将图像上传到数据库 [英] Upload Image into database

问题描述

您好我想将图片上传到数据库







请帮帮我





我创建方法

  public   void  insert_blog（ ref   string  connection_string， ref   Int64  blog_id， ref   string  title， ref   string  intro， ref   string  text， ref  < span class =code-keyword> string  keyword， ref   byte  categotry， out   byte  affected_rows）
 {
 affected_rows =  0 ; 
 
 SqlConnection conn =  new  SqlConnection（connection_string）; 
 SqlCommand cmd =  new  SqlCommand（）; 
 
  if （conn.State == ConnectionState.Closed）
 {
 conn.Open（）; 
} 
 cmd.CommandText =   insert_blog; 
 cmd.Connection = conn; 
 cmd.CommandType = CommandType.StoredProcedure; 
 
 cmd.Parameters.Add（  @ blogid，SqlDbType.BigInt ）.Value = blog_id; 
 cmd.Parameters.Add（  @ intro，SqlDbType.VarChar）.Value =介绍; 
 cmd.Parameters.Add（  @ title，SqlDbType.VarChar）.Value =标题; 
 cmd.Parameters.Add（  @ text，SqlDbType.NVarChar）.Value =文本; 
 cmd.Parameters.Add（  @ keyword，SqlDbType.VarChar）.Value =关键词; 
 cmd.Parameters.Add（  @ category，SqlDbType.TinyInt）.Value = categotry; 
 
 affected_rows = Convert.ToByte（cmd.ExecuteNonQuery（））; 
 
 
 conn.Close（）; 
 conn.Dispose（）; 
} 
 

和

  string  connection = System.Configuration.ConfigurationManager.ConnectionStrings [  博客]。ToString（）; 
 blog_class cla =  new  blog_class（）; 
  string  title = TextBox1_title.Text; 
  string  intro = TextBox1_intro.Text; 
  string  meta = TextBox1_meta_tags.Text; 
  string  \\ temp =  ; 
  Int64  blog_id =  0 ; 
 order_by_tag_photo_ids_wall（ ref  connection， out  blog_id）; 
 blog_id = blog_id +  1 ; 
  byte  cat_id = Convert.ToByte（dropdown1.SelectedValue.ToString（））; 
  byte  n; 
 cla.insert_blog（ ref  connection， ref  blog_id， ref  title， ref  intro， ref  \\ temp， ref  meta， ref  cat_id， out  n）; 
  if （n >   0 ）
 {
 Response.Redirect（ 〜/ user / redirect / edit_blog.aspx ？blogid = + blog_id）; 
} 
  else  
 {
 Label1.Text =  错误发布; 
} 

我如何上传我创建的图片数据库中的字段



img varchar（300）

解决方案

您好，



请查看这些Code Project文章。他们应该帮助您解决问题。使用 varchar（300）不是一个好主意，除非你的图像非常小&该字段定义不允许您以二进制格式存储图像。

• 使用Strored Procedures和C＃从SQL Server存储和检索图像 [ ^ ]
• C＃从数据库保存和加载图像 [ ^ ]
• SQL Database Image Storage&简易缩略图 [ ^ ]
• 保存图像进入SQL Server 2000数据库 [ ^ ]
• 使用Microsoft .NET从SQL Server存储和检索图像 [ ^ ]

问候，

< blockquote>在sql中使用image作为数据类型

 protected void Button1_Click（object sender，EventArgs e）
 {
 if（FileUpload1.PostedFile！= null&& FileUpload 1.PostedFile.FileName！=）
 {
 byte [] img = new byte [FileUpload1.PostedFile.ContentLength]; 
 HttpPostedFile Image = FileUpload1.PostedFile; 
 Image.InputStream.Read（img，0，（int）FileUpload1.PostedFile.ContentLength）; 
 string str =insert img values（@img）; 
 SqlConnection cn = new SqlConnection（data source = localhost; initial catalog = aadarsh; user id = sa; password = cos123）; 
 cn.Open（）; 
 SqlCommand cmd = new SqlCommand（str，cn）; 
 cmd.Parameters.AddWithValue（@ img，img）; 
 SqlDataReader dr = cmd.ExecuteReader（）; 
 cn.Close（）; 
} 
} 
 

