Question

我有一个具有一对多关系的数据库（场地，相关类别），如下所示：

venue_id | venue_name
---------------------
1        | venue1
2        | venue2

和类别

venue_id | category_id
---------------------
1        | 5
2        | 7
1        | 8
2        | 5

我想要显示所有场地category_id为5，但没有category_id的7和8.我尝试使用这样的连接：

SELECT distinct(`venue_to_category`.`venue_id`),`venue_name` 
FROM `venue_to_category` INNER JOIN `venues` 
    ON `venues`.venue_id = `venue_to_category`.venue_id 
WHERE `category_id` != 7
AND `category_id` != 8 
AND `category_id` = 5

但它没有返回正确的结果（实际上我不确定它返回的结果）

Answer 1

你案件中的2件事

获取category_id = 5而非7或8
获取category_id = 5而不是7和8

对于第一个，您可以使用

select
v.venue_id,
v.venue_name
from venues v
join categories c on c.venue_id = v.venue_id
where c.category_id = 5
AND NOT EXISTS
(
  select 1 from categories c1 where v.venue_id = c1.venue_id
  AND c1.category_id in (7,8)
);

第二个

select
v.venue_id,
v.venue_name
from venues v
join categories c on c.venue_id = v.venue_id
where c.category_id = 5
AND NOT EXISTS
(
  select 1 from categories c1 where v.venue_id = c1.venue_id
  AND c1.category_id in (7,8) having count(*) = 2
);

<强> DEMO

Answer 2

执行您网站的搜索功能时。 Match和Against比Like sql语句更好。该字段必须设置为FullText与字段上的术语匹配：

SELECT vc.venue_id, venue_name FROM venue_to_category vc, venues v WHERE MATCH(category_id) AGAINST('-7 -8 +5');

您还可以使用IN BOOLEAN MODE来允许sql语句中的运算符。例如。 ......

MATCH（category_id）反对（＆＃39; -7 -8 + 5＆＃39;在布尔模式中）......

（ - ）减号表示什么都不匹配＆＃39; -7＆＃39; ＆＃39; -8＆＃39;

（+）这个词必须出现在比赛中。

还有许多其他运营商需要使用。有关更多操作员和说明，请参阅this page

Answer 3

尝试

SELECT distinct `venue_to_category`.`venue_id` ,`venue_name`
FROM `venue_to_category` INNER JOIN `venues` 
    ON `venues`.venue_id = `venue_to_category`.venue_id 
WHERE `category_id` <> 7
AND `category_id` <> 8 
AND `category_id` = 5

或

SELECT distinct `venue_to_category`.`venue_id` ,`venue_name`
FROM `venue_to_category` INNER JOIN `venues` 
    ON `venues`.venue_id = `venue_to_category`.venue_id 
WHERE `category_id` NOT IN (7,8)
AND `category_id` = 5

注意：distinct不是函数。而对于你的例子，你可以完全放弃它，因为没有重复的行。

Answer 4

SELECT v.id, v.name
  FROM venues v
  JOIN venue_to_category vc
    ON vc.venue_id=v.id
   AND vc.category_id=5
 WHERE NOT EXISTS (
    SELECT 1 
      FROM venue_to_category vci
     WHERE vci.venue_id = v.id
       AND vci.category_id IN (7,8)  
       );

MySQL One to Many删除特定的＆＃39; many＆＃39;

4 个答案: