blocks|key|1833175|text|在项目中使用生成后操作，并添加命令以复制有问题的DLL。构建后操作以批处理脚本的形式编写。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1833176|输出目录可以引用为$(OutDir)。项目目录为$(ProjDir)。尝试在适用的情况下使用相对路径，以便您可以复制或移动项目文件夹，而不会破坏生成后操作。|offset|length|style|CODE|1833177|entityMap^0|0|9|9|O|A|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@$D|L|E|M|F|G]|$D|N|E|O|F|G]]|9|@]|A|$]]|$1|H|3|-4|5|6|7|P|8|@]|9|@]|A|$]]]|I|$]]

Use a post-build action in your project, and add the commands to copy the offending DLL. The post-build action are written as a batch script.

The output directory can be referenced as <code>$(OutDir)</code>. The project directory is available as <code>$(ProjDir)</code>. Try to use relative pathes where applicable, so that you can copy or move your project folder without breaking the post-build action.

blocks|key|1835410|text|$(OutDir)原来是VS2013中的一个相对路径，所以我必须将它与$(ProjectDir)结合起来才能达到预期的效果：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1835411|xcopy+/y+/d++"$(ProjectDir)External\*.dll"+"$(ProjectDir)$(OutDir)"|code-block|syntax|javascript|1835412|顺便说一句，您可以通过在开头添加“echo”来轻松地调试脚本，并观察构建输出窗口中展开的文本。|1835413|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|$]]

$(OutDir) turned out to be a relative path in VS2013, so I had to combine it with $(ProjectDir) to achieve the desired effect:

<pre><code>xcopy /y /d "$(ProjectDir)External\*.dll" "$(ProjectDir)$(OutDir)"
</code></pre>

BTW, you can easily debug the scripts by adding 'echo ' at the beginning and observe the expanded text in the build output window.

blocks|key|533623|text|当尝试将输出dll从一个C%2B%2B项目复制到同一解决方案中另一个C#项目的发布和调试文件夹时，上面注释部分中的详细信息对我(VS+2013)不起作用。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|533624|我必须添加以下构建后操作(右键单击具有.dll输出的项目)，然后添加属性->配置属性->构建事件->构建后事件->命令行|533625|现在，我添加了这两行代码，将输出dll复制到两个文件夹中：|533626|xcopy+/y+$(TargetPath)+$(SolutionDir)aeiscontroller\bin\Release
xcopy+/y+$(TargetPath)+$(SolutionDir)aeiscontroller\bin\Debug|code-block|syntax|javascript|533627|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|O|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|P|8|@]|9|@]|A|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

The details in the comments section above did not work for me (VS 2013) when trying to copy the output dll from one C++ project to the release and debug folder of another C# project within the same solution.

I had to add the following post build-action (right click on the project that has a .dll output) then properties -> configuration properties -> build events -> post-build event -> command line

now I added these two lines to copy the output dll into the two folders:

<pre><code>xcopy /y $(TargetPath) $(SolutionDir)aeiscontroller\bin\Release
xcopy /y $(TargetPath) $(SolutionDir)aeiscontroller\bin\Debug
</code></pre>

blocks|key|1835336|text|(此答案仅适用于C#，不适用于C%2B%2B，对不起，我误解了原问题)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1835337|我以前经历过像这样的DLL地狱。我的最后一个解决方案是将非托管DLL作为二进制资源存储在托管DLL中，并在程序启动时将它们提取到一个临时文件夹中，并在释放它们时将其删除。|1835338|这应该是.NET或pinvoke基础设施的一部分，因为它非常有用……它使您的托管DLL易于管理，无论是使用Xcopy还是作为更大的Visual+Studio解决方案中的项目引用。一旦你这样做了，你就不必担心后期构建事件了。|1835339|更新：|1835340|我在这里发布了另一个答案https://stackoverflow.com/a/11038376/364818的代码|offset|length|1835341|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/11038376/364818^0|0|0|0|0|C|17|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|U|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|V|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|W|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|X|8|@]|9|@$J|Y|K|Z|1|10]]|A|$]]|$1|L|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|M|$N|$5|O|P|Q|A|$R|S]]]]

(This answer only applies to C# not C++, sorry I misread the original question)

I've got through DLL hell like this before. My final solution was to store the unmanaged DLLs in the managed DLL as binary resources, and extract them to a temporary folder when the program launches and delete them when it gets disposed.

