Question

我有一个包含唯一ID字段的表。另一个字段（REF）包含对另一个数据集的ID字段的引用。现在我必须选择REF指向不存在的数据集的所有数据集。

SELECT * FROM table WHERE ("no dataset with ID=REF exists")

我该怎么做？

Answer 1

3种方式

SELECT * FROM YourTable y WHERE NOT EXISTS 
     (SELECT * FROM OtherTable o WHERE y.Ref = o.Ref)

SELECT * FROM YourTable WHERE Ref NOT IN 
     (SELECT Ref FROM OtherTable WHERE Ref IS NOT NULL)

SELECT y.* FROM YourTable y 
LEFT OUTER JOIN  OtherTable o ON y.Ref = o.Ref
WHERE o.Ref IS NULL

另见Five ways to return all rows from one table which are not in another table

Answer 2

试试这个：

SELECT * FROM TABLE WHERE NOT EXISTS 
     (SELECT * FROM OtherTable WHERE TABLE.Ref = OtherTable.ID)

Answer 3

我认为这应该有用

SELECT * FROM table WHERE id NOT IN (SELECT ref_id FROM ref_table)

或使用JOIN

SELECT table.* 
FROM table LEFT JOIN ref_table ON table.id = ref_table.ref_id
WHERE ref_table.ref_id IS NULL

Answer 4

SELECT 
 table1.* 
FROM 
 table1
 LEFT JOIN table2 ON table1.id = table2.ref
WHERE 
 table2.ref IS NULL

Answer 5

您可以执行以下子查询：

select * from table where somefield not in (select otherfield from sometable where ID=REF)

Answer 6

SELECT * 
FROM table 
WHERE ((SELECT COUNT(*) FROM table2 WHERE table2.id = table.ref) = 0)

Answer 7

类似的东西：

SELECT * FROM table WHERE ID NOT IN(SELECT REF FROM Table2 )

Answer 8

是的，你可以使用

从x中选择*不存在（从y中选择*）

如何在选择中选择

8 个答案: