任何人都可以在MySQL IN子句中解释这个逻辑

时间:2013-03-14 14:51:29

标签: mysql

任何人都可以在MySQL IN子句中解释这个逻辑并帮助我理解这个问题

我有一个用户表,这个表用户属于一个或多个组。 组表主键引用在users表中以逗号(,)分隔值更新,如下所示

Query 1. SELECT * FROM user;
+---------+-----------+-------------------------+-----------+
| user_id | user_name | user_email              | group_id  |
+---------+-----------+-------------------------+-----------+
|       1 | suresh    | xxxx@yyyyyyyyyy.com     | 22        |
|       2 | sundar    | s7sundera@gmail.com     | 2         |
|       3 | tester    | xxxxxxxx@yyyyyyyyyy.com | 1,2,3,4   |
|       4 | gail      | zzzzzz@gmail.com        | 1,2,3,4,5 |
+---------+-----------+-------------------------+-----------+

如果我在MySQL中使用IN子句和组id值为2,我只得到一个结果

Query 2. SELECT * FROM user WHERE group_id IN(2)
+---------+-----------+---------------------+----------+
| user_id | user_name | user_email          | group_id |
+---------+-----------+---------------------+----------+
|       2 | sundar    | s7sundera@gmail.com | 2        |
+---------+-----------+---------------------+----------+

如果我在MySQL中使用IN子句和组ID值为(1,2),我得到了三个结果

Query 3. SELECT * FROM user WHERE group_id IN(1,2)
+---------+-----------+-------------------------+-----------+
| user_id | user_name | user_email              | group_id  |
+---------+-----------+-------------------------+-----------+
|       2 | sundar    | s7sundera@gmail.com     | 2         |
|       3 | tester    | xxxxxxxx@yyyyyyyyyy.com | 1,2,3,4   |
|       4 | gail      | zzzzzz@gmail.com        | 1,2,3,4,5 |
+---------+-----------+-------------------------+-----------+

我想让组ID 2用户喜欢以下输出,但它没有按预期工作

如果我使用此查询,我需要获得查询3结果是否可能?

SELECT * FROM user WHERE group_id IN(2)

4 个答案:

答案 0 :(得分:3)

这个评论太长了,但您需要重新考虑当前的表格设计。您不应将group_id值存储为逗号分隔列表。

您的表格结构应类似于以下内容:

create table user
(
    user_id int,  PK
    user_name varchar(50),
    user_email varchar(100)
);

create table groups
(
    group_id int, PK
    group_name varchar(10)
);

create table user_group
(
    user_id int,
    group_id int
);

user_group表将包含user_idgroup_id的主键,因此您无法获得重复项,然后这些列应该是相应表的外键。此表允许您为每个user_id分配多个组。

然后当您查询表时,查询将是:

select u.user_id, 
  u.user_name,
  u.user_email,
  g.group_id
from user u
inner join user_group ug 
  on u.user_id = ug.user_id
inner join groups g
  on ug.group_id = g.group_id

请参阅SQL Fiddle with Demo

如果您需要用于显示目的,请在逗号分隔列表中显示group_id值,您可以使用GROUP_CONCAT()

select u.user_id, 
  u.user_name,
  u.user_email,
  group_concat(g.group_id order by g.group_id) group_id
from user u
inner join user_group ug 
  on u.user_id = ug.user_id
inner join groups g
  on ug.group_id = g.group_id
group by u.user_id, u.user_name, u.user_email

请参阅SQL Fiddle with Demo

如果您重新设计表格,那么当您搜索它时会变得更容易:

select u.user_id, 
  u.user_name,
  u.user_email,
  g.group_id
from user u
inner join user_group ug 
  on u.user_id = ug.user_id
inner join groups g
  on ug.group_id = g.group_id
where g.group_id in (1, 2)

请参阅SQL Fiddle with Demo

答案 1 :(得分:2)

