Table1 (id, name) Table2 (id, name)
查询:
SELECT name
FROM table2
-- that are not in table1 already
Table1 (id, name) Table2 (id, name)
查询:
SELECT name
FROM table2
-- that are not in table1 already
当前回答
以下是对我最有效的方法。
SELECT *
FROM @T1
EXCEPT
SELECT a.*
FROM @T1 a
JOIN @T2 b ON a.ID = b.ID
这比我试过的其他方法快了一倍多。
其他回答
你可以在mssql中使用EXCEPT或在oracle中使用MINUS,它们是相同的:
http://blog.sqlauthority.com/2008/08/07/sql-server-except-clause-in-sql-server-is-similar-to-minus-clause-in-oracle/
我尝试了以上所有的解决方案,但它们都不适合我。下面的查询对我有用。
SELECT NAME
FROM table_1
WHERE NAME NOT IN
(SELECT a.NAME
FROM table_1 AS a
LEFT JOIN table_2 AS b
ON a.NAME = b.NAME
WHERE any further condition);
你可以使用以下查询结构:
SELECT t1.name FROM table1 t1 JOIN table2 t2 ON t2。Fk_id != t1.id;
表1:
id | name |
---|---|
1 | Amit |
2 | Sagar |
表二:
id | fk_id | |
---|---|---|
1 | 1 | amit@ma.com |
输出:
name |
---|
Sagar |
首先定义表的别名,如t1和t2。 然后得到第二个表的记录。 然后使用where条件匹配记录:
SELECT name FROM table2 as t2
WHERE NOT EXISTS (SELECT * FROM table1 as t1 WHERE t1.name = t2.name)
小心陷阱。如果表1中的字段Name包含null,你就会感到惊讶。 更好的是:
SELECT name
FROM table2
WHERE name NOT IN
(SELECT ISNULL(name ,'')
FROM table1)