Asp.net 将表单中的文档上载到SharePoint文档库

我的想法是，我有一个表单，允许用户将文件上传到我的SharePoint文档库中。我只是不知道在“HasFile”声明之后该怎么做。下面是我的一些小代码片段。我只是不知道如何从fileupload ID将其放入我的文档库。请提供代码

<asp:FileUpload ID="FileUpload1" runat="server" />



<asp:Button ID="Button1" runat="server" Text="Button" onclick="Button1_Click" />

---

可以使用以下方法将文档上载到SharePoint库：

/// <summary>
/// Uploads the specified file into a SharePoint Library
/// </summary>
/// <param name="url">site url</param>
/// <param name="creds">Credentials</param>
/// <param name="listTitle">List Title</param>
/// <param name="fileName">File Name</param>
/// <param name="stream">File Stream</param>
private static void UploadFile(string url,string listTitle, string fileName, Stream stream)
{
    using (var clientContext = new ClientContext(url))
    {


        var list = clientContext.Web.Lists.GetByTitle(listTitle);
        clientContext.Load(list.RootFolder);
        clientContext.ExecuteQuery();
        var fileUrl = String.Format("{0}/{1}", list.RootFolder.ServerRelativeUrl, fileName);

        Microsoft.SharePoint.Client.File.SaveBinaryDirect(clientContext, fileUrl, stream, true);

     }
 } 

用法

/// <summary>
/// Uploads the specified file into a SharePoint Library
/// </summary>
/// <param name="url">site url</param>
/// <param name="creds">Credentials</param>
/// <param name="listTitle">List Title</param>
/// <param name="fileName">File Name</param>
/// <param name="stream">File Stream</param>
private static void UploadFile(string url,string listTitle, string fileName, Stream stream)
{
    using (var clientContext = new ClientContext(url))
    {


        var list = clientContext.Web.Lists.GetByTitle(listTitle);
        clientContext.Load(list.RootFolder);
        clientContext.ExecuteQuery();
        var fileUrl = String.Format("{0}/{1}", list.RootFolder.ServerRelativeUrl, fileName);

        Microsoft.SharePoint.Client.File.SaveBinaryDirect(clientContext, fileUrl, stream, true);

     }
 } 

clientContext.Credentials = credentials;

 if (FileUpload1.HasFile)
 {

     UploadFile(webUrl, listTitle, FileUpload1.FileName, FileUpload1.PostedFile.InputStream);
 }