c＃獲取文件名

Question

private void button1_Click(object sender, EventArgs e)
    {


        OpenFileDialog newOpen = new OpenFileDialog();

       DialogResult result = newOpen.ShowDialog();

       this.textBox1.Text = result + "";

    }

它只是返回“OK”

我究竟做錯了什么？ 我希望將PATH放到文件中並將其顯示在文本框中。

Answer 1

您需要訪問文件名：

 string filename = newOpen.FileName;

或文件名，如果您允許多個文件選擇：

newOpen.FileNames;

參考： OpenFileDialog類

 private void button1_Click(object sender, System.EventArgs e) { Stream myStream = null; OpenFileDialog openFileDialog1 = new OpenFileDialog(); openFileDialog1.InitialDirectory = "c:\\\\" ; openFileDialog1.Filter = "txt files (*.txt)|*.txt|All files (*.*)|*.*" ; openFileDialog1.FilterIndex = 2 ; openFileDialog1.RestoreDirectory = true ; if(openFileDialog1.ShowDialog() == DialogResult.OK) { try { if ((myStream = openFileDialog1.OpenFile()) != null) { using (myStream) { // Insert code to read the stream here. } } } catch (Exception ex) { MessageBox.Show("Error: Could not read file. Error: " + ex.Message); } } } 

Answer 2

ShowDialog方法返回用戶是按OK還是Cancel 。 這是有用的信息，但實際文件名存儲為對話框中的屬性

private void button1_Click(object sender, EventArgs e)
    {
        OpenFileDialog newOpen = new OpenFileDialog();
        DialogResult result = newOpen.ShowDialog();
        if(result == DialogResult.OK) {
              this.textBox1.Text = newOpen.FileName;
        }
    }

Answer 3

您需要讀取OpenFileDialog實例的FileName屬性。 這將為您提供所選文件的路徑。

Answer 4

以下是將現有文件用作默認文件並獲取新文件的示例：

private string open(string oldFile)
    {
        OpenFileDialog newOpen = new OpenFileDialog();
        if (!string.IsNullOrEmpty(oldFile))
        {
          newOpen.InitialDirectory = Path.GetDirectoryName(oldFile);
          newOpen.FileName = Path.GetFileName(oldFile);
        }

        newOpen.Filter = "eXtensible Markup Language File  (*.xml) |*.xml"; //Optional filter
        DialogResult result = newOpen.ShowDialog();
        if(result == DialogResult.OK) {
              return newOpen.FileName;
        }

        return string.Empty;
    }

Path.GetDirectoryName（file）：返回路徑

Path.GetFileName（file）：返回文件名