How do I set the working directory for a Winforms project using dollar sign macros?












2















I am working on a C# app, the contents of it I guess do not matter. Basically, I often use Save/OpenFileDialogs in it. And I use an initial dir property, to make it more convenient, and I set it to the current working directory.



When I have a release build, I pack my application and have all the dirs inside, so it works well. But when I actually debug it in Visual Studio I want to set the working directory which is different from bin/Debug. And so I did. I wrote a literal path, like C:SomethingblahxyzworkingDir. And it works. The problem is I have two machines that I am working on, and this path won't work on the other machine.



I know about the macros like $(ProjectDir) in Visual Studio, so I wanted to use them. Unfortunately when I type i.e. $(ProjectDir) in here:



enter image description here



I get this error:



enter image description here



Do you guys know what may be the issue? I thought this is actually a valid macro, according to the MSDN










share|improve this question




















  • 1





    @dabljues Oh. $(ProjectDir) is definitely a valid macro, but it may not be for that specific field. Also, make sure that directory really exists (I know, sounds obvious). I'll make that my answer and will delete it if wrong.

    – CRAGIN
    Nov 21 '18 at 22:26













  • Oh, right, forgot to add. It is a Windows Forms application. The solution contains two projects (but I guess that doesn't matter since I am setting this for only one of them).

    – dabljues
    Nov 21 '18 at 22:30
















2















I am working on a C# app, the contents of it I guess do not matter. Basically, I often use Save/OpenFileDialogs in it. And I use an initial dir property, to make it more convenient, and I set it to the current working directory.



When I have a release build, I pack my application and have all the dirs inside, so it works well. But when I actually debug it in Visual Studio I want to set the working directory which is different from bin/Debug. And so I did. I wrote a literal path, like C:SomethingblahxyzworkingDir. And it works. The problem is I have two machines that I am working on, and this path won't work on the other machine.



I know about the macros like $(ProjectDir) in Visual Studio, so I wanted to use them. Unfortunately when I type i.e. $(ProjectDir) in here:



enter image description here



I get this error:



enter image description here



Do you guys know what may be the issue? I thought this is actually a valid macro, according to the MSDN










share|improve this question




















  • 1





    @dabljues Oh. $(ProjectDir) is definitely a valid macro, but it may not be for that specific field. Also, make sure that directory really exists (I know, sounds obvious). I'll make that my answer and will delete it if wrong.

    – CRAGIN
    Nov 21 '18 at 22:26













  • Oh, right, forgot to add. It is a Windows Forms application. The solution contains two projects (but I guess that doesn't matter since I am setting this for only one of them).

    – dabljues
    Nov 21 '18 at 22:30














2












2








2








I am working on a C# app, the contents of it I guess do not matter. Basically, I often use Save/OpenFileDialogs in it. And I use an initial dir property, to make it more convenient, and I set it to the current working directory.



When I have a release build, I pack my application and have all the dirs inside, so it works well. But when I actually debug it in Visual Studio I want to set the working directory which is different from bin/Debug. And so I did. I wrote a literal path, like C:SomethingblahxyzworkingDir. And it works. The problem is I have two machines that I am working on, and this path won't work on the other machine.



I know about the macros like $(ProjectDir) in Visual Studio, so I wanted to use them. Unfortunately when I type i.e. $(ProjectDir) in here:



enter image description here



I get this error:



enter image description here



Do you guys know what may be the issue? I thought this is actually a valid macro, according to the MSDN










share|improve this question
















I am working on a C# app, the contents of it I guess do not matter. Basically, I often use Save/OpenFileDialogs in it. And I use an initial dir property, to make it more convenient, and I set it to the current working directory.



When I have a release build, I pack my application and have all the dirs inside, so it works well. But when I actually debug it in Visual Studio I want to set the working directory which is different from bin/Debug. And so I did. I wrote a literal path, like C:SomethingblahxyzworkingDir. And it works. The problem is I have two machines that I am working on, and this path won't work on the other machine.



