或者我可能不需要,我不知道。这是我想要做的。我在一个页面上有一个gridview,当我点击一个项目时,它会获取该ID并将其链接到另一个页面,该页面显示gridview中项目的所有信息。在第二页上,我希望能够将带有一些文本的照片插入到我的数据库中,但我想确保它插入正确的blogID(从第一页点击)。这里是我到目前为止的代码:如何从我的网站中的其他页面访问控件?
EditBlogPosts.aspx
<asp:GridView ID="GridView1" runat="server" AllowSorting="True"
AutoGenerateEditButton="True"
DataSourceID="AccessDataSource1"
AutoGenerateColumns="False" DataKeyNames="ID"
AlternatingRowStyle-BackColor="Gray"
AlternatingRowStyle-CssClass="editGridFormat" RowStyle-CssClass="editGridFormat"
RowStyle-VerticalAlign="Top"
onselectedindexchanged="GridView1_SelectedIndexChanged">
后面的代码:
protected void GridView1_SelectedIndexChanged(object sender, EventArgs e)
{
GridViewRow row = GridView1.SelectedRow;
Response.Redirect("~/EditThisPost.aspx?ID=" + row.Cells[2].Text);
}
EditThisPost.aspx
<asp:FormView ID="Formview1" runat="server" DefaultMode="Insert" DataSourceID="AccessDataSource1" >
<InsertItemTemplate>
<br />
<asp:Label ID="TextforPhotoLabel" runat="server" Text="Put your text to go with your photo here:" /><br />
<asp:TextBox ID="PhotoText" runat="server" Rows="10" Columns="100" /><br /><br />
<asp:FileUpload ID="FileUpload1" runat="server" />
<asp:Label ID="UploadStatusLabel" runat="server" Text="Status: " /><br />
<asp:Button ID="UploadButton" runat="server" OnClick="UploadFile" Text="Insert Item" /><br />
</InsertItemTemplate>
</asp:FormView>
后面的代码(特别关注所在行我声明int blogID):
protected void UploadFile(object sender, EventArgs e)
{
TextBox txtPhotoText = (TextBox)Formview1.FindControl("PhotoText");
FileUpload FileUpload1 = (FileUpload)Formview1.FindControl("FileUpload1");
Label UploadStatusLabel = (Label)Formview1.FindControl("UploadStatusLabel");
if (FileUpload1.HasFile)
{
try
{
if (FileUpload1.PostedFile.ContentType == "image/jpeg")
{
if (FileUpload1.PostedFile.ContentLength < 10240000)
{
string filename = Path.GetFileName(FileUpload1.FileName);
FileUpload1.SaveAs(Server.MapPath("~/photos/PeoplePhotos/") + filename);
UploadStatusLabel.Text = "Upload status: Complete!";
string constr = @"Provider=Microsoft.Jet.OLEDB.4.0;Data Source=|DataDirectory|\TravelJoansDB.mdb;";
string cmdstr = "INSERT INTO BlogEntryItems (BlogID, Picture, PicText1) VALUES (?,?,?)";
int blogID = int.Parse(Request.QueryString["ID"]);
OleDbConnection con = new OleDbConnection(constr);
OleDbCommand com = new OleDbCommand(cmdstr, con);
con.Open();
com.Parameters.AddWithValue("@BlogID", blogID);
com.Parameters.AddWithValue("@Picture", filename);
com.Parameters.AddWithValue("@PicText1", txtPhotoText);
com.ExecuteNonQuery();
con.Close();
}
else
UploadStatusLabel.Text = "Upload status: The file has to be less than 10 MB!";
}
else
UploadStatusLabel.Text = "Upload status: Only JPEG files are accepted!";
}
catch (Exception ex)
{
UploadStatusLabel.Text = "Upload status: The file could not be uploaded. The following error occured: " + ex.Message;
}
}
}
我真的很感谢任何帮助。我想我可以以某种方式获得传递的ID,使“〜\ EditThisPost.aspx?ID =”成为一个有效的链接。但是,如果有更好的方法来做到这一点,或者我的想法甚至不存在,那我该如何实现我所需要的?
嘿,我很确定你是对的。我只想问现在我遇到了一个与插入数据库不同的错误。当我按下按钮时,它给我一个错误,说“多步OLE DB操作产生错误”有什么想法?另外,请看我的更新代码,我添加了一些东西。 – Joseph
没关系,让它工作!你是一个救星! – Joseph