WHERE LIKE 有多个关键字

Question

我有一个相当复杂的查询，它在 4 个表中搜索一个或多个关键字：

select distinct textures.id from textures
left join `category_texture` on `category_texture`.`texture_id` = `textures`.`id`
left join `categories` on `categories`.`id` = `category_texture`.`category_id`
left join `tag_texture` on `tag_texture`.`texture_id` = `textures`.`id`
left join `tags` on `tags`.`id` = `tag_texture`.`tag_id`
left join `size_texture` on `size_texture`.`texture_id` = `textures`.`id`
left join `sizes` on `sizes`.`id` = `size_texture`.`size_id` 
WHERE ( (textures.name LIKE '%artwork%' 
OR categories.name LIKE '%artwork%' 
OR tags.name LIKE '%artwork%'   
OR sizes.name LIKE '%artwork%')  AND  (textures.name LIKE '%super%' 
OR categories.name LIKE '%super%'   
OR tags.name LIKE '%super%' 
OR sizes.name LIKE '%super%')  AND  (textures.name LIKE '%master%' 
OR categories.name LIKE '%master%'  
OR tags.name LIKE '%master%'    
OR sizes.name LIKE '%master%') ) AND `textures`.`is_published` = 1  group by `tags`.`name`, `categories`.`name`, `sizes`.`name`, `textures`.`id`

在这个例子中，artifacts 是一个类别，master 和 super 是标签。问题是，如果我重复任何表格，都找不到结果：

1. 艺术品（类别）+大师（标签）作品
2. 艺术作品
3. artwork + master + super 不起作用 - 它应该打印所有具有艺术品作为类别和 2 个标签（master 和 super）的纹理，但它没有。

提前致谢。

为清楚起见进行编辑：

我的目标是能够在这些表中搜索数据，从而可以组合同一个表的多个实例。例如，如果我搜索“artwork super master”，它应该返回我所有的纹理，类别为“artwork”（因为它是唯一找到这个词的地方）和标签“super”和“master”（两者都是)。

目前使用此查询，我可以在任何这些表中进行搜索，但前提是我的搜索在同一个表中没有找到 2 个以上的内容。因此，搜索类别 + 纹理名称 + 标记 + 大小是可行的，但搜索发现为 2 个标记的内容会失败。

Answer 1

我重新考虑了一下，也许这种方法更好：

SELECT T1.id
FROM 
(  
select textures.id 
from textures
WHERE textures.name LIKE '%texture1%' 
UNION 
select textures.id 
from textures
join category_texture on category_texture.texture_id = textures.id
join categories on categories.id = category_texture.category_id
WHERE categories.name LIKE '%texture1%'
UNION
select textures.id 
from textures
join tag_texture on tag_texture.texture_id = textures.id
join tags on tags.id = tag_texture.tag_id
WHERE tags.name LIKE '%texture1%'
UNION 
select textures.id 
from textures
join size_texture on size_texture.texture_id = textures.id
join sizes on sizes.id = size_texture.size_id 
WHERE sizes.name LIKE '%texture1%'
) AS T1
JOIN 
(  
select textures.id 
from textures
WHERE textures.name LIKE '%category2%' 
UNION 
select textures.id 
from textures
join category_texture on category_texture.texture_id = textures.id
join categories on categories.id = category_texture.category_id
WHERE categories.name LIKE '%category2%'
UNION
select textures.id 
from textures
join tag_texture on tag_texture.texture_id = textures.id
join tags on tags.id = tag_texture.tag_id
WHERE tags.name LIKE '%category2%'
UNION 
select textures.id 
from textures
join size_texture on size_texture.texture_id = textures.id
join sizes on sizes.id = size_texture.size_id 
WHERE sizes.name LIKE '%category2%'
) AS T2
ON T1.id = T2.id
JOIN
(  
select textures.id 
from textures
WHERE textures.name LIKE '%tag3%' 
UNION 
select textures.id 
from textures
join category_texture on category_texture.texture_id = textures.id
join categories on categories.id = category_texture.category_id
WHERE categories.name LIKE '%tag3%'
UNION
select textures.id 
from textures
join tag_texture on tag_texture.texture_id = textures.id
join tags on tags.id = tag_texture.tag_id
WHERE tags.name LIKE '%tag3%'
UNION 
select textures.id 
from textures
join size_texture on size_texture.texture_id = textures.id
join sizes on sizes.id = size_texture.size_id 
WHERE sizes.name LIKE '%tag3%'
) AS T3
ON T1.id = T3.id

只需添加更多/更少JOIN...ON T1.id = Tn.id 部分以匹配您的参数。

这是一个显示它执行的小提琴：http://sqlfiddle.com/#!9/9abe6/1

Answer 2

您的查询尝试执行的操作存在两个问题。首先，参数的数量可能会有所不同，这需要为每个变体使用不同的查询语法（或者一些丑陋的黑客来解决它）。其次，对于每个表，当您想要返回所有行时，您只想返回与参数之一匹配的行除非没有匹配项。

也许这对你有用：

select distinct textures.id 
from textures
join category_texture on category_texture.texture_id = textures.id
join categories on categories.id = category_texture.category_id
join tag_texture on tag_texture.texture_id = textures.id
join tags on tags.id = tag_texture.tag_id
left join size_texture on size_texture.texture_id = textures.id
join sizes on sizes.id = size_texture.size_id 
WHERE ((textures.name LIKE '%artwork%' OR textures.name LIKE '%super%' OR textures.name LIKE '%master%')  
  OR   (0 = select count(*) from textures where (textures.name LIKE '%artwork%' OR textures.name LIKE '%super%' OR textures.name LIKE '%master%') ) )
AND   ((categories.name LIKE '%artwork%' OR categories.name LIKE '%super%' OR categories.name LIKE '%master%') 
  OR   (0 = select count(*) from categories where (categories.name LIKE '%artwork%' OR categories.name LIKE '%super%' OR categories.name LIKE '%master%') ) )
AND   ((tags.name LIKE '%artwork%' OR tags.name LIKE '%super%' OR tags.name LIKE '%master%') 
  OR   (0 = select count(*) from tags where (tags.name LIKE '%artwork%' OR tags.name LIKE '%super%' OR tags.name LIKE '%master%') ) )
AND   ((sizes.name LIKE '%artwork%' OR sizes.name LIKE '%super%' OR sizes.name LIKE '%master%') 
  OR   (0 = select count(*) from name where (sizes.name LIKE '%artwork%' OR sizes.name LIKE '%super%' OR sizes.name LIKE '%master%') ) )
AND textures.is_published = 1   
group by textures.id, tags.name, categories.name, sizes.name, textures.name

编辑 由于您的代码动态生成 SQL，因此我已输入原始示例中的值。

整个查询非常丑陋，但它应该可以工作。 如果您需要进行这样的搜索，则需要重新考虑您的数据结构。