1,2传递给IN运算符时,您需要12;这就是它将返回所有三个结果的原因。如果您的列中包含逗号分隔值,则表示您违反了普通表单;因为每列不应包含多个值。如果要在多值逗号分隔列中查找单个值,则可以使用FIND_IN_SET

规范化架构如下所示:

+---------+-----------+-------------------------+
| user_id | user_name | user_email              |
+---------+-----------+-------------------------+
|       2 | sundar    | s7sundera@gmail.com     |
|       3 | tester    | xxxxxxxx@yyyyyyyyyy.com |
|       4 | gail      | zzzzzz@gmail.com        |
+---------+-----------+-------------------------+

+---------+-----------+
| user_id | group_id  |
+---------+-----------+
|       2 | 2         |
|       3 | 1         |
|       3 | 2         |
|       3 | 3         |
|       3 | 4         |
|       4 | 1         |
|       4 | 2         |
|       4 | 3         |
|       4 | 4         |
|       4 | 5         |
+---------+-----------+

+----------+
| group_id |
+----------+
|        1 |
|        2 |
|        3 |
|        4 |
+----------+

答案 2 :(得分:2)

MySQL不会将逗号分隔列表视为字符串。执行WHERE group_id IN(2)后,它会将group_id转换为INT,因此可以将其与2进行比较。

当转换为INT时,MySQL将停在第一个非数字字符处。

例如,'1,2,3,4,5' IN (2)变为1 IN (2)。这是假的。

您可以尝试使用FIND_IN_SET来执行您想要的操作,但效率不高(因为它不能使用索引;它需要读取每一行以查看是否它匹配)。

WHERE FIND_IN_SET(2, group_id)

要搜索多行,请使用OR

WHERE FIND_IN_SET(1, group_id) OR FIND_IN_SET(2, group_id)

正确的方法是创建一个“链接表”,其中包含每个用户的一行(或多行),显示他们所在的组。

答案 3 :(得分:1)

说明

查询的逻辑是什么?SELECT * FROM user WHERE group_id IN(1,2);

  • 您提供了一个数字列表(1,2)
  • groud_id正在进行数字比较
  • 任何在数字上与第一个逗号匹配1或2的内容都会出现

SUGGESTION

我要呈现给你的东西可能看起来很不正统,但请跟我来......

这是一个查询,它将获得group_ids中包含1和2的每一行:

SELECT user.* FROM
(SELECT * FROM (SELECT id,CONCAT(',',group_id ,',') group_ids
FROM user) U WHERE LOCATE(',2,',group_ids)) U1
INNER JOIN
(SELECT * FROM (SELECT id,CONCAT(',',group_id ,',') group_ids
FROM user) U WHERE LOCATE(',4,',group_ids)) U2
ON U1.id = U2.id
INNER JOIN user ON user.id = U2.id;

以下是创建我们的示例数据的代码

DROP DATABASE IF EXISTS sundar;
CREATE DATABASE sundar;
use sundar
CREATE TABLE user
(
    id int not null auto_increment,
    user_name VARCHAR(30),
    user_email VARCHAR(70),
    group_id VARCHAR(128),
    PRIMARY KEY (id)
);
INSERT INTO user (user_name,user_email,group_id) VALUES
('suresh' , 'xxxx@yyyyyyyyyy.com'     ,'22'),
('sundar' , 's7sundera@gmail.com'     ,'2'),
('tester' , 'xxxxxxxx@yyyyyyyyyy.com' ,'1,2,3,4'),
('gail'   , 'zzzzzz@gmail.com'        ,'1,2,3,4,5');
SELECT * FROM user;