I know about the macros like $(ProjectDir) in Visual Studio, so I wanted to use them. Unfortunately when I type i.e. $(ProjectDir) in here:



enter image description here



I get this error:



enter image description here



Do you guys know what may be the issue? I thought this is actually a valid macro, according to the MSDN







visual-studio visual-studio-2017






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 21 '18 at 22:58









jonsca

8,684115058




8,684115058










asked Nov 21 '18 at 21:53









dabljuesdabljues

1827




1827








  • 1





    @dabljues Oh. $(ProjectDir) is definitely a valid macro, but it may not be for that specific field. Also, make sure that directory really exists (I know, sounds obvious). I'll make that my answer and will delete it if wrong.

    – CRAGIN
    Nov 21 '18 at 22:26













  • Oh, right, forgot to add. It is a Windows Forms application. The solution contains two projects (but I guess that doesn't matter since I am setting this for only one of them).

    – dabljues
    Nov 21 '18 at 22:30














  • 1





    @dabljues Oh. $(ProjectDir) is definitely a valid macro, but it may not be for that specific field. Also, make sure that directory really exists (I know, sounds obvious). I'll make that my answer and will delete it if wrong.

    – CRAGIN
    Nov 21 '18 at 22:26













  • Oh, right, forgot to add. It is a Windows Forms application. The solution contains two projects (but I guess that doesn't matter since I am setting this for only one of them).

    – dabljues
    Nov 21 '18 at 22:30








1




1





@dabljues Oh. $(ProjectDir) is definitely a valid macro, but it may not be for that specific field. Also, make sure that directory really exists (I know, sounds obvious). I'll make that my answer and will delete it if wrong.

– CRAGIN
Nov 21 '18 at 22:26







@dabljues Oh. $(ProjectDir) is definitely a valid macro, but it may not be for that specific field. Also, make sure that directory really exists (I know, sounds obvious). I'll make that my answer and will delete it if wrong.

– CRAGIN
Nov 21 '18 at 22:26















Oh, right, forgot to add. It is a Windows Forms application. The solution contains two projects (but I guess that doesn't matter since I am setting this for only one of them).

– dabljues
Nov 21 '18 at 22:30





Oh, right, forgot to add. It is a Windows Forms application. The solution contains two projects (but I guess that doesn't matter since I am setting this for only one of them).

– dabljues
Nov 21 '18 at 22:30












2 Answers
2






active

oldest

votes


















1














It doesn't appear that you can apply this macro in that dialog. Leave the value blank and go to the directory for the solution, drill down into the folder for the project and you should see [ProjectName].csproj



Add the following StartWorkingDirectory setting to to the PropertyGroup for the Debug config.



The property group is:



<PropertyGroup Condition="'$(Configuration)|$(Platform)' == 'Debug|AnyCPU'">


And add:



<StartWorkingDirectory>$(ProjectDir)</StartWorkingDirectory>


So it ends up looking like:



<PropertyGroup Condition=" '$(Configuration)|$(Platform)' == 'Debug|AnyCPU' ">
<PlatformTarget>AnyCPU</PlatformTarget>
<DebugSymbols>true</DebugSymbols>
<DebugType>full</DebugType>
<Optimize>false</Optimize>
<OutputPath>binDebug</OutputPath>
<StartWorkingDirectory>$(ProjectDir)</StartWorkingDirectory>
<DefineConstants>DEBUG;TRACE</DefineConstants>
<ErrorReport>prompt</ErrorReport>
<WarningLevel>4</WarningLevel>
</PropertyGroup>


The value still won't show up in the dialog, but if you use the working directory in your program, the value will be correct.






share|improve this answer

































    2














    You cannot use that (valid) macro for that field as Visual Studio does not apply macros there. What makes me believe that is that it seems to be interpreting what you entered literally.



    You might be better off by doing something in the code itself to use the project directory. If so, you may want to check to see if the debugger is attached, at least for the purposes of development and "convenience" (e.g. Debugger.IsAttached) and then get the CodeBase property of the executing assembly or the CurrentDirectory. That gives you the convenience you're looking for while debugging. For production, of course, you might want a different strategy.






    share|improve this answer


























    • As I said in the comment of @jonsca answer - for this particular case it's an overkill for me. Thank you for the input, maybe VS team will change that some time

      – dabljues
      Nov 21 '18 at 22:42











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    2 Answers
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    2 Answers
    2






    active

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    active

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    1














    It doesn't appear that you can apply this macro in that dialog. Leave the value blank and go to the directory for the solution, drill down into the folder for the project and you should see [ProjectName].csproj



    Add the following StartWorkingDirectory setting to to the PropertyGroup for the Debug config.



