Question

我有两个MySQL表。除了有一个额外的列之外，它们具有相同的结构。我想找出一些实例的数量，其中只有一列“地址”中的记录是相同的。换句话说，我想找到两个表中出现的地址数。

Answer 1

如果您想要两个表格中的地址数量，请使用union all和count()：

select count(distinct address)
from ((select address from table1) union all
      (select address from table2)
     ) t

如果您想要两个表中都存在的地址：

select address
from ((select address, 1 as in1, 0 as in2 from table1) union all
      (select address, 0 as in1, 1 as in2 from table2)
     ) t
group by address
having sum(in1) > 0 and sum(in2) > 0;

您可以使用以下方式获取计数：

select count(*)
from (select address
      from ((select address, 1 as in1, 0 as in2 from table1) union all
            (select address, 0 as in1, 1 as in2 from table2)
           ) t
      group by address
      having sum(in1) > 0 and sum(in2) > 0
     ) a;

Answer 2

我想你只想加入这两个表并计算记录：

SELECT COUNT(*)
FROM table1 t1
JOIN table2 t2
  ON t1.address = t2.address

如果您想忽略重复项，例如，＆＃39; 123 Main＆＃39; 123在table1中出现一次，但在table2中出现两次，上面的查询会将其计算两次，但如果您只想为＆＃39; 123 Main＆＃39;计算一次，则还要使用DISTINCT：

SELECT DISTINCT COUNT(t1.address)
FROM table1 t1
JOIN table2 t2
  ON t1.address = t2.address

Answer 3

您所描述的是内部连接，对于Mysql，它是默认连接。

这是一个匹配数字的例子。注意我将结果限制为5以便于阅读。该示例应该可以很好地扩展到字符串，前提是您可以找到实际匹配字符串的好方法。

mysql> create temporary table test.t1 ( i int unsigned not null );
Query OK, 0 rows affected (0.05 sec)

mysql> create temporary table test.t2 ( i int unsigned not null );
Query OK, 0 rows affected (0.01 sec)

mysql> insert into test.t1 select floor(rand() * 100) from mysql.user;
Query OK, 125 rows affected (0.02 sec)
Records: 125  Duplicates: 0  Warnings: 0

mysql> select test.t1.i, test.t2.i from test.t1 inner join test.t2 on test.t1.i = test.t2.i limit 5;
+----+----+
| i  | i  |
+----+----+
|  2 |  2 |
|  2 |  2 |
| 24 | 24 |
| 24 | 24 |
| 11 | 11 |
+----+----+
5 rows in set (0.01 sec)

查找两个MySQL表之间的重叠

3 个答案: