Mysql:从表中选择不在另一个表中的行

Question

如何选择一个表中没有出现在另一个表中的所有行？

表格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 |
+-----------+----------+------------+

Table1中不在Table2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)

Answer 1

您需要根据列名进行子选择,而不是*.

例如,如果您有id两个表共有的字段,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子查询语法.

Answer 2

如果您在另一条注释中提到了300列,并且想要在所有列上进行比较(假设列都是相同的名称),则可以使用a NATURAL LEFT JOIN隐式连接两个表之间的所有匹配列名,以便您不必手动输入所有连接条件繁琐:

SELECT            a.*
FROM              tbl_1 a
NATURAL LEFT JOIN tbl_2 b
WHERE             b.FirstName IS NULL

Answer 3

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...

Answer 4

标准的LEFT JOIN可以解决问题,如果连接上的字段被索引,
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

Answer 5

试试这个简单的查询。它完美地工作。

select * from Table1 where (FirstName,LastName,BirthDate) not in (select * from Table2);

Answer 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