    The property group is:



    <PropertyGroup Condition="'$(Configuration)|$(Platform)' == 'Debug|AnyCPU'">


    And add:



    <StartWorkingDirectory>$(ProjectDir)</StartWorkingDirectory>


    So it ends up looking like:



    <PropertyGroup Condition=" '$(Configuration)|$(Platform)' == 'Debug|AnyCPU' ">
    <PlatformTarget>AnyCPU</PlatformTarget>
    <DebugSymbols>true</DebugSymbols>
    <DebugType>full</DebugType>
    <Optimize>false</Optimize>
    <OutputPath>binDebug</OutputPath>
    <StartWorkingDirectory>$(ProjectDir)</StartWorkingDirectory>
    <DefineConstants>DEBUG;TRACE</DefineConstants>
    <ErrorReport>prompt</ErrorReport>
    <WarningLevel>4</WarningLevel>
    </PropertyGroup>


    The value still won't show up in the dialog, but if you use the working directory in your program, the value will be correct.






    share|improve this answer






























      1














      It doesn't appear that you can apply this macro in that dialog. Leave the value blank and go to the directory for the solution, drill down into the folder for the project and you should see [ProjectName].csproj



      Add the following StartWorkingDirectory setting to to the PropertyGroup for the Debug config.



      The property group is:



      <PropertyGroup Condition="'$(Configuration)|$(Platform)' == 'Debug|AnyCPU'">


      And add:



      <StartWorkingDirectory>$(ProjectDir)</StartWorkingDirectory>


      So it ends up looking like:



      <PropertyGroup Condition=" '$(Configuration)|$(Platform)' == 'Debug|AnyCPU' ">
      <PlatformTarget>AnyCPU</PlatformTarget>
      <DebugSymbols>true</DebugSymbols>
      <DebugType>full</DebugType>
      <Optimize>false</Optimize>
      <OutputPath>binDebug</OutputPath>
      <StartWorkingDirectory>$(ProjectDir)</StartWorkingDirectory>
      <DefineConstants>DEBUG;TRACE</DefineConstants>
      <ErrorReport>prompt</ErrorReport>
      <WarningLevel>4</WarningLevel>
      </PropertyGroup>


      The value still won't show up in the dialog, but if you use the working directory in your program, the value will be correct.






      share|improve this answer




























        1












        1








        1







        It doesn't appear that you can apply this macro in that dialog. Leave the value blank and go to the directory for the solution, drill down into the folder for the project and you should see [ProjectName].csproj



        Add the following StartWorkingDirectory setting to to the PropertyGroup for the Debug config.



        The property group is:



        <PropertyGroup Condition="'$(Configuration)|$(Platform)' == 'Debug|AnyCPU'">


        And add:



        <StartWorkingDirectory>$(ProjectDir)</StartWorkingDirectory>


        So it ends up looking like:



        <PropertyGroup Condition=" '$(Configuration)|$(Platform)' == 'Debug|AnyCPU' ">
        <PlatformTarget>AnyCPU</PlatformTarget>
        <DebugSymbols>true</DebugSymbols>
        <DebugType>full</DebugType>
        <Optimize>false</Optimize>
        <OutputPath>binDebug</OutputPath>
        <StartWorkingDirectory>$(ProjectDir)</StartWorkingDirectory>
        <DefineConstants>DEBUG;TRACE</DefineConstants>
        <ErrorReport>prompt</ErrorReport>
        <WarningLevel>4</WarningLevel>
        </PropertyGroup>


        The value still won't show up in the dialog, but if you use the working directory in your program, the value will be correct.






        share|improve this answer















        It doesn't appear that you can apply this macro in that dialog. Leave the value blank and go to the directory for the solution, drill down into the folder for the project and you should see [ProjectName].csproj



        Add the following StartWorkingDirectory setting to to the PropertyGroup for the Debug config.



