如何在选择中进行选择

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 

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

Answer 2

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

Answer 3

我认为这应该工作

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

或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

你可以做一个子查询：

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

Answer 5

试试这个：

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

Answer 6

类似的东西：

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

Answer 7

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

Answer 8

哟耶我从众üliek选择，所以我们把选择在乌拉圭回合选择这样当你查询可以查询。

Answer 9

是的，你可以使用

SELECT * FROM x，其中不存在（SELECT * FROM Y）