让我们创建您的样本

mysql> DROP DATABASE IF EXISTS sundar;
Query OK, 1 row affected (0.00 sec)

mysql> CREATE DATABASE sundar;
Query OK, 1 row affected (0.00 sec)

mysql> use sundar
Database changed
mysql> CREATE TABLE user
    -> (
    ->     id int not null auto_increment,
    ->     user_name VARCHAR(30),
    ->     user_email VARCHAR(70),
    ->     group_id VARCHAR(128),
    ->     PRIMARY KEY (id)
    -> );
Query OK, 0 rows affected (0.04 sec)

mysql> INSERT INTO user (user_name,user_email,group_id) VALUES
    -> ('suresh' , 'xxxx@yyyyyyyyyy.com'     ,'22'),
    -> ('sundar' , 's7sundera@gmail.com'     ,'2'),
    -> ('tester' , 'xxxxxxxx@yyyyyyyyyy.com' ,'1,2,3,4'),
    -> ('gail'   , 'zzzzzz@gmail.com'        ,'1,2,3,4,5');
Query OK, 4 rows affected (0.00 sec)
Records: 4  Duplicates: 0  Warnings: 0

mysql>

这就是它的样子

mysql> SELECT * FROM user;
+----+-----------+-------------------------+-----------+
| id | user_name | user_email              | group_id  |
+----+-----------+-------------------------+-----------+
|  1 | suresh    | xxxx@yyyyyyyyyy.com     | 22        |
|  2 | sundar    | s7sundera@gmail.com     | 2         |
|  3 | tester    | xxxxxxxx@yyyyyyyyyy.com | 1,2,3,4   |
|  4 | gail      | zzzzzz@gmail.com        | 1,2,3,4,5 |
+----+-----------+-------------------------+-----------+
4 rows in set (0.00 sec)

mysql>

同样,这是一个混乱的查询,可以得到你想要的东西:

SELECT user.* FROM
(SELECT * FROM (SELECT id,CONCAT(',',group_id ,',') group_ids
FROM user) U WHERE LOCATE(',1,',group_ids)) U1
INNER JOIN
(SELECT * FROM (SELECT id,CONCAT(',',group_id ,',') group_ids
FROM user) U WHERE LOCATE(',2,',group_ids)) U2
ON U1.id = U2.id
INNER JOIN user ON user.id = U2.id;

这里执行:

mysql> SELECT user.* FROM
    -> (SELECT * FROM (SELECT id,CONCAT(',',group_id ,',') group_ids
    -> FROM user) U WHERE LOCATE(',1,',group_ids)) U1
    -> INNER JOIN
    -> (SELECT * FROM (SELECT id,CONCAT(',',group_id ,',') group_ids
    -> FROM user) U WHERE LOCATE(',2,',group_ids)) U2
    -> ON U1.id = U2.id
    -> INNER JOIN user ON user.id = U2.id;
+----+-----------+-------------------------+-----------+
| id | user_name | user_email              | group_id  |
+----+-----------+-------------------------+-----------+
|  3 | tester    | xxxxxxxx@yyyyyyyyyy.com | 1,2,3,4   |
|  4 | gail      | zzzzzz@gmail.com        | 1,2,3,4,5 |
+----+-----------+-------------------------+-----------+
2 rows in set (0.00 sec)

mysql>

好的,如何寻找(2,4)

mysql> SELECT user.* FROM
    -> (SELECT * FROM (SELECT id,CONCAT(',',group_id ,',') group_ids
    -> FROM user) U WHERE LOCATE(',2,',group_ids)) U1
    -> INNER JOIN
    -> (SELECT * FROM (SELECT id,CONCAT(',',group_id ,',') group_ids
    -> FROM user) U WHERE LOCATE(',4,',group_ids)) U2
    -> ON U1.id = U2.id
    -> INNER JOIN user ON user.id = U2.id;
+----+-----------+-------------------------+-----------+
| id | user_name | user_email              | group_id  |
+----+-----------+-------------------------+-----------+
|  3 | tester    | xxxxxxxx@yyyyyyyyyy.com | 1,2,3,4   |
|  4 | gail      | zzzzzz@gmail.com        | 1,2,3,4,5 |
+----+-----------+-------------------------+-----------+
2 rows in set (0.00 sec)

mysql>

看起来很有效。

试一试!!!