嘗試將圖像插入到我的數據庫中。所有節省了同樣的事情plsss的幫助，而我試圖插入圖片到我的數據庫這一切的時候，我失去了

-2

的時間節省了這一點：嘗試將圖像插入到我的數據庫中。所有節省了同樣的事情plsss的幫助，而我試圖插入圖片到我的數據庫這一切的時候，我失去了

0xFFD8FFE145AB4578696600004D4D002A00000008000A011200030000000100000000021300030000000100010000011A0005000000010000018E011B0005000000010000019601280003000000010002000001320002000000140000019E010F000200000064000001B201100002000000640000021687690004000000010000

即使圖像是空值。

請幫忙。

C＃代碼：

if (FileUpload1.PostedFile.FileName!="") 
{ 
    Byte[] image; 
    Stream s = FileUpload1.PostedFile.InputStream; 

    BinaryReader br = new BinaryReader(s); 
    image = br.ReadBytes((Int32)s.Length); 

    SqlCommand NewUser = new SqlCommand("INSERT INTO [User] Values (@username,@password,@name,@lastname,@location,@profesion,@email,@gender,@money,@pro,@xp,@lv,@m1,@m2,@m3,@m4,@m5,@d1,@d2,@d3,@d4,@d5,@im,@phone);", c); 
    NewUser.Connection = c; 

    NewUser.Parameters.AddWithValue("@username", txtuser.Text); 
    NewUser.Parameters.AddWithValue("@password", txtpass.Text); 
    NewUser.Parameters.AddWithValue("@name", txtFName.Text); 
    NewUser.Parameters.AddWithValue("@lastname", txtLName.Text); 
    NewUser.Parameters.AddWithValue("@location", ddlcountry.SelectedItem.Text); 
    NewUser.Parameters.AddWithValue("@profesion", txtprofession.Text); 
    NewUser.Parameters.AddWithValue("@email", txtemail.Text); 
    NewUser.Parameters.AddWithValue("@gender", rbgendere.SelectedItem.Text); 
    NewUser.Parameters.AddWithValue("@money", 0); 
    NewUser.Parameters.AddWithValue("@pro", DBNull.Value); 
    NewUser.Parameters.AddWithValue("@xp", 0); 
    NewUser.Parameters.AddWithValue("@lv", 1); 
    NewUser.Parameters.AddWithValue("@m1", 0); 
    NewUser.Parameters.AddWithValue("@m2", 0); 
    NewUser.Parameters.AddWithValue("@m3", 0); 
    NewUser.Parameters.AddWithValue("@m4", 0); 
    NewUser.Parameters.AddWithValue("@m5", 0); 
    NewUser.Parameters.AddWithValue("@d1", 0); 
    NewUser.Parameters.AddWithValue("@d2", 0); 
    NewUser.Parameters.AddWithValue("@d3", 0); 
    NewUser.Parameters.AddWithValue("@d4", 0); 
    NewUser.Parameters.AddWithValue("@d5", 0); 
    NewUser.Parameters.AddWithValue("@im", image); 
    NewUser.Parameters.AddWithValue("@phone", PhoneNumber.Text); 

    Session["CurentUserid"] = txtuser.Text; 

    c.Open(); 
    int row = NewUser.ExecuteNonQuery(); 
    c.Close(); 

    if (row > 0) 
    { 
     LabelError.Text = "success"; 
    } 

    Session["Conect"] = (bool)true; 
    Response.Redirect("Finish Had Member.aspx", true);  
} 
else 
{ 
    SqlCommand NewUser = new SqlCommand("INSERT INTO [User] Values (@username,@password,@name,@lastname,@location,@profesion,@email,@gender,@money,@pro,@xp,@lv,@m1,@m2,@m3,@m4,@m5,@d1,@d2,@d3,@d4,@d5,@im,@phone);", c); 
    NewUser.Connection = c; 

