blocks|key|1557155|text|您可以向连接表中添加新列，但是该表将成为一个新实体，因此您的模型现在将由三个实体where和两个一对多关系组成，而不是两个实体和单个多对多关系。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1557156|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

You can add new column to the junction table but the table will become a new entity so your model will now consist of three entities where and two one-to-many relations instead of two entities and single many-to-many relation.

blocks|key|2732056|text|正如Ladislav+Mrnka所说，如果您将新列添加到连接表中，将会有一个新的实体“在中间”，这将使导航变得更加困难。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2732057|如果您希望避免这种情况，但仍然可以像往常一样进行导航，则可以保留连接表并添加一个新表，就像添加了order列的连接表一样。当您需要订单信息时，您只需连接此表即可获取并使用它。|2732058|当然，这个新的表需要一些维护。也就是说，您可以为连接表的junction%2Border创建delete+on+cascade。并使用触发器(哦，这可不好！)为每个新创建的关系创建具有默认顺序的新行。因此，在您的业务逻辑中处理这一点会更可取。|2732059|我知道这太棘手了，但没有什么神奇的解决方案...只要选择你觉得更舒服的就行了。|2732060|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|L|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|M|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

As Ladislav Mrnka states, if you add the new column to the junction table, there will be a new entity "in the middle" that will make navigation much harder.

If you want to avoid this, but still be able to make the navigation as usual, you can keep the junction table and add a new table, just like the junction, with the order column added. When you need the order info, you can just join this table to get it and use it.

This new table will, of course, require some maintenance. I.e. you can create a delete on cascade for the junction+order to the junction table. And use a trigger (ooops, that's not good!) to create a new row with default order for each new created relation. So, it would be much more advisable to handle this in you business logic.

I know it's too tricky, but there's no magic solution... just choose what is more comfortable to you.

blocks|key|1557204|text|由于您需要按table1行而不是全局地对table2结果进行排序，因此您有三种不太好的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1557205|使用Ladislav建议的方法(使用外观较差的模型)-+add+order列，并添加桥entity.|1557206|The方法JotaBe建议(使用外观较差的模式)-添加额外的表并维护both.|unordered-list-item|offset|length|style|BOLD|1557207|
|1557208|If上下文仅用于读取代码(不需要更改关系)，并且您不介意在每次从DB更新后手动更改EDMX，那么您可以修改emdx并将关系表的SSDL定义更改为SQL查询，例如|1557209|
<>F215>|1557210|​|1557211|<EntitySet+Name="AS_TO_BS"+EntityType="BlaBla.Store.AS_TO_BS">
++++<DefiningQuery>
++++++++SELECT+ID1,+ID2+ORDER+BY+ORDERVALUE
++++++++FROM+AS_TO_BS
++++</DefiningQuery>
</EntitySet>|code-block|syntax|javascript|1557212|1557213|而不是：|1557214|1557215|<EntitySet+Name="AS_TO_BS"+EntityType="BlaBla.Store.AS_TO_BS"
++++store:Type="Tables"+Schema="MY_SCHEMA"+/>|1557216|1557217|看看你是否可以放宽你的要求，如果不能，那就选择三种解决方案中的一种。|1557218|编辑：|1557219|另一个想法是：|1557220|1557221|使用视图复制关系表，然后将关系映射到视图(只读)，将订单实体映射到表(可写)。|1557222|1557223|entityMap^0|0|0|P|D|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1G|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1H|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|1I|8|@$G|1J|H|1K|I|J]]|9|@]|A|$]]|$1|K|3|L|5|6|7|1L|8|@]|9|@]|A|$]]|$1|M|3|N|5|F|7|1M|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|1N|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1O|8|@]|9|@]|A|$]]|$1|S|3|T|5|U|7|1P|8|@]|9|@]|A|$V|W]]|$1|X|3|R|5|6|7|1Q|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|6|7|1R|8|@]|9|@]|A|$]]|$1|10|3|R|5|6|7|1S|8|@]|9|@]|A|$]]|$1|11|3|12|5|U|7|1T|8|@]|9|@]|A|$V|W]]|$1|13|3|R|5|6|7|1U|8|@]|9|@]|A|$]]|$1|14|3|15|5|6|7|1V|8|@]|9|@]|A|$]]|$1|16|3|17|5|6|7|1W|8|@]|9|@]|A|$]]|$1|18|3|19|5|6|7|1X|8|@]|9|@]|A|$]]|$1|1A|3|-4|5|6|7|1Y|8|@]|9|@]|A|$]]|$1|1B|3|1C|5|F|7|1Z|8|@]|9|@]|A|$]]|$1|1D|3|-4|5|6|7|20|8|@]|9|@]|A|$]]|$1|1E|3|-4|5|6|7|21|8|@]|9|@]|A|$]]]|1F|$]]

