CODEWITHSUNDEEP
c#web-servicesopenfiledialog
Ebikeneser
Ebikeneser

Reputation: 2374

Add prompt after user selects file

I have added an open file dialog box to my client application so that the used can select a specific file they want to send to the web service.

However the file gets sent the moment the file has been selected, whereas I would like to have a secondary prompt e.g. "Send - 'file name' Button Yes. Button No." to pop up after they have selected the file.

This would be incase the user selected the wrong file they would have a chance to see which one they selected.

So far I have the following code -

private void button1_Click(object sender, EventArgs e)
    {

        //Read txt File
        openFileDialog1.Filter = "Text Files (*.txt)|*.txt|All Files (*.*)|*.*";
        openFileDialog1.FilterIndex = 1;
        if (openFileDialog1.ShowDialog() == DialogResult.OK)
        {
            StreamReader myReader = new StreamReader(openFileDialog1.FileName);

            myReader.Close();

            string csv = File.ReadAllText(openFileDialog1.FileName);

I need the prompt to come up after they have selected the file but not sure how to do this so any input would be greatly appreciated.

Upvotes: 1

Views: 257

Answers (4)

Sujay Ghosh
Sujay Ghosh

Reputation: 2868

A sample of the code modified.

if (openFileDialog1.ShowDialog() == DialogResult.OK)
{
     DialogResult dr = MessageBox.Show(message, caption, MessageBoxButtons.YesNo);

     if(dr == DialogResult.Yes )
        StreamReader myReader = new StreamReader(openFileDialog1.FileName);
        // more code
     else
        // do something else

Upvotes: 1

jeroenh
jeroenh

Reputation: 26792

You can use a message box:

 if (MessageBox.Show(string.Format("Upload {0}, are you sure?", openFileDialog1.FileName), "Please Confirm", MessageBoxButtons.YesNo) == DialogResult.Yes)
 {
     // ...
 }

Upvotes: 1

Sam Holder
Sam Holder

Reputation: 32954

you need to add the second check manually after the first dialog:

private void button1_Click(object sender, EventArgs e)
{

    //Read txt File
    openFileDialog1.Filter = "Text Files (*.txt)|*.txt|All Files (*.*)|*.*";
    openFileDialog1.FilterIndex = 1;
    if (openFileDialog1.ShowDialog() == DialogResult.OK)
    {
        if (MessageBox.Show("Message", "Title",MessageBoxButtons.YesNo)==DialogResult.Yes)
        {
            StreamReader myReader = new StreamReader(openFileDialog1.FileName);
            myReader.Close();
            string csv = File.ReadAllText(openFileDialog1.FileName);

etc etc

Information on MessageBox.Show. You can get information on the possible results/options from here.

You could ensure that the user sees the file to be uploaded by making the message something like:

"Are you sure you want to upload " + openFileDialog1.FileName;

Upvotes: 3

Nathan
Nathan

Reputation: 6216

MessageBox.Show(...) is the method you're looking for.

Upvotes: 2

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