    NewUser.Parameters.AddWithValue("@username", txtuser.Text); 
    NewUser.Parameters.AddWithValue("@password", txtpass.Text); 
    NewUser.Parameters.AddWithValue("@name", txtFName.Text); 
    NewUser.Parameters.AddWithValue("@lastname", txtLName.Text); 
    NewUser.Parameters.AddWithValue("@location", ddlcountry.SelectedItem.Text); 
    NewUser.Parameters.AddWithValue("@profesion", txtprofession.Text); 
    NewUser.Parameters.AddWithValue("@email", txtemail.Text); 
    NewUser.Parameters.AddWithValue("@gender", rbgendere.SelectedItem.Text); 
    NewUser.Parameters.AddWithValue("@money", 0); 
    NewUser.Parameters.AddWithValue("@pro", DBNull.Value); 
    NewUser.Parameters.AddWithValue("@xp", 0); 
    NewUser.Parameters.AddWithValue("@lv", 1); 
    NewUser.Parameters.AddWithValue("@m1", 0); 
    NewUser.Parameters.AddWithValue("@m2", 0); 
    NewUser.Parameters.AddWithValue("@m3", 0); 
    NewUser.Parameters.AddWithValue("@m4", 0); 
    NewUser.Parameters.AddWithValue("@m5", 0); 
    NewUser.Parameters.AddWithValue("@d1", 0); 
    NewUser.Parameters.AddWithValue("@d2", 0); 
    NewUser.Parameters.AddWithValue("@d3", 0); 
    NewUser.Parameters.AddWithValue("@d4", 0); 
    NewUser.Parameters.AddWithValue("@d5", 0); 
    NewUser.Parameters.AddWithValue("@im", DBNull.Value); 
    NewUser.Parameters.AddWithValue("@phone", PhoneNumber.Text); 

    Session["CurentUserid"] = txtuser.Text; 

    c.Open(); 
    NewUser.ExecuteNonQuery(); 
    c.Close(); 

    Session["Conect"] = (bool)true; 
    Response.Redirect("Finish Had Member.aspx", true); 
}

來源

2016-06-07 Oz Cohen

最小，完整和可驗證示例如何http://stackoverflow.com/help/mcve？ – DVK

你是正確的伴侶，但我不知道是這個問題，我搜索這麼難plss幫助.... –

第1步是[停止使用'AddWithValue']（http://blogs.msmvps.com/jcoehoorn/blog/ 2014/05/12//）使用-addwithvalue-已經可以-WE-停。這可能是也可能不是問題，但你正在用你的數據玩俄羅斯輪盤賭。 –

我有一個工作示例的變化相應

HTML頁面的

<div> 
     <table> 
      <tr> 
       <td> 
        Image Name: 
       </td> 
       <td> 
        <asp:TextBox ID="txtImageName" runat="server"></asp:TextBox> 
       </td> 
      </tr> 
      <tr> 
       <td> 
        Upload Image: 
       </td> 
       <td> 
        <asp:FileUpload ID="fileuploadImage" runat="server" /> 
       </td> 
      </tr> 
      <tr> 
       <td> 
       </td> 
       <td> 
        <asp:Button ID="btnUpload" runat="server" Text="Upload" OnClick="btnUpload_Click" /> 
       </td> 
      </tr> 
     </table> 
    </div> 
    <div> 
     <asp:GridView ID="gvImages" CssClass="Gridview" runat="server" AutoGenerateColumns="False" 
      HeaderStyle-BackColor="#7779AF" HeaderStyle-ForeColor="white"> 
      <Columns> 
       <asp:BoundField HeaderText="Image Name" DataField="imagename" /> 
       <asp:TemplateField HeaderText="Image"> 
        <ItemTemplate> 
         <asp:Image ID="Image1" runat="server" ImageUrl='<%# "Handler.ashx?ImID="+ Eval("ImageID") %>' 
          Height="150px" Width="150px" /> 
        </ItemTemplate> 
       </asp:TemplateField> 
      </Columns> 
     </asp:GridView> 
    </div>

