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Upload and Download Files From DataGridView Cells

By , 25 May 2011
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This post helps to upload any file into a datagridview cell and also download it from the datagridview cell. In the attached project file, I have declared a model variable Dictionary<int, byte[]> _myAttachments which stores the attachments as Byte array, and key stores the rowIndex.

Dictionary<int, byte[]> _myAttachments = new Dictionary<int, byte[]>();

Select the attachment cell and click on the ‘Upload File’ button. It will pop up a FileDialog for the user to select the required file which is to be uploaded.

The following routine uploads the user selected file and shows the file name in the provided dataGridViewCell.

/// <summary>
/// Upload Attachment at the provided DataGridViewCell
/// </summary>
/// <param name="dgvCell"></param>
private void UploadAttachment(DataGridViewCell dgvCell)
    using (OpenFileDialog fileDialog = new OpenFileDialog())
        //Set File dialog properties
        fileDialog.CheckFileExists = true;
        fileDialog.CheckPathExists = true;
        fileDialog.Filter = "All Files|*.*";
        fileDialog.Title = "Select a file";
        fileDialog.Multiselect = false;

        if (fileDialog.ShowDialog() == DialogResult.OK)
            FileInfo fileInfo = new FileInfo(fileDialog.FileName);
            byte[] binaryData = File.ReadAllBytes(fileDialog.FileName);
            dataGridView1.Rows[dgvCell.RowIndex].Cells[1].Value = fileInfo.Name;

            if (_myAttachments.ContainsKey(dgvCell.RowIndex))
                _myAttachments[dgvCell.RowIndex] = binaryData;
                _myAttachments.Add(dgvCell.RowIndex, binaryData);

The DownloadAttachment routine can be called on cell double click or the ‘Download’ button click event.

/// <summary>
/// Download Attachment from the provided DataGridViewCell
/// </summary>
/// <param name="dgvCell"></param>
private void DownloadAttachment(DataGridViewCell dgvCell)
    string fileName = Convert.ToString(dgvCell.Value);

    //Return if the cell is empty
    if (fileName == string.Empty)

    FileInfo fileInfo = new FileInfo(fileName);
    string fileExtension = fileInfo.Extension;

    byte[] byteData = null;

    //show save as dialog
    using (SaveFileDialog saveFileDialog1 = new SaveFileDialog())
        //Set Save dialog properties
        saveFileDialog1.Filter = "Files (*" + fileExtension + ")|*" + fileExtension;
        saveFileDialog1.Title = "Save File as";
        saveFileDialog1.CheckPathExists = true;
        saveFileDialog1.FileName = fileName;

        if (saveFileDialog1.ShowDialog() == DialogResult.OK)
            byteData = _myAttachments[dgvCell.RowIndex];
            File.WriteAllBytes(saveFileDialog1.FileName, byteData);


This article, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

About the Author

Balu Sathish
Software Developer Calpine Technologies
India India
No Biography provided

Comments and Discussions

Questiondatagried PinmemberMember 101908545-Oct-13 15:26 
QuestionCan we open the file from DataGridView? PinmemberTAN THIAM HUAT19-Jul-12 22:44 
GeneralMy vote of 3 Pinmembersharry2kool27-Jun-12 19:55 

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