HTML如下:<HTML>
<title>上传图片</title>
<BODY bgColor="#ffffff">
<FORM id="Form1" encType="multipart/form-data" RUNAT="server">
<TABLE id="Table1" cellSpacing="0" cellPadding="0" width="700" align="left" border="0" RUNAT="server">
<TR>
<TD>上传图片(选择你要上传的图片)</TD>
<TD><INPUT id="UP_FILE" style="WIDTH: 320px" type="file" accept="text/*" name="UP_FILE" RUNAT="server">
</TD>
</TR>
<TR>
<TD><FONT face="宋体"></FONT></TD>
<TD><asp:textbox id="txtDescription" RUNAT="server" MAINTAINSTATE="false" WIDTH="239"></asp:textbox></TD>
</TR>
<TR>
<TD><asp:label id="txtMessage" RUNAT="server" MAINTAINSTATE="false" FORECOLOR="red"></asp:label></TD>
<TD><asp:button id="Button1" onclick="Button_Submit" RUNAT="server" WIDTH="239" TEXT="Upload Image"></asp:button></TD>
</TR>
</TABLE>
</FORM>
</BODY>
</HTML>

解决方案 »

  1.   

    用dataset试试
    SqlConnection con = new SqlConnection("Server=Darkover;uid=sa;pwd=Password1;database=northwind");
    SqlDataAdapter da = new SqlDataAdapter("Select * From MyImages", con);
    SqlCommandBuilder MyCB = new SqlCommandBuilder(da);
    DataSet ds = new DataSet("MyImages");da.MissingSchemaAction = MissingSchemaAction.AddWithKey;
    FileStream fs = new FileStream(@"C:\winnt\Gone Fishing.BMP", FileMode.OpenOrCreate, FileAccess.Read);

    byte[] MyData= new byte[fs.Length];
    fs.Read(MyData, 0, System.Convert.ToInt32(fs.Length));

    fs.Close();

    da.Fill(ds,"MyImages");

    DataRow myRow;
    myRow=ds.Tables["MyImages"].NewRow();myRow["Description"] = "This would be description text";
    myRow["imgField"] = MyData;
    ds.Tables["MyImages"].Rows.Add(myRow);
    da.Update(ds, "MyImages");con.Close();