string strtxtPath =@"d:\f\fd.mp3;
string strzipPath = @"d:\fd.rar";
System.Diagnostics.Process Process1 = new System.Diagnostics.Process();
Process1.StartInfo.FileName = @"C:\Program Files\winrar\WinRAR.exe";
Process1.StartInfo.CreateNoWindow = true;
//Process1.StartInfo.Arguments = "x -p123456 " + strzipPath + " " +strtxtPath ;
Process1.StartInfo.Arguments = "x - " + strzipPath + " " +strtxtPath ;
Process1.Start();
不成功
??
如果strtxtPath="d:\1\",这样就解压成功,问题是我想直接解压缩成一个指定路径,指定文件名。这样要怎么写参数啊》?》??
string strzipPath = @"d:\fd.rar";
System.Diagnostics.Process Process1 = new System.Diagnostics.Process();
Process1.StartInfo.FileName = @"C:\Program Files\winrar\WinRAR.exe";
Process1.StartInfo.CreateNoWindow = true;
//Process1.StartInfo.Arguments = "x -p123456 " + strzipPath + " " +strtxtPath ;
Process1.StartInfo.Arguments = "x - " + strzipPath + " " +strtxtPath ;
Process1.Start();
不成功
??
如果strtxtPath="d:\1\",这样就解压成功,问题是我想直接解压缩成一个指定路径,指定文件名。这样要怎么写参数啊》?》??
帮忙啊!