entityMap|blocks|key|ets5e|text|你需要每一件产品的巫师...所以你应该在productID.also上做group+by，你可以在'select‘和'group+by’中添加你需要的其他产品列|type|unstyled|depth|inlineStyleRanges|entityRanges|data|67o4m|SELECT
++++products.product_id,products.name
FROM
++++products
++++INNER+JOIN
++++product_categories
++++ON+
++++++++products.product_id+=+product_categories.product_id
++++INNER+JOIN
++++categories
++++ON+
++++++++product_categories.category_id+=+categories.category_id
WHERE
++++categories.is_private+=+1
GROUP+BY
++++products.product_id,products.name
HAVING
++++COUNT+(categories.category_id)+>+3|code-block|syntax|javascript|6muk9^0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|I|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|J|9|@]|A|@]|B|$F|G]]|$2|H|4|-4|6|7|8|K|9|@]|A|@]|B|$]]]]

you need every product witch .... so you should do group by on productID.also you can add other columns of products that you need in both 'select' and 'group by'
<pre><code>SELECT
 products.product_id,products.name
FROM
 products
 INNER JOIN
 product_categories
 ON 
 products.product_id = product_categories.product_id
 INNER JOIN
 categories
 ON 
 product_categories.category_id = categories.category_id
WHERE
 categories.is_private = 1
GROUP BY
 products.product_id,products.name
HAVING
 COUNT (categories.category_id) &gt; 3
</code></pre>

entityMap|blocks|key|cn01k|text|要获取所有产品列，请执行以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|7rr22|SELECT+products.*
FROM+products+p
WHERE+product_id+in+(SELECT+pc.product_id
+++++++++++++++++++++++FROM+product_categories+pc
++++++++++++++++++++++INNER+JOIN+categories+c+ON+pc.category_id+=+c.category_id
++++++++++++++++++++++WHERE+c.is_private+=+1
++++++++++++++++++++++GROUP+BY+c.category_id
+++++++++++++++++++++HAVING+COUNT(c.category_id)+>+3)|code-block|syntax|javascript|bndgp^0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|I|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|J|9|@]|A|@]|B|$F|G]]|$2|H|4|-4|6|7|8|K|9|@]|A|@]|B|$]]]]

To get all columns of products:
<pre><code>SELECT products.*
FROM products p
WHERE product_id in (SELECT pc.product_id
 FROM product_categories pc
 INNER JOIN categories c ON pc.category_id = c.category_id
 WHERE c.is_private = 1
 GROUP BY c.category_id
 HAVING COUNT(c.category_id) &gt; 3)
</code></pre>

entityMap|blocks|key|enotk|text|看看这个！|type|unstyled|depth|inlineStyleRanges|entityRanges|data|b9m60|select+*+from+products+where+product_id+in+
(select+pc.product_id+from+product_categories+inner+join+categories+c+on+
pc.category_id=c.category_id+where+c.is_private=1+
group+by+c.category_id+having+count(c.category_id)>3)|code-block|syntax|javascript|9d95j^0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|I|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|J|9|@]|A|@]|B|$F|G]]|$2|H|4|-4|6|7|8|K|9|@]|A|@]|B|$]]]]

Check this out!
<pre><code>select * from products where product_id in 
(select pc.product_id from product_categories inner join categories c on 
pc.category_id=c.category_id where c.is_private=1 
group by c.category_id having count(c.category_id)&gt;3)

</code></pre>

entityMap|blocks|key|96un3|text|如果按products.product_id聚合，查询将是正确的--根据数据库的不同，可能需要在SELECT中包含其他列。但是假设product_id在该表中是唯一的，那么它就是有效的SQL。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|11c96|如果您只需要产品If，则不需要products表：|a0qg8|select+pc.product_id
from+product_categories+pc+join
+++++categories+c
+++++on+pc.category_id+=+c.category_id
where+is_private+=+1
group+by+pc.product_id
having+count(*)+>+3;|code-block|syntax|javascript|5jikn^0|3|J|1C|6|1T|A|0|F|8|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|O|9|@$A|P|B|Q|C|D]|$A|R|B|S|C|D]|$A|T|B|U|C|D]]|E|@]|F|$]]|$2|G|4|H|6|7|8|V|9|@$A|W|B|X|C|D]]|E|@]|F|$]]|$2|I|4|J|6|K|8|Y|9|@]|E|@]|F|$L|M]]|$2|N|4|-4|6|7|8|Z|9|@]|E|@]|F|$]]]]

Your query would be correct if you aggregated by <code>products.product_id</code> -- well, depending on the database, you might need to include other columns in the <code>SELECT</code>. But it is valid SQL assuming that <code>product_id</code> is unique in that table.
If you only want the product ids you don't need the <code>products</code> table:
<pre><code>select pc.product_id
from product_categories pc join
 categories c
 on pc.category_id = c.category_id
where is_private = 1
group by pc.product_id
having count(*) &gt; 3;
</code></pre>

<img src="https://i.imgur.com/vv6BBzX.png" alt="" />
What's the SQL to find the list of products that belong to more than 3 private categories.
I tried this:
<pre class="lang-sql prettyprint-override"><code>SELECT
 products.*
FROM
 products
 INNER JOIN
 product_categories
 ON 
 products.product_id = product_categories.product_id
 INNER JOIN
 categories
 ON 
 product_categories.category_id = categories.category_id
WHERE
 categories.is_private = 1
GROUP BY
 categories.category_id
HAVING
 COUNT(categories.category_id) &gt; 3
</code></pre>
Thanks!

How to list products that belong to more than 3 private categories

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

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 ​ ? ​ 查找属于3个以上私有类别的产品列表的SQL是什么。 我试过这个： SELECT    products.*FROM    products    INNER JOIN    product_categories    ON         products.product_id = product_cat...

问如何列出属于3个以上私有类别的产品
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问如何列出属于3个以上私有类别的产品EN