我正在用C#开发wpf应用程序。我能够运行以下exe文件
public static void GenerateCsvFile(string fileName)
{
System.Diagnostics.Process process = new System.Diagnostics.Process();
System.Diagnostics.ProcessStartInfo startInfo = new System.Diagnostics.ProcessStartInfo();
startInfo.WindowStyle = System.Diagnostics.ProcessWindowStyle.Hidden;
startInfo.FileName = @"C:\ndfd\degrib\bin\degrib.exe";
startInfo.Arguments = fileName + " -C -msg 1 -Csv";
process.StartInfo = startInfo;
process.Start();
System.Diagnostics.Process process1 = new System.Diagnostics.Process();
System.Diagnostics.ProcessStartInfo startInfo1 = new System.Diagnostics.ProcessStartInfo();
startInfo1.WindowStyle = System.Diagnostics.ProcessWindowStyle.Hidden;
startInfo1.FileName = @"C:\ndfd\degrib\bin\degrib.exe";
startInfo1.Arguments = fileName + " -C -msg all -nMet -Csv";
process1.StartInfo = startInfo1;
process1.Start();
}
上面的代码为我成功生成了csv文件。但.csv文件是根据fileName在不同位置生成的。意味着.csv文件每次都在不同的文件夹中生成。我可以强制exe在特定文件夹中生成.csv文件吗?能否请您提供我可以解决上述问题的任何代码或链接?
编辑:用户选择zip文件并对表单进行汇总。 App.ApplicationPath是硬编码路径。以下是我的代码
private void ShowPointsButton_Click(object sender, RoutedEventArgs e)
{
ZipHelper.UnZip(FileNameTextBox.Text, App.ApplicationPath, safeFileName, 9999999);
}
public static void UnZip(string SrcFile, string DstFile, string safeFileName, int bufferSize)
{
//ICSharpCode.SharpZipLib.Zip.UseZip64.Off;
FileStream fileStreamIn = new FileStream(SrcFile, FileMode.Open, FileAccess.Read);
ZipInputStream zipInStream = new ZipInputStream(fileStreamIn); ;
//if (SrcFile.Contains(".bz2"))
//{
//BZip2InputStream zipInStream = new BZip2InputStream(fileStreamIn);
//}
//else
//{
// zipInStream = new ZipInputStream(fileStreamIn);
//}
string rootDirectory = string.Empty;
if (safeFileName.Contains(".zip"))
{
rootDirectory = safeFileName.Replace(".zip", string.Empty);
}
else
{
rootDirectory = safeFileName;
}
Directory.CreateDirectory(App.ApplicationPath + rootDirectory);
while (true)
{
ZipEntry entry = zipInStream.GetNextEntry();
if (entry == null)
break;
if (entry.Name.Contains("/"))
{
string[] folders = entry.Name.Split('/');
string lastElement = folders[folders.Length - 1];
var folderList = new List<string>(folders);
folderList.RemoveAt(folders.Length - 1);
folders = folderList.ToArray();
//string folderPath = "";
//foreach (string str in folders)
//{
//folderPath = folderPath + @"\" + str;
//string blackslash = folderPath.Substring(0, 1);
//if (blackslash == "\\")
//{
// folderPath = folderPath.Remove(0, 1);
//}
//if (!Directory.Exists(App.ApplicationPath + rootDirectory + "/" + folderPath))
//{
// Directory.CreateDirectory(App.ApplicationPath + rootDirectory + "/" + folderPath);
//}
//}
if (!string.IsNullOrEmpty(lastElement))
{
//folderPath = folderPath + @"\" + lastElement;
//string blackslash = folderPath.Substring(0, 1);
//if (blackslash == "\\")
//{
// folderPath = folderPath.Remove(0, 1);
//}
WriteToFile(DstFile + rootDirectory + @"\" + lastElement, bufferSize, zipInStream, rootDirectory, entry);
}
}
else
{
WriteToFile(DstFile + rootDirectory + @"\" + entry.Name, bufferSize, zipInStream, rootDirectory, entry);
}
}
zipInStream.Close();
fileStreamIn.Close();
}
private static void WriteToFile(string DstFile, int bufferSize, ZipInputStream zipInStream, string rootDirectory, ZipEntry entry)
{
WriteFileContents(DstFile, bufferSize, zipInStream);
if (DstFile.Contains(".grb"))
{
Utility.GenerateCsvFile(DstFile);
}
//if(DstFile.Contains(".csv"))
//{
// WriteFileContents(@"D:\Documents" + rootDirectory + @"\" + entry.Name, bufferSize, zipInStream);
//}
}
private static void WriteFileContents(string DstFile, int bufferSize, ZipInputStream zipInStream)
{
FileStream fileStreamOut = new FileStream(DstFile, FileMode.OpenOrCreate, FileAccess.Write);
int size;
byte[] buffer = new byte[bufferSize];
do
{
size = zipInStream.Read(buffer, 0, buffer.Length);
fileStreamOut.Write(buffer, 0, size);
} while (size > 0);
fileStreamOut.Close();
}
在上面的代码中,参见Utility.GenerateCsvFile(DstFile);我想在位置'DstFile'生成.csv文件。简而言之,我解压缩文件的文件夹,在我想要.exev文件的同一文件夹中。例如,考虑有D:/ XYZ文件夹,我在其中解压缩我的zip文件。在此文件夹中有test.grib文件。我想为test.grib运行exe并生成.csv文件。我希望将这些.csv文件写入XYZ文件夹。
答案 0 :(得分:1)
processStartInfo.WorkingDirectory = @“你的目录”;
请试试这个。
答案 1 :(得分:0)
文件名需要指向正确的目录和文件,例如c:\ somedir \ mycsvfile.csv
答案 2 :(得分:0)
string folder = "c:\temp";
string fileName = "c:\somedir\blah\file.csv";
string outputFilePath = Path.Combine(folder, new FileInfo(fileName).Name);
那应该为您提供c:\ temp \ file.csv的文件路径。这是你追求的那种东西吗?