Due to your requirement of sorting table2 results per table1 row and not globally, you have three non-elegant solutions:

<ul>
<li>The approach Ladislav suggested (with the bad looking model) - add order column, add bridge entity.</li>
<li>The approach JotaBe suggested (with the bad looking schema) - add an additional table and maintain both.</li>
<li>If the context is used only for reading (no need to change relationships) and you don't mind changing the EDMX manually after every update from DB, then you could hack the emdx and change the SSDL definition of the relationship table to an SQL query e.g.</li>
</ul>

<code>

<pre><code>&lt;EntitySet Name="AS_TO_BS" EntityType="BlaBla.Store.AS_TO_BS"&gt;
 &lt;DefiningQuery&gt;
 SELECT ID1, ID2 ORDER BY ORDERVALUE
 FROM AS_TO_BS
 &lt;/DefiningQuery&gt;
&lt;/EntitySet&gt;
</code></pre>

</code>

Instead of:

<code>

<pre><code>&lt;EntitySet Name="AS_TO_BS" EntityType="BlaBla.Store.AS_TO_BS"
 store:Type="Tables" Schema="MY_SCHEMA" /&gt;
</code></pre>

</code>

See if you can relax your requirements, if not then settle on one of the three solutions.

Edit:

Another idea:

<ul>
<li>Use a view to duplicate the relationship table, then map the relationship to the view (as read only) and the order entity to the table (writable).</li>
</ul>

blocks|key|1557235|text|感谢大家对我的问题的良好回答。我现在对不同解决方案的利弊更有信心了。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1557236|我最终做的是:事实证明，只需向关系表添加一个排序列并不会影响模型，从DB更新仍然有效，并且表仍然映射为多对多关系。然后，我创建了一个从关系表获取排序列的存储过程，以及另一个更新指定记录的排序索引的存储过程。|1557237|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

Thank you all for the good answers to my question. I now feel more confident about the pros and cons of the different solutions.

What I ended up doing was this: As it turns out, just adding a sort column to the relation-table doesn't affect the model, update from DB still works and the table still gets mapped as a many-to-many relation. Then I created a stored procedure that fetches the sort column from the relation-table and another stored procedure to update the sort-index of a specified record.

I have two entities with a many-to-many relationship and I'm looking for a way to be able to sort the result from the tables. 
In other words, when I get a row from table1 and all the corresponding records from table2 I want to be able to have a stored sort order for table2 that's specific for that row in table1.

My first thought was to add a sort column to the table that represents the relation, but to my knowledge there is no way of accessing the new column in the relation.

Does anybody have any suggestions on how to accomplish this?

Entity-Framework: Sort on Many-to-Many

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。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋

腾讯云代码助手

云原生构建

TAPD 敏捷项目管理

Cloud Studio

SDK中心

API中心

命令行工具

涵盖代码开发、场景应用、自动测试全流程，助你从零构建专属AI助手

一站式MCP教程库，解锁AI应用新玩法

我有两个具有多对多关系的实体，我正在寻找一种能够对表中的结果进行排序的方法。换句话说，当我从table1获得一行，并从table2获得所有相应的记录时，我希望能够有一个针对table2的特定于table1中该行的存储排序顺序。我的第一个想法是向表示关系的表中添加一个排序列，但据我所知，没有办法访问关系中的新列。有没有人...

问Entity-Framework:多对多排序
EN

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问Entity-Framework:多对多排序EN