C# Unity容器Ctor注入

发布于 2024-12-09 05:18:53 字数 534 浏览 0 评论 0原文

假设我们已经：

class A
{
   ILogger myLog;
   A(ILogger log)
   {
      this.myLog = log;
   }
 ...
}

并且我们之前已经在unity容器中注册了ILogger接口，例如

 container.RegisterType<ILogger, SomeLogger>();

这里是SomeLogger类：

class SomeLogger : ILogger
{
   string myString;
   SomeLogger(string test)
   {
     myString = test;
   }
 ...
}

现在，unity如何为类A创建SomeLogger的实例而不将字符串传递给SomeLogger的ctor？假设 SomeLogger 没有其他 ctor。在哪里可以指定映射 SomeLogger 类型的 ctor 的参数？

原文

Say we have:

class A
{
   ILogger myLog;
   A(ILogger log)
   {
      this.myLog = log;
   }
 ...
}

And we have registered the ILogger interface before in the unity container, e.g.

 container.RegisterType<ILogger, SomeLogger>();

And here the SomeLogger class:

class SomeLogger : ILogger
{
   string myString;
   SomeLogger(string test)
   {
     myString = test;
   }
 ...
}

Now, how can unity create an instance of SomeLogger for class A without passing a string to the ctor of SomeLogger?
Suppose there is no other ctor for SomeLogger. Where can one specify the parameter(s) for ctor of the mapped SomeLogger type?

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七堇年 2024-12-16 05:18:53

您可以在配置中执行此操作：

<?xml version="1.0"?>
<configuration>
    <configSections>
        <section name="unity" type="Microsoft.Practices.Unity.Configuration.UnityConfigurationSection, Microsoft.Practices.Unity.Configuration" />
    </configSections>
    <unity>
        <containers>
            <container>
                <types>
                    <type type="[Namespace].ILogger, [AssemblyName]" mapTo="[Namespace].SomeLogger, [AssemblyName]">
                        <constructor>
                            <param name="test">
                                <value value="MyDesiredValue" />
                            </param>
                        </constructor>
                    </type>
                </types>
            </container>
        </containers>
  </unity>
</configuration>

这也以声明方式注册您的类型，因此

container.RegisterType<ILogger, SomeLogger>();

不再需要调用。

-道格

You can do that in your configuration:

<?xml version="1.0"?>
<configuration>
    <configSections>
        <section name="unity" type="Microsoft.Practices.Unity.Configuration.UnityConfigurationSection, Microsoft.Practices.Unity.Configuration" />
    </configSections>
    <unity>
        <containers>
            <container>
                <types>
                    <type type="[Namespace].ILogger, [AssemblyName]" mapTo="[Namespace].SomeLogger, [AssemblyName]">
                        <constructor>
                            <param name="test">
                                <value value="MyDesiredValue" />
                            </param>
                        </constructor>
                    </type>
                </types>
            </container>
        </containers>
  </unity>
</configuration>

This also declaratively registers your type, so the

container.RegisterType<ILogger, SomeLogger>();

call is no longer necessary.

-Doug

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画离情绘悲伤 2024-12-16 05:18:53

你可以这样做：

container.RegisterType<ILogger, SomeLogger>(new InjectionConstructor("myStringValue"));

You can do this:

container.RegisterType<ILogger, SomeLogger>(new InjectionConstructor("myStringValue"));

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从﹋此江山别 2024-12-16 05:18:53

您也可以在注册代码中执行此操作，如下所示：

UnityContainer.RegisterType<ILogger, SomeLogger>();
UnityContainer.Configure<InjectedMembers>()
              .ConfigureInjectionFor<SomeLogger>(new InjectionConstructor("TestString"));

you could also do this in the registration code as follows:

UnityContainer.RegisterType<ILogger, SomeLogger>();
UnityContainer.Configure<InjectedMembers>()
              .ConfigureInjectionFor<SomeLogger>(new InjectionConstructor("TestString"));

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