Question

我正在尝试为图片设置评级应用。我想显示当前登录用户尚未评级的图像。

这包括两个表，我已经成功地加入了表格但是额外的条件：

必须是图片
排除当前用户已评级的所有图片均无效。

它显示结果，好像两个AND条件根本不存在。

SELECT 
   ver_data_media.id, ver_data_media.value, ver_data_media.ref, 
   ver_data_media.type, ver_data_rating.a
FROM 
   ver_data_media
LEFT JOIN 
   ver_data_rating ON ver_data_media.ref = ver_data_rating.a
                   AND ver_data_media.type = 'image'
                   AND ver_data_rating.a != $current_user->ID

非常感谢任何建议，谢谢弗兰克。

Answer 1

您可以尝试以下方法吗？选择当前用户的记录，然后显示评级为空/空的图像......也许。如果你向我们展示了你的表格模式，那就更好了......这是一个更有效的答案。

SELECT ver_data_media.id, ver_data_media.value, 
ver_data_media.ref, ver_data_media.type, ver_data_rating.a
FROM ver_data_media
LEFT JOIN ver_data_rating
ON ver_data_media.ref = ver_data_rating.a
WHERE ver_data_media.type = 'image'
AND ver_data_rating.a is null
AND id = $current_user -- shouldn't this be like userid = $current_user, 
                        -- where is this ID coming from? what is its relation to this table?

否则最可能的问题是此行上<验证您要验证的数据类型，ver_data_rating.a != $current_user->ID不同。我想最好的建议是将强制转换一种数据类型同时放入ver_Data_rating.a和$current_user。

例如您可以使用以下任一方法进行测试。

CAST(ver_data_rating.a AS INT) != CAST($current_user as INT)
CAST(ver_data_rating.a AS INT) <> = CAST($current_user as INT)

更新：使用推荐的表格架构和示例查询

看起来，您的查询连接的方式，关系确实存在一些问题。请查看此示例，并告知我们您的结果。我列出了两种方法，可以从特定用户/用户的媒体表中获取未分级 media records。变量用于存储用户ID。

<强> SQLFIDDLE DEMONSTRATION

示例数据：

-- media table
ID  VALUE   REF     TYPE
1   300     15      image
2   100     25      art
3   500     35      cartoon
4   200     35      image
5   100     25      image
6   200     15      image
7   400     15      image
8   250     15      image
9   500     15      image
10  180     15      art

-- rating table
RID     MID     UID     SOMETHING
100     1       15      xyz
101     2       25      abc
102     5       25      efg
103     3       35      abc

逐步查询：

set @user_id:=15;

-- find all media records for @user, type image
select * from media
where ref = @user_id
and type = 'image'
;

-- find all rated reords for @user
select mid from rating
 where uid = @user_id
;

-- media records that needs to be rated by @user
-- using IN clause
select * from media
where id not in 
(select mid from rating
 where uid = @user_id)
and type = 'image'
and ref = @user_id
;

-- media records that needs to be rated by @user
-- using INNER JOIN
select m.*
from media m
inner join (
select mid from rating
 where uid = @user_id) as x
on m.id <> x.mid
and m.type = 'image'
and m.ref = @user_id
;

上面逐步查询的结果：

ID  VALUE   REF     TYPE
1   300     15      image
6   200     15      image
7   400     15      image
8   250     15      image
9   500     15      image

来自user = 15下的评级表的

评级记录ID，类型=图像

MID
1

使用IN子句：user = 15的未分级媒体记录，type = image`

ID  VALUE   REF     TYPE
6   200     15      image
7   400     15      image
8   250     15      image
9   500     15      image

使用INNER JOIN：用户= 15，type = image

的未评级媒体记录

ID  VALUE   REF     TYPE
6   200     15      image
7   400     15      image
8   250     15      image
9   500     15      image

Answer 2

左连接基本上是左外连接的缩写。您需要在JOIN条件之外将其他过滤器分隔到WHERE子句中。您的查询是说从ver_data_media获取所有记录。除此之外，给我ver_data_ratings符合我的条件或否则没有ver_data_ratings。很难翻译成英语:)我也认为，你所比较的列不正确或应该重命名。

试试这个......

SELECT ver_data_media.id, ver_data_media.value, ver_data_media.ref, ver_data_media.type, ver_data_rating.a
FROM ver_data_media
LEFT OUTER JOIN ver_data_rating
ON ver_data_media.id= ver_data_rating.media_id
WHERE ver_data_media.type = 'image'
AND ver_data_rating.user_id <> $current_user->ID

内部联接与额外条件

2 个答案:

更新：使用推荐的表格架构和示例查询