1

我有一个调用MSBuild和构建三个 Visual Studio 解决方案的批处理文件：

call "C:\Program Files (x86)\Microsoft Visual Studio 10.0\VC\vcvarsall.bat"

MSBuild solution0.sln  /property:Configuration=Release
MSBuild solution1.sln  /property:Configuration=Release  
MSBuild solution2.sln  /property:Configuration=Release

这工作正常。但是，我想提示用户选择要构建的程序版本。根据用户输入，我们构建了一组特定的解决方案。

我的问题是，如果我PATH在调用后修改变量vcvarsall，我将无法再调用MSBuild。

call "C:\Program Files (x86)\Microsoft Visual Studio 10.0\VC\vcvarsall.bat"
set /P version="Enter the version number [1] or [2]:"

IF %version% == 1 (
    set PATH="%PATH%;%PATH_TO_VERSION1_LIBS%"
    MSBuild solution0_v1.sln  /property:Configuration=Release
    MSBuild solution1_v1.sln  /property:Configuration=Release   
    MSBuild solution2_v1.sln  /property:Configuration=Release
)

IF %version% == 2 (
    set PATH="%PATH%;%PATH_TO_VERSION2_LIBS%"
    MSBuild solution0_v2.sln  /property:Configuration=Release
    MSBuild solution1_v2.sln  /property:Configuration=Release   
    MSBuild solution2_v2.sln  /property:Configuration=Release
)

我收到以下错误：

Setting environment for using Microsoft Visual Studio 2010 x86 tools.
ERROR: Cannot determine the location of the VS Common Tools folder.

这很令人费解，因为定义了环境变量VS100COMNTOOLS。

4

1 回答 1

3

这可能是解析错误，请尝试：

IF "%version%"=="1" set "PATH=%PATH%;%PATH_TO_VERSION1_LIBS%"
IF "%version%"=="1" (
    MSBuild solution0_v1.sln  /property:Configuration=Release
    MSBuild solution1_v1.sln  /property:Configuration=Release   
    MSBuild solution2_v1.sln  /property:Configuration=Release
)

IF "%version%"=="2" set "PATH=%PATH%;%PATH_TO_VERSION2_LIBS%"
IF "%version%"=="2" (
    MSBuild solution0_v2.sln  /property:Configuration=Release
    MSBuild solution1_v2.sln  /property:Configuration=Release   
    MSBuild solution2_v2.sln  /property:Configuration=Release
)
于 2013-07-05T15:35:29.393 回答