sql - 使用 if 条件合并两个 sql 查询

Question

我想合并两个查询。

我有 Query1 返回一些 ID。

然后我有 Query2，它首先检查表中是否存在一些 Id。如果否，则结果将是 Query1 返回的 Id。

如果是，那么我想要 Query1 和 Query2 返回的常见 Id，例如 Query1 和 Query2 返回的 Id 的交集。

那么我怎样才能在一个查询中做到这一点。

Answer 1

这可以使用 DECODE 和 CASE 语句的组合来实现

SELECT DISTINCT BT1.ID
  FROM BUFFER_TABLE_1 BT1, BUFFER_TABLE_2 BT2
 WHERE DECODE((SELECT COUNT(BT1.ID)
                FROM BUFFER_TABLE_1 BT1, BUFFER_TABLE_2 BT2
               WHERE BT1.ID = BT2.ID),
              0,
              'TRUE',
              (SELECT CASE
                        WHEN BT1.ID = BT2.ID THEN
                         'TRUE'
                        ELSE
                         'FALSE'
                      END
                 FROM DUAL)) = 'TRUE'
 ORDER BY BT1.ID

主要部分在解码中使用，如果两个表中的匹配 ID 计数为零，则从第一个表（BUFFER_TABLE_1）返回所有 ID。但是，如果此计数大于 0，则两个表中的 ID 都匹配，并且仅返回公共 ID

希望能帮助到你

Answer 2

(For people who have already read my answer: It has been hugely re-edited because of the comments of @nawfal that I want to preserve)

First I want to clarify that I understood your problem correct.

You have the Parts:

1. select id from table1 That produces Result1
2. select id from table2 where <some condition> That produces Result2
3. select count(id) > 0 from table2 where <some condition> That decides what result to use

If 3. returns more than 1 then you want the rows that are returned by 1. AND 2. (but not the ones that are only returned by 1. OR only by 2.)

If 3. returns 0 then you want the results of 1.

-> The solution is to have 3. in a view and select it in a where clause of a union statement TWICE.

Like this:

SELECT t1.id FROM table1 t1 WHERE 
       (SELECT COUNT(t2.id) from table2 t2 where <some condition>) = 0
UNION
select t2.id from table2 t2 WHERE <some condition> 
       AND t2.id IN (SELECT t1.id FROM table1 t1)
       AND (SELECT COUNT(t2.id) from table2 t2 where <some condition>) > 0

one of the two parts will always be empty (because query 3. can't be = 0 and > 0 at the same time)