Question

如何更新此选项以选择所有重复项？

SELECT address FROM list  
GROUP BY address HAVING count(id) > 1

目前，我认为它只是重新发布了重复的地址。我想要所有的副本。

Answer 1

Select * from list
where address in (
  select address from list group by address
  having count(*) > 1);

查看我运行的示例查询：

mysql> select * from flights;
+--------+-------------+
| source | destination |
+--------+-------------+
|      1 |           2 |
|      3 |           4 |
|      5 |           6 |
|      6 |           1 |
|      2 |           4 |
|      1 |           3 |
|      5 |           2 |
|      6 |           3 |
|      6 |           5 |
|      6 |           4 |
+--------+-------------+
10 rows in set (0.00 sec)

mysql> select * from flights where source in 
       (select source from flights group by source having count(*) > 1);
+--------+-------------+
| source | destination |
+--------+-------------+
|      1 |           2 |
|      5 |           6 |
|      6 |           1 |
|      1 |           3 |
|      5 |           2 |
|      6 |           3 |
|      6 |           5 |
|      6 |           4 |
+--------+-------------+
8 rows in set (0.00 sec)

Answer 2

如果我是正确的，那么您正在寻找包含重复项的实际行 - 因此，如果您有三行具有相同的地址，则返回所有三行。

以下是如何操作：

SELECT * FROM list
WHERE address in (
    SELECT address FROM list GROUP BY address HAVING count(id) > 1
);

除非您的地址是“文本”字段，或者您的地址表中有多于几千个重复字段，否则这通常应该有效。

Answer 3

你在找这个吗？

SELECT * FROM list
WHERE id IN (
    SELECT id FROM list
    GROUP BY address HAVING count(id) > 1
);

选择所有重复项

3 个答案: