我想选择满足所选LID的所有LID。我有一个名为LNZ的表,例如:
+------------+
| LNZ |
+-----+------+
| NID | LID |
+-----+------+
| 1 | 1 |
| 1 | 2 |
| 1 | 3 |
| 2 | 1 |
| 2 | 3 |
| 3 | 1 |
| 4 | 1 |
| 4 | 2 |
+-----+-------+
我想要的是,在这种情况下选择包含NID 1和4的所有条目。但它可能超过两个NID被选中。
对于NID 1和4,输出应为:
+------------+
| LNZ |
+-----+------+
| NID | LID |
+-----+------+
| 1 | 1 |
| 1 | 2 |
| 4 | 1 |
| 4 | 2 |
+-----+-------+
如果我这样做
SELECT NID, LID
FROM lnz
WHERE NID = 1 OR NID = 4;
我得到了错误的结果:
+------------+
| LNZ |
+-----+------+
| NID | LID |
+-----+------+
| 1 | 1 |
| 1 | 2 |
| 1 | 3 |
| 4 | 1 |
| 4 | 2 |
+-----+-------+
它适用于以下SQL语句,但它不具有所需的NID
SELECT T1.NID, T1.LID
FROM LNZ AS T1
JOIN LNZ AS T2 ON T1.LID = T2.LID
WHERE T1.NID = 1 AND T2.NID = 4
OR T2.NID = 1 AND T1.NID = 4;
我现在的问题是......如何更改此声明,使其与所选NID的数量一起变化?
随时问你是否不知道我的意思。
答案 0 :(得分:2)
如果您尝试
SELECT * FROM LNZ
WHERE NID IN (1,2,4)
AND LID IN (
select LID from LNZ
WHERE NID IN (1,2,4)
GROUP BY LID
HAVING COUNT(*) = 3
)
在您用于构建查询的语言中,您动态地将(1,2,4)设置为您想要的NID值并设置计数(基于您想要的NID数量,在我的示例中)它是3),它应该工作
答案 1 :(得分:0)
SELECT T1.NID, T1.LID
FROM LNZ AS T1
JOIN (SELECT LID
FROM lnz
WHERE NID = 1 OR NID = 4
group by lid
having count(*)>1) sub ON t1.lid=sub.lid
WHERE (T1.NID = 1 OR T1.NID = 4)
答案 2 :(得分:0)
@NemanjaPerovic提供了很好的解决方案,这个解决方案看起来很像,不同的是你可以在这里计算NID,而不是语言/应用程序级别。像这样:
select * from LNZ
where NID in(1,2,4)
and
lid in(
select lid from LNZ
where NID in(1,2,4)
group by lid
having count(*) = (select count(distinct NID) from LNZ where NID in(1,2,4) )
)