = 5 union all select count(*) from b where b1 = x and b2 = y and b3 = 5 ) as n5 ,
    SELECT sum(q1) from
    (SELECT CASE WHEN COUNT(1) > 0 THEN 1 ELSE 0 END as q1 FROM p1 where p11 = x and p12 = y union all
    SELECT CASE WHEN COUNT(1) > 0 THEN 1 ELSE 0 END as q1 FROM p2 where p21 = x and p22 = y )  as q1
  ); 
END;

'x'和'y'是我的输入参数，其值将传递给函数abcd_insert（）。 a，b，p1和p2是同一模式中的表。当我在运行时将'x'和'y'传递给函数时，它会抛出错误。有人可以告诉我这里我做错了什么。

Answer 1

您需要为列命名，而不仅仅是派生表：

SELECT SUM(foo) as foo
FROM ((SELECT CASE WHEN COUNT(1) > 0 THEN 1 ELSE 0 END) as foo FROM a.a1 where p1=100) UNION ALL
      . . .
     ) x

Postgresql函数用于插入内的嵌套选择

1 个答案: