C#进程.开始文件名和传递参数 [英] C# process.Start filename and passing arguments

问题描述

我需要打开文件test.mdb. 路径必须是从相对于C#程序exe所在的任何目录构建的fullpath.

I need to open the file test.mdb. The path must be fullpath built from whatever directory it is located in relative to the C# program exe

我需要传递Arguments行，该行将像这样/cmd"MyArgument"一样批量传递 为了使/cmd开关起作用，该路径必须包含已安装的MSACCESS.EXE的完整路径.

I need to pass the Arguments line which in batch would be passed like this /cmd "MyArgument" For the /cmd switch to work the path must include the full path to the installed MSACCESS.EXE

我只是不知道如何将完整路径传递给MSACCESS> EXE，然后将完整路径传递给test.mdb，然后是参数.

I just can't work out how to pass the full path to MSACCESS>EXE followed by the fullpath to test.mdb followed by the Arguments.

要尝试提供帮助，我将发布以下有效的批处理字符串，但我需要C#

To try to help I'm posting the following batch string which works but I need C#

"C:\ Program Files \ Microsoft Office \ Office10 \ MSACCESS.EXE""C:\ Documents and Settings \ User \ Test示例Folder \ test.mdb"/cmd"MyArgument"

"C:\Program Files\Microsoft Office\Office10\MSACCESS.EXE" "C:\Documents and Settings\User\Test Examples Folder\test.mdb" /cmd "MyArgument"

要阐明两个重要点:

test.mdb全路径必须由启动的C#exe自动确定，它将是C#exes路径，但带有test.mdb.

The test.mdb fullpath must be automatically determined by the launching C# exe it will be the C# exes path but with test.mdb.

如果C#exe程序可以自动检查哪个版本(例如Office 10，Office 12等)并使用它，那么MSACCESS.EXE路径必须是MSACCESS.EXE安装版本的完整路径.

And the MSACCESS>EXE path must be the full path to the installed version of MSACCESS.EXE if the C# exe program can automatically check which version i.e. Office 10, Office 12 etc and use that this would be excellent.

到目前为止，这是我的代码:

This is my code so far:

var filePath = @"test.mdb";

Process process = new Process();
process.StartInfo.FileName = filePath;
process.StartInfo.Arguments = "/cmd " + "\"MyArgument\"";
process.StartInfo.WindowStyle = ProcessWindowStyle.Hidden;
process.Start();

推荐答案

我认为您不必担心检测Office版本，只是这样对您有用:

I don't think you need to worry about detecting the office version, just this should work for you:

        string filepath = '"' + Directory.GetCurrentDirectory() + "\\test.mdb" + '"';
        string acc_cmd_arg = "HELLO";

        using (System.Diagnostics.Process process = new System.Diagnostics.Process() )  {
            process.StartInfo.FileName = "msaccess.exe";
            process.StartInfo.Arguments = filepath + " /cmd " + acc_cmd_arg;
            process.Start(); 
        }

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