加入表格以引入另一个字段

Ame*_*ico -4 sql oracle

我目前有这个查询:

select f.chainid,count(f.player_uuid) as Favorites
from deals_player_favorite f
group by f.chainid
order by 2 desc
Run Code Online (Sandbox Code Playgroud)

结果如下:

CHAINID            FAVORITES
25                 2771
2207               2282
3940               1954
Run Code Online (Sandbox Code Playgroud)

等等...

我有另一个名为deals_deals的表,它还包含CHAINID字段.从这个表中,我想加入一个名为VENUE的字段,这样每个CHAIN ID都有一个Venue描述,输出看起来像这样

CHAINID    VENUE           FAVORITES    
25        Amazon.com        2771
2207       Walmart          2282
3940       CVS              1954
Run Code Online (Sandbox Code Playgroud)

等等...

如何将场地字段正确加入查询,使用CHAIN id作为deals_deals表和deals_player_favorite表中的键.

我尝试了一个内连接,导致了太多的结果.

deals_deals表的字段为CHAINID和VENUE ..

deals_player_favorite表的字段为CHAINID和PLAYER_UUID,但不包括deals_deals表的所有CHAINID,只包含player_uuid访问过的那些CHAINID.

样本数据:

deals_deals表

VENUE        CHAINID
Walmart         235
Aeropostale     1467
Checker's       881
Run Code Online (Sandbox Code Playgroud)

deals_player_favorite表

PLAYER_UUID         CHAINID
23rjior23-32fjdf     235
Run Code Online (Sandbox Code Playgroud)

请记住,deals_player_favorite只包含已被点击的特定CHAINID,而不是所有的chainids ....

Lam*_*mak 6

SELECT F.chainid, V.Venue, COUNT(f.player_uuid) as Favorites
FROM deals_player_favorite F
INNER JOIN Venues V
    ON F.chainid = V.chainid
GROUP BY F.chainid, V.Venue
ORDER BY COUNT(f.player_uuid) DESC
Run Code Online (Sandbox Code Playgroud)

  • @Stuave如果你错过了那么你可能会得到错误的答案. (3认同)
  • @Stuave那么请发布你需要"JOIN"的两个表的表定义,这样我们就可以发布正确的答案 (3认同)