Question

我有2个表，奖励和awards_user，并且在mysql查询上有LEFT JOIN的问题没有得到像我想要的结果

奖励：

awardid  | name   | link  
1        | award1 | photo link    
2        | award2 | photo link   
3        | award3 | photo link   
4        | award4 | photo link

awards_user：

awardid  | username
2        | mark      
2        | jon
3        | mark    
1        | jon

我试着得到这样的结果：

award name  |  award photo  |  Users
award1      |      link     |  jon
award2      |      link     |  mark,jon
award3      |      link     |  mark
award4      |      link     |  -------

这是我的尝试：

 SELECT au.username 
      , au.userid 
      , a.link 
      , a.name 
      , a.awardid 
   FROM awards a
   LEFT 
   JOIN awards_user au
     ON au.awardid = a.awardid
  WHERE a.forumid = 22
  GROUP 
     BY a.awardid 
      , au.username
  ORDER 
     BY a.awardid DESC

解决问题：

$all_awards = $db->query_read(" SELECT awards.name as name, awards.link as link, 
       GROUP_CONCAT(DISTINCT awards_user.username) AS username
   FROM awards LEFT JOIN awards_user ON (awards_user.awardid = awards.awardid)
   where awards.forumid = '".$_REQUEST['forumid']."'
   GROUP BY awards.awardid, awards.name, awards.link
  ");

Answer 1

GROUP_CONCAT完全符合您的要求：

SELECT
       a.name AS name
     , a.link AS photo
     , GROUP_CONCAT(DISTINCT au.username) AS users
FROM awards AS a
     LEFT JOIN awards_user AS au ON (au.awardid = a.awardid)
GROUP BY
       a.awardid
     , a.name
     , a.link
ORDER BY
     a.awardid DESC

此处有关MySQL GROUP_CONCAT的更多信息：http://dev.mysql.com/doc/refman/5.0/en/group-by-functions.html#function_group-concat

顺便说一下，你甚至可以在group_concat中对用户名进行排序或选择不同的分隔符。只需阅读MySQL手册和 google ＆＃34; mysql group_concat示例＆＃34; ：）

Answer 2

并在表名上添加``，因为你的命名约定非常粗略可能与mysql保留字匹配，并再次实践安全过滤你的输入 $_REQUEST['forumid']

Answer 3

试试这个：

SELECT awards.name as award_name, award.link as award_photo, 
       GROUP_CONCAT(DISTINCT awards_user.username AS users
   FROM awards LEFT JOIN awards_user ON (awards_user.awardid = awards.awardid)
   GROUP BY awards.awardid, awards.name, awards.link;

关于mysql查询的LEFT JOIN没有得到我想要的结果

3 个答案: