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问题描述
如何选择一个表中未出现在另一表中的所有行?
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 |
+-----------+----------+------------+
也许类似的东西应该起作用:
Maybe something like this should work:
SELECT * FROM Table1 WHERE * NOT IN (SELECT * FROM Table2)
推荐答案
如果您在另一条注释中提到有300列,并且想要在所有列上进行比较(假设这些列的名称相同),则可以使用NATURAL LEFT JOIN
隐式联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联接条件了:
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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