This should be part of the .NET or pinvoke infrastructure, since it is so useful.... It makes your managed DLL easy to manage, both using Xcopy or as a Project reference in a bigger Visual Studio solution. Once you do this, you don't have to worry about post-build events.

UPDATE:

I posted code here in another answer <a href="https://stackoverflow.com/a/11038376/364818">https://stackoverflow.com/a/11038376/364818</a>

blocks|key|1941357|text|在project.csproj文件中添加内置副本：|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|1941358|++<Project>
++++...
++++<Target+Name="AfterBuild">
++++++<Copy+SourceFiles="$(ProjectDir)..\..\Lib\*.dll"+DestinationFolder="$(OutDir)Debug\bin"+SkipUnchangedFiles="false"+/>
++++++<Copy+SourceFiles="$(ProjectDir)..\..\Lib\*.dll"+DestinationFolder="$(OutDir)Release\bin"+SkipUnchangedFiles="false"+/>
++++</Target>
++</Project>|code-block|syntax|javascript|1941359|entityMap^0|1|E|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Add builtin COPY in project.csproj file:

<pre><code> &lt;Project&gt;
 ...
 &lt;Target Name="AfterBuild"&gt;
 &lt;Copy SourceFiles="$(ProjectDir)..\..\Lib\*.dll" DestinationFolder="$(OutDir)Debug\bin" SkipUnchangedFiles="false" /&gt;
 &lt;Copy SourceFiles="$(ProjectDir)..\..\Lib\*.dll" DestinationFolder="$(OutDir)Release\bin" SkipUnchangedFiles="false" /&gt;
 &lt;/Target&gt;
 &lt;/Project&gt;
</code></pre>

blocks|key|1835428|text|xcopy+/y+/d++"$(ProjectDir)External\*.dll"+"$(TargetDir)"|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1835429|您还可以参考相对路径，下一个示例将在位于项目文件夹上一级的文件夹中找到DLL。如果在一个解决方案中有多个使用DLL的项目，这会将DLL的源代码放在一个公共区域，当您将其中任何一个设置为“启动项目”时，都可以访问它们。|unstyled|1835430|xcopy+/y+/d++"$(ProjectDir)..\External\*.dll"+"$(TargetDir)"|1835431|/y选项无需确认即可复制。/d选项检查目标中是否存在文件，如果存在，则仅在源的时间戳比目标的时间戳新的情况下进行复制。|offset|length|style|CODE|1835432|我发现至少在较新版本的Visual+Studio中，例如VS2109，$(ProjDir)是未定义的，必须使用$(ProjectDir)。|1835433|如果在xcopy中省略了目标文件夹，则应默认将其放在输出目录中。这对于理解为什么$(OutDir)本身没有帮助是很重要的。|1835434|至少在Visual+Studio的最新版本中，$(OutDir)被定义为输出文件夹的相对路径，如bin/x86/Debug。单独使用它作为目标将从项目输出文件夹开始创建一组新的文件夹。例如：…+bin/x86/Debug/bin/x86/Debug。|1835435|将其与项目文件夹相结合，应该会让您找到合适的位置。例如：$(ProjectDir)$(OutDir)。|1835436|但是，$(TargetDir)将在一个步骤中提供输出目录。|1835437|Microsoft's+list+of+MSBuild+macros+for+current+and+previous+versions+of+Visual+Studio|1835438|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.microsoft.com/en-us/cpp/build/reference/common-macros-for-build-commands-and-properties^0|0|0|0|0|2|D|2|0|Z|A|1J|D|0|3|5|14|9|0|N|9|1C|D|2N|T|0|S|M|0|3|C|0|0|2D|0|0^^$0|@$1|2|3|4|5|6|7|18|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|19|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|1A|8|@]|9|@]|A|$B|C]]|$1|I|3|J|5|F|7|1B|8|@$K|1C|L|1D|M|N]|$K|1E|L|1F|M|N]]|9|@]|A|$]]|$1|O|3|P|5|F|7|1G|8|@$K|1H|L|1I|M|N]|$K|1J|L|1K|M|N]]|9|@]|A|$]]|$1|Q|3|R|5|F|7|1L|8|@$K|1M|L|1N|M|N]|$K|1O|L|1P|M|N]]|9|@]|A|$]]|$1|S|3|T|5|F|7|1Q|8|@$K|1R|L|1S|M|N]|$K|1T|L|1U|M|N]|$K|1V|L|1W|M|N]]|9|@]|A|$]]|$1|U|3|V|5|F|7|1X|8|@$K|1Y|L|1Z|M|N]]|9|@]|A|$]]|$1|W|3|X|5|F|7|20|8|@$K|21|L|22|M|N]]|9|@]|A|$]]|$1|Y|3|Z|5|F|7|23|8|@]|9|@$K|24|L|25|1|26]]|A|$]]|$1|10|3|-4|5|F|7|27|8|@]|9|@]|A|$]]]|11|$12|$5|13|14|15|A|$16|17]]]]