代碼背後

protected void btnUpload_Click(object sender, EventArgs e) 
    { 
     if (fileuploadImage.HasFile) 
     { 
      //getting length of uploaded file 
      int length = fileuploadImage.PostedFile.ContentLength; 
      //create a byte array to store the binary image data 
      byte[] imgbyte = new byte[length]; 
      //store the currently selected file in memeory 
      HttpPostedFile img = fileuploadImage.PostedFile; 
      //set the binary data 
      img.InputStream.Read(imgbyte, 0, length); 
      string imagename = txtImageName.Text; 
      //use the web.config to store the connection string 
      SqlConnection connection = new SqlConnection(strcon); 
      connection.Open(); 
      SqlCommand cmd = new SqlCommand("INSERT INTO Image (ImageName,Image) VALUES (@imagename,@imagedata)", connection); 
      cmd.Parameters.Add("@imagename", SqlDbType.VarChar, 50).Value = imagename; 
      cmd.Parameters.Add("@imagedata", SqlDbType.Image).Value = imgbyte; 
      int count = cmd.ExecuteNonQuery(); 
      connection.Close(); 
      if (count == 1) 
      { 
       BindGridData(); 
       txtImageName.Text = string.Empty; 
       ScriptManager.RegisterStartupScript(this, this.GetType(), "alertmessage", "javascript:alert('" + imagename + " image inserted successfully')", true); 
      } 
     } 
    } 
    private void BindGridData() 
    { 
     SqlConnection connection = new SqlConnection(strcon); 
     SqlCommand command = new SqlCommand("SELECT imagename,ImageID from [Image]", connection); 
     SqlDataAdapter daimages = new SqlDataAdapter(command); 
     DataTable dt = new DataTable(); 
     daimages.Fill(dt); 
     gvImages.DataSource = dt; 
     gvImages.DataBind(); 
     gvImages.Attributes.Add("bordercolor", "black"); 
    }

右鍵點擊你的項目添加新的Generic Handler命名爲 - > Handler.ashx

與此

string strcon = System.Configuration.ConfigurationManager.ConnectionStrings["ConnectionString"].ConnectionString; 
    public void ProcessRequest(HttpContext context) 
    { 
     string imageid = context.Request.QueryString["ImID"]; 
     System.Data.SqlClient.SqlConnection connection = new System.Data.SqlClient.SqlConnection(strcon); 
     connection.Open(); 
     System.Data.SqlClient.SqlCommand command = new System.Data.SqlClient.SqlCommand("select Image from Image where ImageID=" + imageid, connection); 
     System.Data.SqlClient.SqlDataReader dr = command.ExecuteReader(); 
     dr.Read(); 
     context.Response.BinaryWrite((Byte[])dr[0]); 
     connection.Close(); 
     context.Response.End(); 
    }

在網頁替換ProcessRequest配置在我的情況下添加此下<configuration> -> <connectionStrings>

這裏數據庫名稱是testDB

<add name="ConnectionString" connectionString="data source=.\SQLEXPRESS;database=testDB;trusted_Connection=yes"/>

創建使用此查詢的SQL中的表格

CREATE TABLE [dbo].[Image](
    [ImageId] [int] IDENTITY(1,1) NOT NULL, 
    [ImageName] [varchar](50) NULL, 
    [Image] [image] NULL 
)

希望這將幫助你

來源

2016-06-07 17:36:37 Ansari

嘗試

command.Parameters.Add("@im", SqlDbType.VarBinary); 
command.Parameters["@ID"].Value = image;

它相匹配的SQL列類型。
根據Ansari的答案圖片是正確的類型。

來源

2016-06-07 17:58:58 Paparazzi

嘗試將圖像插入到我的數據庫中。所有節省了同樣的事情plsss的幫助，而我試圖插入圖片到我的數據庫這一切的時候，我失去了

回答

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