如何传递目录路径作为进程的命令行？ [英] How to pass the Directory path as command line for the process?

问题描述

iam使用regasm.exe来生成tlb文件并注册程序集。但是tlb在.NET根目录本身的路径。所以像这样

缓冲区包含c：\windows \Microsoft.Net\framework\v2.0.57\RegAsm.exe

  if（！CreateProcessW（buffer，LC：\\Program Files\\Test\\Test.dll / codebase / tlb / silent，NULL，NULL，FALSE，0，NULL，NULL，（LPSTARTUPINFOW）& pi，& pi））
  

但我认为它不会采取完整的路径，因为有一个空白b / w Program和Files.as期望当我运行命令，它也显示无法找到输入程序集c：\program。

通常在命令提示符下，我们可以给出

RegAsm.exe c：\program file \Test\test.dll/ codebase / tlb
这个粗体字符我必须作为命令行传递，但它有双引号的双quptes。

解决方案

您需要在字符串中用引号将路径括起来，您可以通过使用反斜杠转义引号字符来执行此操作。因此，CreateProcessW的第二个参数是：

  L\C：\Program Files\Test\ Test.dll\/ codebase / tlb / silent
  

您将在命令提示符处使用的引号。

iam using regasm.exe to generate tlb file and register the assembly programatically.But the path of tlb in .NET root directory itself. so do like this

buffer contains c:\windows\Microsoft.Net\framework\v2.0.57\RegAsm.exe

if(!CreateProcessW(buffer,L" C:\\Program Files\\Test\\Test.dll  /codebase /tlb /silent" ,NULL, NULL,FALSE, 0,NULL,NULL,(LPSTARTUPINFOW)&si,&pi ) )

But i think it wont take full path since there is a blank b/w Program and Files.as expected the when i run the command it also shows unable to locate input assembly c:\program.

normally at command prompt we can give as

RegAsm.exe "c:\program files\Test\test.dll" /codebase /tlb this bold characters i have to pass as command line but it have Double quptes with in double quotes. so i was strucked.

How can i fix it

解决方案

You need to enclose the path in quotes within the string, which you can do by escaping the quote character by preceding it with a backslash. So, your 2nd parameter to CreateProcessW would be:

L"\"C:\Program Files\Test\Test.dll\" /codebase /tlb /silent"

This would give you the command line with quotes that you would use at the command prompt.

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