如何仅将文件名保存到gridview，但下载文件时我需要路径

Question

我终于知道了如何使用路径下载文件。但我想知道如何才能将文件名仅保留在网格视图上。虽然我需要完整的下载路径。

在调试时，我发现我不能只在文件上传时保留文件名。由于它被带到下载部分。 如果我保留文件名，则文件名被带到下载部分，文件不下载。

谁能帮帮我

代码

 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 = true;

            if (fileDialog.ShowDialog() == System.Windows.Forms.DialogResult.OK)
            {
                string strfilename = fileDialog.FileName;
                cncInfoDataGridView.Rows[dgvCell.RowIndex].Cells[1].Value = strfilename;
            }
        }
    }

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

        if (!string.IsNullOrEmpty(fileName))
        {

            byte[] objData;

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

            //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)
                {
                    string s = cncInfoDataGridView.Rows[dgvCell.RowIndex].Cells[1].Value.ToString();
                    objData = File.ReadAllBytes(s);
                    File.WriteAllBytes(saveFileDialog1.FileName, objData);
                }
            }
        }
    }

}

Answer 1

你的问题不清楚。但如果我说对了。为什么不使用列作为名称，使用隐藏/0 宽度列作为文件路径。它是一个网格，因此您可能有很多列。另外，我认为下面的代码应该返回完整路径，除非您修剪并且看不到您修剪的位置。

            string strfilename = fileDialog.FileName;

要获取文件名，您可以使用路径

   string filenameOnly= System.IO.Path.GetFileName(strfilename);

上面应该返回你的文件名，只有你可以检查here以便更好地理解。

添加另一列并将宽度设置为 0 。下面的例子

  DataGridViewColumn column = dataGridView.Columns[0];
    column.Width = 0;
cncInfoDataGridView.Columns.Add(column);

将您的文件路径保存在新列宽度为 0 宽度并在下载期间检索。

这里问了一个类似的问题。 How to extract file name from file path name?

Answer 2

Dictionary<int, byte[]> _myAttachments;
private void btnUpload_Click(object sender, EventArgs e)
    {
        try
        {
            //Throw error if attachment cell is not selected.
            //make sure user select only single cell
            if (dataGridView1.SelectedCells.Count == 1 && dataGridView1.SelectedCells[0].ColumnIndex == 1)
            {
                UploadAttachment(dataGridView1.SelectedCells[0]);
            }
            else
                MessageBox.Show("Select a single cell from Attachment column", "Error uploading file", MessageBoxButtons.OK, MessageBoxIcon.Error);
        }
        catch (Exception ex)
        {
            MessageBox.Show(ex.Message, "Error uploading file", MessageBoxButtons.OK, MessageBoxIcon.Error);
        }
    }
    private void btnDownload_Click(object sender, EventArgs e)
    {
        //Throw error if attachment cell is not selected.
        //make sure user select only single cell
        //and the cell have a value in it
        if (dataGridView1.SelectedCells.Count == 1 && dataGridView1.SelectedCells[0].ColumnIndex == 1 && dataGridView1.SelectedCells[0].Value != null)
        {
            DownloadAttachment(dataGridView1.SelectedCells[0]);
        }
        else
            MessageBox.Show("Select a single cell from Attachment column", "Error uploading file", MessageBoxButtons.OK, MessageBoxIcon.Error);
    }
    private void dataGridView1_CellDoubleClick(object sender, DataGridViewCellEventArgs e)
    {
        //Throw error if attachment cell is not selected.
        //make sure user select only single cell
        //and the cell have a value in it
        if (dataGridView1.SelectedCells.Count == 1 && dataGridView1.SelectedCells[0].ColumnIndex == 1 && dataGridView1.SelectedCells[0].Value != null)
        {
            DownloadAttachment(dataGridView1.SelectedCells[0]);
        }
        else
            MessageBox.Show("Select a single cell from Attachment column", "Error uploading file", MessageBoxButtons.OK, MessageBoxIcon.Error);
    }
    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;
                else
                    _myAttachments.Add(dgvCell.RowIndex, binaryData);
            }
        }
    }
    private void DownloadAttachment(DataGridViewCell dgvCell)
    {
        string fileName = Convert.ToString(dgvCell.Value);

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

        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);
            }
        }
    }