如何选择一个表中没有出现在另一个表中的所有行?
表1:
+-----------+----------+------------+
| FirstName | LastName | BirthDate |
+-----------+----------+------------+
| Tia | Carrera | 1975-09-18 |
| Nikki | Taylor | 1972-03-04 |
| Yamila | Diaz | 1972-03-04 |
+-----------+----------+------------+
表2:
+-----------+----------+------------+
| FirstName | LastName | BirthDate |
+-----------+----------+------------+
| Tia | Carrera | 1975-09-18 |
| Nikki | Taylor | 1972-03-04 |
+-----------+----------+------------+
表1中不在表2中的行的示例输出:
+-----------+----------+------------+
| FirstName | LastName | BirthDate |
+-----------+----------+------------+
| Yamila | Diaz | 1972-03-04 |
+-----------+----------+------------+
也许这样的事情应该有效:
SELECT * FROM Table1 WHERE * NOT IN (SELECT * FROM Table2)
答案 0 :(得分:158)
您需要根据列名进行子选择,而不是*
。
例如,如果您有两个表共有的id
字段,则可以执行以下操作:
SELECT * FROM Table1 WHERE id NOT IN (SELECT id FROM Table2)
有关更多示例,请参阅MySQL subquery syntax。
答案 1 :(得分:85)
如果您在另一条评论中提到了300列,并且想要在所有列上进行比较(假设列的名称都相同),则可以使用NATURAL LEFT JOIN
隐式连接所有匹配列两个表之间的名称,这样您就不必繁琐地手动输入所有连接条件:
SELECT a.*
FROM tbl_1 a
NATURAL LEFT JOIN tbl_2 b
WHERE b.FirstName IS NULL
答案 2 :(得分:33)
SELECT *
FROM Table1 AS a
WHERE NOT EXISTS (
SELECT *
FROM Table2 AS b
WHERE a.FirstName=b.FirstName AND a.LastName=b.Last_Name
)
EXISTS
会帮助你...
答案 3 :(得分:31)
标准LEFT JOIN可以解决问题,如果连接上的字段已编入索引,
也应该更快
SELECT *
FROM Table1 as t1 LEFT JOIN Table2 as t2
ON t1.FirstName = t2.FirstName AND t1.LastName=t2.LastName
WHERE t2.BirthDate Is Null
答案 4 :(得分:6)
尝试:
SELECT * FROM table1
LEFT OUTER JOIN table2
ON table1.FirstName = table2.FirstName and table1.LastName=table2.LastName
WHERE table2.BirthDate IS NULL
答案 5 :(得分:1)
尝试这个简单的查询。它运作得很好。
select * from Table1 where (FirstName,LastName,BirthDate) not in (select * from Table2);
答案 6 :(得分:0)
这在Oracle中对我有用:
SELECT a.*
FROM tbl1 a
MINUS
SELECT b.*
FROM tbl2 b;
答案 7 :(得分:-4)
SELECT a.* FROM
FROM tbl_1 a
MINUS
SELECT b.* FROM
FROM tbl_2 b