获取应用程序的启动地址 [英] get startup adr of application

问题描述

嗨...

在vs2008(C#)中:

我有:
1.一个主要项目(Windows应用程序)
2. WCF库
3. WCF主机
4. WCF客户端(Windows应用程序)

为了使用创建的客户测试我的服务，我在主项目中调用我的客户表单，并在通过此表单获得服务的输入要求后，通过调用我的服务合同方法开始对服务做出响应.

在此服务中，我需要创建一个带有用于请求的接收参数的正确连接字符串，并且我需要我的应用程序启动路径(D:\ myProj.dll)才能访问exe文件，但是使用diffrent后，不幸的是各种返回此路径的代码，正是在调用服务时，我得到了我不需要的其他路径(myproj，bin.debug或c:\ programfiles \ visual studio9 \ common7 \ IDE)，您可以帮我吗?

有关其他信息，我不得不说我是通过命令添加此路径的，以便对我的服务库和客户端的构建事件进行.


请帮助我访问此路径(我的项目的启动路径-> D:\ myProj.dll)

提前谢谢

hi ...

in vs2008 (C#):

I have :
1. a main project (windows app)
2. WCF Library
3. WCF Host
4. WCF Client (Windows app)

for testing my service with created client, I call my client form in my main proj and after getting the input requirement of service by this form , I start making response of service by calling my service contract method .

in this service I need to create a proper connection string with recieved parameter for request and I need my application startup path (D:\myProj.dll) to get access to an exe file , Unfortunately after using diffrent kinds of codes that return this path, exactly at point of calling my service, I get another paths I don''t need (myproj,bin.debug or c:\programfiles\visual studio9\common7\IDE) can U help me ?

for additional info I have to say that I added this path by a command to post-build event of my service library and client.


plz help me to get access to this path (startup path of my project -> D:\myProj.dll)

thank you in advance

推荐答案

因此，启动路径，而不是"adr"，对吗?

如果某个程序集中有一些代码，则此调用可以为您提供执行中的程序集:

So, start-up path, not "adr", right?

If you have some code in some assembly, this call can give you the executing assembly:

string assemblyLocation = System.Reflection.Assembly.GetExecutingAssembly().Location;

自然地，您可以获取任何位置基本上，通过这种方式，您可以找到某个程序集的主要可执行模块(即文件)的位置. (即使VS仅支持创建单模块程序集，程序集本身也不必是单个文件；该文件也是模块，而不是程序集.)

现在，您可以提取目录路径:

Naturally, you can get the location of any other assembly, for example, the entry assembly, using System.Reflection.Assembly.GetEntryAssembly(); etc. Basically, this way you find a location of a main executable module of some assembly, which is a file. (An assembly itself does not have to be a single file, even though VS only support creation of single-module assemblies; the file is also a module, not an assembly.)

Now, you can extract the directory path:

string exePath = System.IO.Path.GetDirectoryName(assemblyLocation);

请参阅:
http://msdn.microsoft.com/en-us/library/system.reflection. assembly.aspx [^ ]，
http://msdn.microsoft.com/en-us/library/system.io. path.aspx [^ ].

我不明白您怎么能获得对exe文件的访问权限"，这与WCF有什么关系，但是看来这是您的问题. :-)

—SA

Please see:
http://msdn.microsoft.com/en-us/library/system.reflection.assembly.aspx[^],
http://msdn.microsoft.com/en-us/library/system.io.path.aspx[^].

I don''t understand how can you "get access to an exe file", and how is this related to WCF, but it looks like this is your problem. :-)

—SA

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