Вопрос

I am trying to open a file which a user can set. In other words it will never be a set path or file. so when the user has selected the file they want to open this button below will open it. I have declared l1 and p1 as public strings.

    public void button4_Click(object sender, EventArgs e)
    {

         DialogResult result = openFileDialog1.ShowDialog();
         if (result == DialogResult.OK)
         {

             l1 = Path.GetFileName(openFileDialog1.FileName);
             p1 = System.IO.Path.GetDirectoryName(openFileDialog1.FileName);

         }


    public void button2_Click(object sender, EventArgs e)
    {
    //p1 = directory path for example "C:\\documents and settings\documents"
    //l1 = filename

        Process.Start(p1+l1);
    }

So just to review I want to open up the file just using the directory path and filename. Is this possible? I can just have p1 there and it will open an explorer showing me that directory. Thank you for looking.

Это было полезно?

Решение

Yes it will work, but I would recommend you update your code this instead:

var path = Path.Combine(p1, l1);
Process.Start(path);

Другие советы

You shouldn't use string concatenation to combine a directory and filename. In your case the resulting string will look like this:

C:\documents and settings\documentsfilename
                                  ^^
                             this is wrong

Instead use Path.Combine.

string path = Path.Combine(p1, l1);
Process.Start(path);

Why don't you simply do this: -

public void button4_Click(object sender, EventArgs e)
{
    string fileNameWithPath;
    DialogResult result = openFileDialog1.ShowDialog();
    if (result == DialogResult.OK)
    {
        fileNameWithPath = openFileDialog1.FileName;
    }
}

public void button2_Click(object sender, EventArgs e)
{
    Process.Start(fileNameWithPath);
}
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