        The property group is:



        <PropertyGroup Condition="'$(Configuration)|$(Platform)' == 'Debug|AnyCPU'">


        And add:



        <StartWorkingDirectory>$(ProjectDir)</StartWorkingDirectory>


        So it ends up looking like:



        <PropertyGroup Condition=" '$(Configuration)|$(Platform)' == 'Debug|AnyCPU' ">
        <PlatformTarget>AnyCPU</PlatformTarget>
        <DebugSymbols>true</DebugSymbols>
        <DebugType>full</DebugType>
        <Optimize>false</Optimize>
        <OutputPath>binDebug</OutputPath>
        <StartWorkingDirectory>$(ProjectDir)</StartWorkingDirectory>
        <DefineConstants>DEBUG;TRACE</DefineConstants>
        <ErrorReport>prompt</ErrorReport>
        <WarningLevel>4</WarningLevel>
        </PropertyGroup>


        The value still won't show up in the dialog, but if you use the working directory in your program, the value will be correct.







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Nov 21 '18 at 23:02

























        answered Nov 21 '18 at 22:35









        jonscajonsca

        8,684115058




        8,684115058

























            2














            You cannot use that (valid) macro for that field as Visual Studio does not apply macros there. What makes me believe that is that it seems to be interpreting what you entered literally.



            You might be better off by doing something in the code itself to use the project directory. If so, you may want to check to see if the debugger is attached, at least for the purposes of development and "convenience" (e.g. Debugger.IsAttached) and then get the CodeBase property of the executing assembly or the CurrentDirectory. That gives you the convenience you're looking for while debugging. For production, of course, you might want a different strategy.






            share|improve this answer


























            • As I said in the comment of @jonsca answer - for this particular case it's an overkill for me. Thank you for the input, maybe VS team will change that some time

              – dabljues
              Nov 21 '18 at 22:42
















            2














            You cannot use that (valid) macro for that field as Visual Studio does not apply macros there. What makes me believe that is that it seems to be interpreting what you entered literally.



            You might be better off by doing something in the code itself to use the project directory. If so, you may want to check to see if the debugger is attached, at least for the purposes of development and "convenience" (e.g. Debugger.IsAttached) and then get the CodeBase property of the executing assembly or the CurrentDirectory. That gives you the convenience you're looking for while debugging. For production, of course, you might want a different strategy.






            share|improve this answer


























            • As I said in the comment of @jonsca answer - for this particular case it's an overkill for me. Thank you for the input, maybe VS team will change that some time

              – dabljues
              Nov 21 '18 at 22:42














            2












            2








            2







            You cannot use that (valid) macro for that field as Visual Studio does not apply macros there. What makes me believe that is that it seems to be interpreting what you entered literally.



            You might be better off by doing something in the code itself to use the project directory. If so, you may want to check to see if the debugger is attached, at least for the purposes of development and "convenience" (e.g. Debugger.IsAttached) and then get the CodeBase property of the executing assembly or the CurrentDirectory. That gives you the convenience you're looking for while debugging. For production, of course, you might want a different strategy.






            share|improve this answer















            You cannot use that (valid) macro for that field as Visual Studio does not apply macros there. What makes me believe that is that it seems to be interpreting what you entered literally.



            You might be better off by doing something in the code itself to use the project directory. If so, you may want to check to see if the debugger is attached, at least for the purposes of development and "convenience" (e.g. Debugger.IsAttached) and then get the CodeBase property of the executing assembly or the CurrentDirectory. That gives you the convenience you're looking for while debugging. For production, of course, you might want a different strategy.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Nov 21 '18 at 22:40

























            answered Nov 21 '18 at 22:35









            CRAGINCRAGIN

            9,65323271




            9,65323271













            • As I said in the comment of @jonsca answer - for this particular case it's an overkill for me. Thank you for the input, maybe VS team will change that some time

              – dabljues
              Nov 21 '18 at 22:42



















            • As I said in the comment of @jonsca answer - for this particular case it's an overkill for me. Thank you for the input, maybe VS team will change that some time

              – dabljues
              Nov 21 '18 at 22:42

















            As I said in the comment of @jonsca answer - for this particular case it's an overkill for me. Thank you for the input, maybe VS team will change that some time

            – dabljues
            Nov 21 '18 at 22:42





            As I said in the comment of @jonsca answer - for this particular case it's an overkill for me. Thank you for the input, maybe VS team will change that some time

            – dabljues
            Nov 21 '18 at 22:42


















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