Question

大家好，首先抱歉我的英语，但我希望你能理解我... 我在VS2008-Smart Device Project-WinCE 5.0 Project上做了一个项目。我需要创建一个文本文件到DATA文件夹，它位于程序的主目录下。

有我的代码，没有错误信息，但它没有创建文本文件。我的目录总是返回null。该代码有什么问题？

if (Form2.dosya_adi != null)
{
    string cfile = Form2.chosenfile;
    path = Path.GetDirectoryName(Assembly.GetExecutingAssembly().GetName().CodeBase) + "\\DATA\\" + cfile+ ".txt";

}
else 
{
    path = Path.GetDirectoryName(Assembly.GetExecutingAssembly().GetName().CodeBase)+"\\DATA";
}
try
{

    StreamReader Read_File= File.OpenText(path);//Dosyayı açmaya çalış olmaz ise catch bloğuna geç
    ReadFile.Close();
}
catch
{

    StreamWriter Write_File= File.CreateText(path+ i.ToString()+".txt");// yeni dosya oluştur.
    Write_File.Close();
}`

并且包含listbox和listbox的form2显示了是否存在文件的目录..

string path = Path.GetDirectoryName(Assembly.GetExecutingAssembly().GetName().CodeBase);
path = path + "\\DATA";
DirectoryInfo di = new DirectoryInfo(path);
FileInfo[] rgFiles = di.GetFiles();
foreach (FileInfo fi in rgFiles)
{
    listBox1.Items.Add(fi.Name);
}


private void button1_Click(object sender, EventArgs e)
{
    if (listBox1.SelectedItem != null)
    {
        chosen_file = listBox1.GetItemText(listBox1.SelectedItem);
        Form1 form1 = new Form1();
        form1.Show();
        this.Hide();
    }
    else 
    {
        MessageBox.Show("HATA:Hiçbir Değer Seçilmedi!"); // That means error:no value chosen!
    }
}

Answer 1

您的代码未正确准备path变量。在一种情况下，它包含文件名，在另一种情况下，它只包含路径名：

if (Form2.dosya_adi != null)
{
    string cfile = Form2.chosenfile;
    path = Path.GetDirectoryName(Assembly.GetExecutingAssembly().GetName().CodeBase) + "\\DATA\\" + cfile+ ".txt";

}
else 
{
    path = Path.GetDirectoryName(Assembly.GetExecutingAssembly().GetName().CodeBase)+"\\DATA";
}

然后尝试打开文件 - 即使变量首先不包含文件名。如果这导致异常，则尝试创建文件，但实际上，您并未向路径添加路径分隔符。所以我认为你的catch块应该是：

StreamWriter Write_File= File.CreateText(path + "\\" + i.ToString() + ".txt");// yeni dosya oluştur.
Write_File.Close();

WinCE 5.0如何获取目录

1 个答案: