WebForm 无法使用文件列表填充按钮命令参数？

Question

我已经有很长时间没有使用 WebForms 了，但我不得不更新一个旧网站。我正在尝试制作一个页面，允许用户将文件下载到他们的机器上。我可以获得目录中的文件列表并将其显示在 table 中，但是，我试图将文件名分配给按钮的命令参数，以便我知道名称要在按钮事件中获取的文件。我似乎无法找到从文件夹对象中获取值以填充命令参数的正确方法。没有错误，它只是空白。有什么建议吗？

var listFiles = dir.GetFiles();

<% 
    foreach(var file in listFiles) { %>
    <tr>
        <td>&nbsp;</td>
        <td><%= file.Name %></td>
        <td><%= file.Length %></td>
        <td><%= file.CreationTime.ToString("MM/dd/yyyy HH:mm") %></td>
        <td>
            <asp:LinkButton ID="btnDownload" runat="server" 
                CommandArgument='<%#Eval("file.Name") %>'
                CommandName="DownloadTechFile" 
                OnCommand="DownloadFile"
                ToolTip="Downloaded file">
                <span aria-hidden="true" class="fa fa-download"></span>
            </asp:LinkButton>
        </td>
    </tr>
<% } %>


protected void DownloadFile(object sender, CommandEventArgs e)
        {
            Alert.Hide();

            var fileName = e.CommandArgument.ToString();

            var fileFullPath = Path.Combine(FileFolderPath, fileName);

            if (string.IsNullOrWhiteSpace(fileName) == false)
            {
                WebClient req = new WebClient();
                HttpResponse response = HttpContext.Current.Response;
                try
                {
                    response.Clear();
                    response.ClearContent();
                    response.ClearHeaders();
                    response.Buffer = true;
                    response.AddHeader("Content-Disposition", "attachment;filename=" + fileName);
                    byte[] data = req.DownloadData(fileFullPath);
                    //byte[] data = req.DownloadData(Server.MapPath(fileFullPath));
                    response.BinaryWrite(data);
                    response.End();

                    Alert.Show(AlertType.Success, "File downloaded.");

                }
                catch (Exception ex)
                {
                    logger.Error("Error downloading file from {0}. {1} | {2} | {3}", fileFullPath, ex.Message, ex.StackTrace, ex.InnerException);
                    Alert.Show(AlertType.Error, string.Format("Error trying to download file: {0}", fileName));
                }
            }
        }

更新： 这是我想到的，以防其他人可以使用它。

 <asp:GridView ID="gvFiles" runat="server" AutoGenerateColumns="false" CssClass="table borderless" HeaderStyle-CssClass="fileheader">
            <Columns>
                <asp:BoundField DataField="FileName" HeaderText="File Name" />
                <asp:BoundField DataField="FileSize" HeaderText="File Size" />
                <asp:BoundField DataField="Date" HeaderText="Created On" />
                <asp:TemplateField HeaderText="">
                    <ItemTemplate>
                        <asp:LinkButton ID="btnDownload" runat="server"
                            CommandArgument='<%# Eval("FileName") %>'
                            OnClick="DownloadFile"
                            ToolTip="Downloaded file">
                            <span aria-hidden="true" class="fa fa-download"></span>
                        </asp:LinkButton>
                    </ItemTemplate>
                </asp:TemplateField>
            </Columns>
        </asp:GridView>


void LoadGrid()
        {
            // create a table for the files, populate, and then bind.
            DataTable dtFiles = new DataTable();
            dtFiles.Columns.Add("Date", typeof(string));
            dtFiles.Columns.Add("FileSize", typeof(string));
            dtFiles.Columns.Add("FileName", typeof(string));

            DirectoryInfo fileDirectory = new DirectoryInfo(FileFolderPath);

            foreach (FileInfo file in fileDirectory.GetFiles("*.txt"))
            {
                DataRow dr = dtFiles.NewRow();
                dr["Date"] = file.CreationTime.ToString("MM/dd/yyyy HH:mm");
                dr["FileSize"] = Utility.GetBytesReadable(file.Length);
                dr["FileName"] = file.Name;
                dtFiles.Rows.Add(dr);
            }

            dtFiles.DefaultView.Sort = "Date DESC";
            gvFiles.DataSource = dtFiles;
            gvFiles.DataBind();
        }

        protected void DownloadFile(object sender, EventArgs e)
        {
            var fileName = (sender as LinkButton).CommandArgument;

            var fileFullPath = Path.Combine(FileFolderPath, fileName);

            string mineType = MimeMapping.GetMimeMapping(fileFullPath);

            if (string.IsNullOrWhiteSpace(fileName) == false)
            {
                byte[] binFile = File.ReadAllBytes(fileFullPath);
                Response.ContentType = mineType;
                Response.AppendHeader("Content-Disposition", "attachment; filename=" + fileName);
                Response.BinaryWrite(binFile);
                Response.End();
            }
        }

Answer 1

你可以说使用 gridview，这样说：

        <asp:GridView ID="GridView1" runat="server" AutoGenerateColumns="false" CssClass="table">
            <Columns>
                <asp:BoundField DataField="File Name" HeaderText="File"  />
                <asp:BoundField DataField="Date" HeaderText="Date" />
                <asp:BoundField DataField="File Size" HeaderText="Size" />
                <asp:TemplateField HeaderText="Select"  ItemStyle-HorizontalAlign="Center">
                    <ItemTemplate>
                        <asp:HyperLink ID="HyperLink1" runat="server" CssClass="btn btn-default"
                            NavigateUrl='<%# Eval("Path") %>'  >Down Load</asp:HyperLink>
                    </ItemTemplate>
                </asp:TemplateField>
            </Columns>
        </asp:GridView>

后面的代码可以是这样的：

    protected void Page_Load(object sender, EventArgs e)
    {
        if (!IsPostBack)
            LoadGrid();
    }

    void LoadGrid()
    {
        // create a table for the files
        DataTable MyTable = new DataTable();

        MyTable.Columns.Add("Date", typeof(string));
        MyTable.Columns.Add("File Size", typeof(string));
        MyTable.Columns.Add("File Name", typeof(string));
        MyTable.Columns.Add("Path", typeof(string));

        string strURLFolder = "~/Content/Animals/";
        string strFolder = Server.MapPath(strURLFolder);

        DirectoryInfo MyDir = new DirectoryInfo(strFolder);
        FileInfo[] MyFiles = MyDir.GetFiles("*.*");


        foreach (FileInfo MyFile in MyDir.GetFiles("*.*"))
        {
            DataRow OneRow = MyTable.NewRow();
            OneRow["Date"] = MyFile.LastAccessTime;
            OneRow["File Size"] = (MyFile.Length / 1024).ToString() + " KB";
            OneRow["File Name"] = MyFile.Name;
            OneRow["Path"] = strURLFolder + MyFile.Name;

            MyTable.Rows.Add(OneRow);

        }

        MyTable.DefaultView.Sort = "Date";
        GridView1.DataSource = MyTable;
        GridView1.DataBind();

    }

现在我们明白了：

现在，我想您可以用一个按钮替换“hyper-link”，也许可以将文件作为字节流读取，然后将其发送给客户端，但上面的内容应该让您从这里开始。

编辑：用户指出他们不想对文件使用 hyper-link（简单 link）。

所以，我们可以这样做：

我们将平面简按钮放入网格中（为什么在您现在声明不想使用 link 时建议、提示、谈论“link”？-我们要为这里的混乱颁奖吗？）。

如果你不想要一个link，那显然我们不需要也不想关心，说说link对吧?????

好的，让我们删除超级 link，然后将平面简按钮放入网格视图中，如下所示：

        <asp:GridView ID="GridView1" runat="server" AutoGenerateColumns="false" CssClass="table">
            <Columns>
                <asp:BoundField DataField="File Name" HeaderText="File"  />
                <asp:BoundField DataField="Date" HeaderText="Date" />
                <asp:BoundField DataField="File Size" HeaderText="Size" />
                <asp:TemplateField HeaderText="Select"  ItemStyle-HorizontalAlign="Center">
                    <ItemTemplate>
                        <asp:Button ID="cmdDownLoad" runat="server" Text="Download" cssclass="btn"
                            OnClick="cmdDownLoad_Click" />
                    </ItemTemplate>
                </asp:TemplateField>
            </Columns>
        </asp:GridView>

所以，代码像以前一样加载，现在我们 get/have 这个：

    protected void cmdDownLoad_Click(object sender, EventArgs e)
    {
        Button btn = sender as Button;
        GridViewRow gRow = btn.NamingContainer as GridViewRow;

        string sFileOnly = gRow.Cells[0].Text;
        string sFile = Server.MapPath("~/Content/Animals/" + sFileOnly);

        string sMineType = MimeMapping.GetMimeMapping(sFile);

        byte[] binFile = File.ReadAllBytes(sFile);
        Response.ContentType = sMineType;
        Response.AppendHeader("Content-Disposition", "attachment; filename=" + sFileOnly);
        Response.BinaryWrite(binFile);
        Response.End();

    }

所以，我们应该看到这样说：

WebForm 无法使用文件列表填充按钮命令参数？

WebForm can't populate button command argument with list of files?

c#

asp.net

webforms

linkbutton

asplinkbutton

编辑：用户指出他们不想对文件使用 hyper-link（简单 link）。