Question

您好我想使用对话框表单选择文本文件，而不是必须使用给定的路径。我该怎么做？

我想用opendialog替换opentext？我已经尝试但是我得到了流的错误我想使用streamreader ....

    private void button2_Click(object sender, EventArgs e)
    {

        using (StreamReader reader = File.OpenText("c:\\myparts.txt"))
        {
            label3.Text = "Ready to Insert";
            textBox7.Text = reader.ReadLine();
            textBox8.Text = reader.ReadLine();
            textBox9.Text = reader.ReadLine();
            textBox10.Text = reader.ReadLine();
}

Answer 1

你想要这样的东西吗？

OpenFileDialog dlg = new OpenFileDialog();

if (dlg.ShowDialog() == DialogResult.OK)
{
    using (var reader = File.OpenText(dlg.FileName))
    {
        ...
    }
}

Answer 2

private void button2_Click(object sender, EventArgs e)
{
        string fileToOpen = "";

        OpenFileDialog dialog = new OpenFileDialog();
        dialog.Title = "Browse for file...";
        dialog.RestoreDirectory = true;

        DialogResult result = dialog.ShowDialog();

        if (result == DialogResult.OK)
        {
            fileToOpen = dialog.FileName;
        }

        using (StreamReader reader = File.OpenText(fileToOpen))
        {
            label3.Text = "Ready to Insert";
            textBox7.Text = reader.ReadLine();
            textBox8.Text = reader.ReadLine();
            textBox9.Text = reader.ReadLine();
            textBox10.Text = reader.ReadLine();
        }
}

Answer 3

我想用 opendialog 替换opentext？我试过但是我得到了流错误我想使用streamreader ....

解决方案1：您可以将Stream返回的openFileDialog.OpenFile()分配给StreamReader

试试这个：

     if (openFileDialog1.ShowDialog() == DialogResult.OK)
        {
            using (var reader = new StreamReader(openFileDialog1.OpenFile()))
            {
                label3.Text = "Ready to Insert";
                textBox7.Text = reader.ReadLine();
                textBox8.Text = reader.ReadLine();
                textBox9.Text = reader.ReadLine();
                textBox10.Text = reader.ReadLine();
            }
        }

解决方案2 ：您可以直接将openFileDialog().FileName属性Path参数分配给File.OpenText()方法，如下所示：

      if (openFileDialog1.ShowDialog() == DialogResult.OK)
        {
            using (var reader = new StreamReader(openFileDialog1.OpenText(openFileDialog1.FileName)))
            {
                label3.Text = "Ready to Insert";
                textBox7.Text = reader.ReadLine();
                textBox8.Text = reader.ReadLine();
                textBox9.Text = reader.ReadLine();
                textBox10.Text = reader.ReadLine();
            }
        }

解决方案3：如果您想将文件内容分配给多个文本框

试试这个：

int startCount=7;
int endCount=10;
string preText="textBox";
if (openFileDialog1.ShowDialog() == DialogResult.OK)
{
   String fileName=openFileDialog1.FileName;
   foreach(var line in File.ReadLines(fileName))
   {
     ((TextBox) (this.Controls.Find(preText+startCount,true)[0])).Text=line;
     if(startCount==endCount)
       break;

       startCount++;
    }
}

注1：所有TextBoxControl都应以preText值启动。
注2：在上述解决方案中，您可以根据自己的要求更改startCount和endCount。

例如，如果要将文件contenet分配到从textBox3到textBox23的20个Textbox控件，则需要更改上述代码中的参数，如下所示：

preText="textBox";
startCount = 3;
endCount = 23;

将opentext更改为opendialog

3 个答案: