blocks|key|2346275|text|我对CI不太了解，但作为SQL查询，这可能有效(将SQL和php混合在一起的伪代码)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2346276|SELECT+id
FROM+cards
WHERE+id+NOT+IN
(
++++SELECT+card+FROM+tag_link+WHERE+tag+IN+$tags
)|code-block|syntax|javascript|2346277|子查询返回包含特定标记的所有卡片ids。然后，主查询返回所有其他卡片ids。(请注意，子查询会导致大型表和/或复杂查询中的性能问题)|2346278|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I don't know too much about CI, but as an SQL query, this might work (pseudo code mixing SQL and php):

<pre><code>SELECT id
FROM cards
WHERE id NOT IN
(
 SELECT card FROM tag_link WHERE tag IN $tags
)
</code></pre>

The subquery returns all card ids that include a certain tag. The main query then returns all the other card ids. (Note that subqueries can cause performance issues in very large tables and/or complicated queries)

blocks|key|211692|text|使用TheWolf的答案并将其转换为active_record。记住，CI不支持子查询(尽管可以使用库)。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|211693|++$this->db->select('id');
++$this->db->where('id+NOT+IN+(SELECT+card+FROM+tag_link+WHERE+tag+IN+$tags)',+NULL,+FALSE);
++$query+=+$this->db->get('cards');|code-block|syntax|javascript|211694|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/users/2679333/thewolf^0|2|7|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

Using <a href="https://stackoverflow.com/users/2679333/thewolf">TheWolf</a> 's answer and converting it to active_record. Remember sub-queries are not supported by CI (although library can be used)

<pre><code> $this-&gt;db-&gt;select('id');
 $this-&gt;db-&gt;where('id NOT IN (SELECT card FROM tag_link WHERE tag IN $tags)', NULL, FALSE);
 $query = $this-&gt;db-&gt;get('cards');
</code></pre>

blocks|key|210725|text|首先，您的表结构应该有明确的名称链接|type|unstyled|depth|inlineStyleRanges|entityRanges|data|210726|cards:
-------
id++%7C+name
1+++%7C+alpha
2+++%7C+beta

tags:
------
id++%7C+name
1+++%7C+a
2+++%7C+b

tag_link:
---------
id++%7C+card_id+++%7C+tag_id
1+++%7C+1+++++++++%7C+1
2+++%7C+2+++++++++%7C+1
3+++%7C+2+++++++++%7C+2|code-block|syntax|javascript|210727|然后，您可以设置一个和条件，以避免其他结果。并提供避险卡。否则回答问题就没用了。|210728|function+search($tag_id,$card_id)+{
++++return+$this->db
++++++++->select('card.id')++++
++++++++->from('tag_link')
++++++++->join('card','card.id+=+tag_link.card_id','INNER')
++++++++->where_not_in('tag_link.tag_id',$_tag)
++++++++->where_not_in('tag_link.card_id',$card_id)
++++++++->get()
++++++++->result_array()
}|210729|这将生成此查询。|210730|SELECT
++card.id
FROM+tag_link
++INNER+JOIN+card
++++ON+card.id+=+tag_link.card_id
WHERE+tag_link.tag_id+NOT+IN(2)
++++AND+tag_link.card_id+NOT+IN(2)|210731|正如您所看到的，您正在避免卡2，其他明智的，您一定会得到所有的卡是链接到任何身份，无论如何，考虑到较少的tag_id你提供。|210732|这里是用来测试的小提琴|offset|length|210733|entityMap|0|LINK|mutability|MUTABLE|url|http://sqlfiddle.com/#!2/d2f04/1^0|0|0|0|0|0|0|0|0|2|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|14|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|16|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|18|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|19|8|@]|9|@$S|1A|T|1B|1|1C]]|A|$]]|$1|U|3|-4|5|6|7|1D|8|@]|9|@]|A|$]]]|V|$W|$5|X|Y|Z|A|$10|11]]]]

First of all your table structure should have clear names link this

<pre><code>cards:
-------
id | name
1 | alpha
2 | beta

tags:
------
id | name
1 | a
2 | b

tag_link:
---------
id | card_id | tag_id
1 | 1 | 1
2 | 2 | 1
3 | 2 | 2
</code></pre>

Then you can put an and condition to avoid other results. and provide the avoiding card. Otherwise answering question is useless.

<pre><code>function search($tag_id,$card_id) {
 return $this-&gt;db
 -&gt;select('card.id') 
 -&gt;from('tag_link')
 -&gt;join('card','card.id = tag_link.card_id','INNER')
 -&gt;where_not_in('tag_link.tag_id',$_tag)
 -&gt;where_not_in('tag_link.card_id',$card_id)
 -&gt;get()
 -&gt;result_array()
}
</code></pre>

This will generate this query.

<pre><code>SELECT
 card.id
FROM tag_link
 INNER JOIN card
 ON card.id = tag_link.card_id
WHERE tag_link.tag_id NOT IN(2)
 AND tag_link.card_id NOT IN(2)
</code></pre>

As you see you are avoiding card 2 other wise you are bound to get all the cards that are linked to any id anyway regards less of the tag_id you provide. 
<a href="http://sqlfiddle.com/#!2/d2f04/1" rel="nofollow">Here</a> is the fiddle to test

I got 3 tables in mysql:

cards:

<pre><code>id | name
1 | alpha
2 | beta
</code></pre>

tags:

<pre><code>id | name
1 | a
2 | b
</code></pre>

tag_link:

<pre><code>id | card | id
1 | 1 | 1
2 | 2 | 1
3 | 2 | 2
</code></pre>

I would like to retrieve all cards which DON'T include a certain tag. CI model:

<pre><code>function search($_tag) {
 $this-&gt;db-&gt;select('card.id');
 $this-&gt;db-&gt;join('tag_link', 'card.id = tag_link.card');

 $this-&gt;db-&gt;where_not_in('tag_link.tag', $_tag);

 $this-&gt;db-&gt;group_by('card.id');
 $query = $this-&gt;db-&gt;get('card');
 return $query;
}
</code></pre>

For tag '2' this returns card '1' as expected. However, card '2' is wrongly returned as well due to the one entry in tag_link connecting card '2' with tag '1'.

I thought about getting a first array of hits using the above function and then subtract another array in php containing all the cards including the tag I'm not interested in.
However this solution feels very clumsy. What's the most efficient approach for this problem?

Thanks,
singultus

how to exclude db entries associated with certain tags using ci active records

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。

页面排版不美观，阅读体验差。

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问如何使用ci活动记录排除与某些标记关联的db条目
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问如何使用ci活动记录排除与某些标记关联的db条目EN