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In my application i upload photo of employee using OpenFileDialog and display photo in picture box. i want photo should be store in local system for example in "E:\" drive.below is my code of photo upload.

C#
private void btnphotoupload_Click(object sender, EventArgs e)
        {
            // upload image and display in picture box

            OpenFileDialog open = new OpenFileDialog();
            // image filters
            open.Filter = "Image Files(*.jpg; *.jpeg; *.gif; *.bmp)|*.jpg; *.jpeg; *.gif; *.bmp";
            if (open.ShowDialog() == DialogResult.OK)
            {
                // display image in picture box
                pictureBox1.Image = new Bitmap(open.FileName);

                // image file path
                // textBox1.Text = open.FileName;

            }
           // var = open.FileName;
            Data.imagelocation = open.FileName;
           

        }
Posted
Comments
Vishal Sharma 20-Jul-13 13:13pm
   
Bro what is Data here in Data.imagelocation = open.FileName;

Not really sure what your issue is but SaveFileDialog could be used:
http://msdn.microsoft.com/en-us/library/system.windows.forms.savefiledialog.aspx#Y2964[^]
and saving the picture:
http://msdn.microsoft.com/en-us/library/9t4syfhh.aspx[^]
   
See this:
C#
private void button1_Click(object sender, EventArgs e)
{
   pictureBox1.Image.Save("C:\\Temp\\Test.bmp");
}
   
Comments
Espen Harlinn 11-Sep-12 10:19am
   
5'ed!
Hi,

try this,

C#
private void btnphotoupload_Click(object sender, EventArgs e)
    {
        // upload image and display in picture box

        OpenFileDialog open = new OpenFileDialog();
        // image filters
        open.Filter = "Image Files(*.jpg; *.jpeg; *.gif; *.bmp)|*.jpg; *.jpeg; *.gif; *.bmp";
        if (open.ShowDialog() == DialogResult.OK)
        {
            BitMap btMap = new Bitmap(open.FileName);

            // display image in picture box
            pictureBox1.Image = btMap;
            
           //Save the image in new location
           btMap.Save(@"E:\Images\Image.jgp");
            
        }
       // var = open.FileName;
        Data.imagelocation = open.FileName;
    }


Hope it works
   
Comments
Vishal Sharma 20-Jul-13 13:28pm
   
what is Data here
Karthik Harve 22-Jul-13 6:26am
   
OP is using that. OP can answer to your question.

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