<pre><code>xcopy /y /d "$(ProjectDir)External\*.dll" "$(TargetDir)"
</code></pre>

You can also refer to a relative path, the next example will find the DLL in a folder located one level above the project folder. If you have multiple projects that use the DLL in a single solution, this places the source of the DLL in a common area reachable when you set any of them as the Startup Project.

<pre><code>xcopy /y /d "$(ProjectDir)..\External\*.dll" "$(TargetDir)"
</code></pre>

The <code>/y</code> option copies without confirmation.
The <code>/d</code> option checks to see if a file exists in the target and if it does only copies if the source has a newer timestamp than the target.

I found that in at least newer versions of Visual Studio, such as VS2109, <code>$(ProjDir)</code> is undefined and had to use <code>$(ProjectDir)</code> instead. 

Leaving out a target folder in <code>xcopy</code> should default to the output directory. That is important to understand reason <code>$(OutDir)</code> alone is not helpful.

<code>$(OutDir)</code>, at least in recent versions of Visual Studio, is defined as a relative path to the output folder, such as <code>bin/x86/Debug</code>. Using it alone as the target will create a new set of folders starting from the project output folder. Ex: <code>… bin/x86/Debug/bin/x86/Debug</code>.

Combining it with the project folder should get you to the proper place. Ex: <code>$(ProjectDir)$(OutDir)</code>.

However <code>$(TargetDir)</code> will provide the output directory in one step.

<a href="https://docs.microsoft.com/en-us/cpp/build/reference/common-macros-for-build-commands-and-properties" rel="nofollow noreferrer">Microsoft's list of MSBuild macros for current and previous versions of Visual Studio</a>

blocks|key|533619|text|我也有类似的问题。在我的项目中，有几个外部DLL。因此，我在项目中创建了一个名为"lib“的新文件夹，并将所有外部dll复制到此文件夹中。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|533620|533621|添加对这些DLL的引用。|ordered-list-item|533622|转到References>dll属性并更改以下属性DLL|533623|533624|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|L|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|M|8|@]|9|@]|A|$]]|$1|F|3|G|5|E|7|N|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|P|8|@]|9|@]|A|$]]]|J|$]]

I had a similar question. In my project, there were couple of external DLLs. So I created a new folder in the project called &quot;lib&quot; and copied all the external dlls to this folder.
<ol>
<li>Add a reference to these DLLs.</li>
<li>Go to Project References&gt;dll properties and change the following properties
<a href="https://i.stack.imgur.com/IBXka.png" rel="nofollow noreferrer">enter image description here</a></li>
</ol>

I have a Visual Studio C++ project that relies on an external DLL file. How can I make Visual Studio copy this DLL file automatically into the output directory (debug/release) when I build the project?

How to make Visual Studio copy a DLL file to the output directory?

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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我有一个依赖于外部C++文件的Visual Studio项目。如何使Visual Studio在生成项目时自动将此DLL文件复制到输出目录(debug/release)？

问如何让Visual Studio将DLL文件复制到输出目录？
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