如何 SELECT none 匹配行从一个 table 到另一个？

Question

Mysql v5.7 我知道之前有人问过这个问题并回答过这个问题，但一般查询在我的数据库中不起作用，我把我的头发拉下来了。我有两个 tables，只有一列包含电子邮件：1) 已发送（41110 行 [有重复项]）2) 块（81132 [有重复项]）

mysql> show create table blocks;
+--------+------------------------------------------------------------------------------------------------+
| Table  | Create Table                                                                                   |
+--------+------------------------------------------------------------------------------------------------+
| blocks | CREATE TABLE `blocks` (
  `email` varchar(200) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1 |
+--------+------------------------------------------------------------------------------------------------+

mysql> show create table sent;
+-------+-----------------------------------------------------------------------------------------------+
| Table | Create Table                                                                                  |
+-------+-----------------------------------------------------------------------------------------------+
| sent  | CREATE TABLE `sent` (
  `emails` varchar(111) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1 |
+-------+-----------------------------------------------------------------------------------------------+

我想获取行作为 "Sent" - "Blocks" 或我想从 "Sent" table 中获取那些不在 "Blocks" table

中的行

已发送table:

mysql> select * from sent limit 10;
+-----------------------------------------------------+
| emails                                              |
+-----------------------------------------------------+
| "52@mail.marketplace" |
| "1cstains@coden.com"                       |
| "a.aldhizer@ldplastics.com"                      |
| "a.antosca@tdx.com"                          |
| "a.balvanz@hrrd-paper.com"                     |
| "a.bandal@medpace.com"                          |
| "a.bass@paettethanol.com"                         |
| "a.bodwalk@bedfndustries.com"                   |
| "a.bravo@sdfhemie.com"                        |
| "a.burton@evdarenterprises.com"                 |
+-----------------------------------------------------+
10 rows in set (0.00 sec)

块

mysql> select * from blocks limit 10;    
+---------------------------------------------------+
| email                                             |
+---------------------------------------------------+
| ""                                                |
| 0f6a88be0a4b45628be38ae08c8fdd71@mail.marketplace |
| 1cstains@coden.com                                |
| 1cstains@coden.com                                |
| 1cstains@coden.com                                |
| 3161foodmgr@mcare.com                     |
| 4b00fce87e5b423c942f5b19f27c3a13@mail.marketplace |
| 52d05d98b59e44b0816401d2cd0411f0@mail.marketplace |
| 52d05d98b59e44b0816401d2cd0411f0@mail.marketplace |
| 6357a2fd35114418a93e0ccda6edd6f4@mail.marketplace |
+---------------------------------------------------+
10 rows in set (0.00 sec)

发送的预期结果table

+-----------------------------------------------------+
| emails                                              |
+-----------------------------------------------------+
| "52@mail.marketplace" |
| "a.aldhizer@ldplastics.com"                      |
| "a.antosca@tdx.com"                          |
| "a.balvanz@hrrd-paper.com"                     |
| "a.bandal@medpace.com"                          |
| "a.bass@paettethanol.com"                         |
| "a.bodwalk@bedfndustries.com"                   |
| "a.bravo@sdfhemie.com"                        |
| "a.burton@evdarenterprises.com"                 |
+-----------------------------------------------------+

这是我使用的查询，它 returns 匹配行 :/

select sent.emails from sent where sent.emails NOT IN ( select email from blocks )

Answer 1

请使用minus运算符。

select sent.emails from sent 
minus
select blocks.email from blocks

Answer 2

select sent.emails 
from sent 
     left join blocks on sent.emails = blocks.email
where blocks.email is null

如果问题是电子邮件中的 " 个字符（正如提到的 @Jens），您可以更改连接以使用 subbstring 跳过第一个和最后一个字符

来剪切 " 个字符

left join blocks on SUBSTRING(sent.emails,1,len(sent.emails)-2) = blocks.email

如何 SELECT none 匹配行从一个 table 到另一个？

How to SELECT none matching rows from one table to another?

mysql

sql

innodb