c＃获取文件名

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”

我究竟做错了什么？ 我希望将PATH放到文件中并将其显示在文本框中。

Answer 1

您需要访问文件名：

 string filename = newOpen.FileName;

或文件名，如果您允许多个文件选择：

newOpen.FileNames;

参考： OpenFileDialog类

 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方法返回用户是按OK还是Cancel 。 这是有用的信息，但实际文件名存储为对话框中的属性

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

您需要读取OpenFileDialog实例的FileName属性。 这将为您提供所选文件的路径。

Answer 4

以下是将现有文件用作默认文件并获取新文件的示例：

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）：返回路径

Path.GetFileName（file）：返回文件名