Mysql:从表中选择不在另一个表中的行
如何选择一个表中没有出现在另一个表中的所有行?
How to select all rows in one table that do not appear on another?
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:
Example output for rows in Table1 that are not in Table2:
+-----------+----------+------------+
| FirstName | LastName | BirthDate |
+-----------+----------+------------+
| Yamila | Diaz | 1972-03-04 |
+-----------+----------+------------+
也许这样的事情应该可行:
Maybe something like this should work:
SELECT * FROM Table1 WHERE * NOT IN (SELECT * FROM Table2)
推荐答案
如果您在另一条评论中提到有 300 列,并且您想对所有列进行比较(假设所有列的名称相同),您可以使用 NATURAL LEFT JOIN
隐式连接两个表之间所有匹配的列名,这样您就不必手动输入所有连接条件:
If you have 300 columns as you mentioned in another comment, and you want to compare on all columns (assuming the columns are all the same name), you can use a NATURAL LEFT JOIN
to implicitly join on all matching column names between the two tables so that you don't have to tediously type out all join conditions manually:
SELECT a.*
FROM tbl_1 a
NATURAL LEFT JOIN tbl_2 b
WHERE b.FirstName IS NULL
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