blocks|key|4043845|text|大多数人认为模型与表类是一致的。事实上，它们使模型扩展了一个基表类。对于大多数简单的CRUD类型的工作，这或多或少都很好。但是，当您跨多个表或事务脚本进行更复杂的更新时，您会发现像您所询问的那样的尴尬情况。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4043846|事实上，a+model不是一个表。a+model是对应用程序业务的某些逻辑部分的封装，而不是将其存储在数据库中的实现。|offset|length|style|BOLD|4043847|我在几年前就开始编写Zend_Db_Table的代码，在编写手册时，我非常小心地避免将表类称为模型。|4043848|因此，您可能需要引入一个真正的Model类层(它不扩展基类)，它知道如何对一个或多个数据库表执行特定于应用程序的操作，但它不是应用程序的控制器。|4043849|Controller+->+Model+->+Table(s)|code-block|syntax|javascript|4043850|如果您有从多个控制器调用的类似业务任务，则此体系结构是避免重复代码的好方法。它有助于将应用程序功能和持久性细节从控制器中解耦出来，这在代码维护、可测试性、封装等方面都有好处。|4043851|查看免费的迷你书Domain-Driven+Design+Quickly+(基于Eric+Evans的书籍)，或者我的书SQL+Antipatterns:+Avoiding+the+Pitfalls+of+Database+Programming中的“魔豆”一章。|4043852|entityMap|0|LINK|mutability|MUTABLE|url|http://www.infoq.com/minibooks/domain-driven-design-quickly|1|http://pragprog.com/book/bksqla/sql-antipatterns^0|0|4|D|0|0|0|0|0|8|S|0|1O|1R|1|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|15|8|@$D|16|E|17|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|18|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|19|8|@]|9|@]|A|$]]|$1|L|3|M|5|N|7|1A|8|@]|9|@]|A|$O|P]]|$1|Q|3|R|5|6|7|1B|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|1C|8|@]|9|@$D|1D|E|1E|1|1F]|$D|1G|E|1H|1|1I]]|A|$]]|$1|U|3|-4|5|6|7|1J|8|@]|9|@]|A|$]]]|V|$W|$5|X|Y|Z|A|$10|11]]|12|$5|X|Y|Z|A|$10|13]]]]

Most people think of a Model as coincident with a table class. In fact, they make models extend a base table class. For most simple CRUD type work, this is more or less fine. But as you get into more complex updates across multiple tables, or transaction scripts, you find awkward situations like the one you are asking about.

In fact, a model is not a table. A model is an encapsulation of some logical part of your application's business, not the implementation of storing it in a database.

I worked on the code of Zend_Db_Table quite a bit a few years ago, and when I was writing the manual, I was careful not to refer to table classes as Models.

So you may need to introduce a true Model class layer (which extends no base class), that knows how to do certain application-specific operations against one or more database tables, yet it is not your application's Controller.

<pre><code>Controller -&gt; Model -&gt; Table(s)
</code></pre>

This architecture is a great way to avoid repeating code if you have similar business tasks that are called from multiple controllers. It helps to decouple application functions and persistence details from your Controllers, which has benefits in code maintenance, testability, encapsulation, etc.

Check out the free mini-book <a href="http://www.infoq.com/minibooks/domain-driven-design-quickly" rel="nofollow">Domain-Driven Design Quickly</a> (based on Eric Evans books), or the chapter "Magic Beans" in my book, <a href="http://pragprog.com/book/bksqla/sql-antipatterns" rel="nofollow">SQL Antipatterns: Avoiding the Pitfalls of Database Programming</a>.

blocks|key|4043868|text|Zend_Db_Table有一个名为"RefrenceMap“的重要属性，您可以在其中定义一个表如何链接到另一个表的逻辑检查下面的url以读取进一步的http://framework.zend.com/manual/en/zend.db.table.relationships.html|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|4043869|entityMap|0|LINK|mutability|MUTABLE|url|http://framework.zend.com/manual/en/zend.db.table.relationships.html^0|23|1W|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

Zend_Db_Table has an important property called "RefrenceMap" in which you define logic how your one table links to the other check the below url to read further 
<a href="http://framework.zend.com/manual/en/zend.db.table.relationships.html" rel="nofollow">http://framework.zend.com/manual/en/zend.db.table.relationships.html</a>

I have a model (call it Model_A) that I use to CRUD on a particular table defined by $name var. I also need to update another table (call it Model_B with different $name var) if the first operation in the other model is successful.

I'm trying to update table represented in Model_B from an action in class Model_A. I'm doing this by creating an instance of Model_B inside an action in Model_A. The update action fails fails because it seems to try to write to the table $name from Model_A, instead of the table $name from the instance of Model_B.

I suppose I could pass the result of the first update back to the controller and then go to the second model, but I'd rather do both updates at the same time.

Apart from an error in my coding, is there any obvious reason why my second update is not finding the right table.. and is this a reasonable aproach for what I'm wanting to do?

Zend framework - updating two tables from the same model

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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我有一个模型(称为Model_A)，用于在$name var定义的特定表上进行CRUD。如果另一个模型中的第一个操作成功，我还需要更新另一个表(称为具有不同$name变量的Model_B )。我正在尝试从类Model_A中的操作更新以Model_B表示的表。我通过在Model_A中的操作中创建Model_B的实例来完成...

问Zend框架-更新同一模型中的两个表
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