blocks|key|1040663|text|我将id_carrier解释为承运人的实体标识：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1040664|select+ec.name+as+CustomerName,+eca.name+as+CarrierName
from+customer+c+join
+++++entity+ec
+++++on+c.id_entity+=+e.id_entity+join
+++++entity+eca
+++++on+c.id_carrier+=+eca.id_entity|code-block|syntax|javascript|1040665|这消除了加入到载体表的需要，除非您需要从那里获得更多的信息。|1040666|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 am interpreting id_carrier as being the entity id for carrier:

<pre><code>select ec.name as CustomerName, eca.name as CarrierName
from customer c join
 entity ec
 on c.id_entity = e.id_entity join
 entity eca
 on c.id_carrier = eca.id_entity
</code></pre>

This eliminates the need to join back to the carrier table, unless you need additional information from there.

blocks|key|881745|text|您需要对entity表加入两次，每种类型一次。使用它们的别名(下面是ca,+cu)来引用SELECT列表中的每个别名。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|881746|SELECT
++/*+Carrier+info+*/
++ca.id_entity+AS+carrier_id_entity,
++ca.name+AS+carrier_name,
++ca.date+AS+carrier_date,
++carrier.carrier_stuff,
++/*+Customer+info+*/
++cu.id_entity_as+AS+customer_id_entity,
++cu.name+AS+customer_name,
++cu.date+AS+customer_date,
++customer.discount,
++customer.payment
FROM
++carrier+
++JOIN+customer+ON+customer.carrier_id+=++carrier.id_entity
++/*+JOIN+first+against+entity+for+carrier+*/
++JOIN+entity+ca+ON+carrier.id_entity+=+ca.id_entity
++/*+and+again+between+customer+and+entity+*/
++JOIN+entity+cu+ON+customer.id_entity+=+cu.id_entity|code-block|syntax|javascript|881747|entityMap^0|4|6|Y|6|18|6|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|U|8|@]|D|@]|E|$]]]|L|$]]

You need to join twice against the <code>entity</code> table, once for each type. Use their aliases (<code>ca, cu</code> below) to reference each in the <code>SELECT</code> list.

<pre><code>SELECT
 /* Carrier info */
 ca.id_entity AS carrier_id_entity,
 ca.name AS carrier_name,
 ca.date AS carrier_date,
 carrier.carrier_stuff,
 /* Customer info */
 cu.id_entity_as AS customer_id_entity,
 cu.name AS customer_name,
 cu.date AS customer_date,
 customer.discount,
 customer.payment
FROM
 carrier 
 JOIN customer ON customer.carrier_id = carrier.id_entity
 /* JOIN first against entity for carrier */
 JOIN entity ca ON carrier.id_entity = ca.id_entity
 /* and again between customer and entity */
 JOIN entity cu ON customer.id_entity = cu.id_entity
</code></pre>

blocks|key|881749|text|两次加入entity并给出name列别名。如果不需要额外的carrier字段，则删除最后一个联接。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|881750|select+customer_entity.name+as+customer_name,+customer.credit,+--+additional+customer+fields...
++++++++carrier_entity.name+as+carrier_name,+carrier.carrier_stuff+--+additional+carrier+fields...
++++++++from+entity+as+customer_entity
++++++++inner+join+customer+on+(customer.id_entity+=+customer_entity.id_entity)
++++++++inner+join+entity+as+carrier_entity+on+(customer.id_carrier+=+carrier_entity.id_entity)
++++++++inner+join+carrier+on+(carrier.id_entity+=+carrier_entity.id_entity)|code-block|syntax|javascript|881751|entityMap^0|4|6|D|4|T|7|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|U|8|@]|D|@]|E|$]]]|L|$]]

Join <code>entity</code> twice and give the <code>name</code> column aliases. Remove the last join if additional <code>carrier</code> fields are not needed.

<pre><code>select customer_entity.name as customer_name, customer.credit, -- additional customer fields...
 carrier_entity.name as carrier_name, carrier.carrier_stuff -- additional carrier fields...
 from entity as customer_entity
 inner join customer on (customer.id_entity = customer_entity.id_entity)
 inner join entity as carrier_entity on (customer.id_carrier = carrier_entity.id_entity)
 inner join carrier on (carrier.id_entity = carrier_entity.id_entity)
</code></pre>

I don't know if it is correct, but look the following schema:

<img src="https://i.stack.imgur.com/8qEJJ.png" alt="enter image description here">

An Entity can be either a Carrier or a Customer. Every customer must be associated with a Carrier, this Carrier can be the own Customer.

I'm stuck, tried without success to perform a single query that returns the NAME of the customer and the NAME of his Carrier. Is there a way to make such operation ?

Thanks

SQL - Select the same column, twice

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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我不知道它是否正确，但请查看以下模式：​​实体可以是载体，也可以是客户。每个客户必须与承运人相关联，这个承运人可以是自己的客户。我被困住了，试图执行一个返回客户的名称和他的运营商的名称的查询，但没有成功。有没有办法做这样的手术？谢谢

问SQL -选择同一列，两次
EN

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问SQL -选择同一列，两次EN