SqlConnection con = new SqlConnection（@Data Source = LUVLY; Initial Catalog = practice;用户ID = sa;密码= lenovo @ 345）;

con.Open（）;

string cmdstr =select * from testing;

SqlCommand cmd = new SqlCommand（cmdstr，con）;

SqlDataAdapter da = new SqlDataAdapter（cmd）;

DataTable dt = new DataTable（）;

da.Fill（dt）;

GridView1.DataSource = dt;



GridView1.DataBind（）;

Hi I want to upload a image into database



Please help me out


I create a method

public void insert_blog(ref string connection_string, ref Int64 blog_id, ref string title, ref string intro, ref string text, ref string keyword, ref byte categotry, out byte affected_rows)
   {
       affected_rows = 0;

       SqlConnection conn = new SqlConnection(connection_string);
       SqlCommand cmd = new SqlCommand();

       if (conn.State == ConnectionState.Closed)
       {
           conn.Open();
       }
       cmd.CommandText = "insert_blog";
       cmd.Connection = conn;
       cmd.CommandType = CommandType.StoredProcedure;

       cmd.Parameters.Add("@blogid", SqlDbType.BigInt).Value = blog_id;
       cmd.Parameters.Add("@intro", SqlDbType.VarChar).Value = intro;
       cmd.Parameters.Add("@title", SqlDbType.VarChar).Value = title;
       cmd.Parameters.Add("@text", SqlDbType.NVarChar).Value = text;
       cmd.Parameters.Add("@keyword", SqlDbType.VarChar).Value = keyword;
       cmd.Parameters.Add("@category", SqlDbType.TinyInt).Value = categotry;

       affected_rows = Convert.ToByte(cmd.ExecuteNonQuery());


       conn.Close();
       conn.Dispose();
   }

and

string connection = System.Configuration.ConfigurationManager.ConnectionStrings["blog"].ToString();
        blog_class cla = new blog_class();
        string title = TextBox1_title.Text;
        string intro = TextBox1_intro.Text;
        string meta = TextBox1_meta_tags.Text;
        string emp = "";
        Int64 blog_id = 0;
        order_by_tag_photo_ids_wall(ref connection, out blog_id);
        blog_id = blog_id + 1;
        byte cat_id = Convert.ToByte(dropdown1.SelectedValue.ToString());
        byte n;
        cla.insert_blog(ref connection, ref blog_id, ref title, ref intro, ref emp, ref meta, ref cat_id, out n);
        if (n > 0)
        {
            Response.Redirect("~/user/redirect/edit_blog.aspx?blogid=" + blog_id);
        }
        else
        {
            Label1.Text = "Error Occured";
        }

How I upload Image I create a field in database

img varchar(300)

解决方案

Hello,

Please have a look at these Code Project articles. They should help you resolve the issue. Using varchar(300) is not a good idea unless your images are pretty small & that field definition will not allow you to store images in binary format.

• Storing and Retrieving Images from SQL Server Using Strored Procedures and C#[^]
• C# Save and Load Image from Database[^]
• SQL Database Image Storage & Easy Thumbnails[^]
• Save An Image Into SQL Server 2000 Database[^]
• Storing and Retrieving Images from SQL Server using Microsoft .NET[^]

Regards,

use image as a data type in sql

protected void Button1_Click(object sender, EventArgs e)
{
    if (FileUpload1.PostedFile != null && FileUpload1.PostedFile.FileName != "")
    {
        byte[] img = new byte[FileUpload1.PostedFile.ContentLength];
        HttpPostedFile Image = FileUpload1.PostedFile;
        Image.InputStream.Read(img, 0, (int)FileUpload1.PostedFile.ContentLength);
        string str = "insert img values(@img)";
        SqlConnection cn = new SqlConnection("data source=localhost;initial catalog=aadarsh;user id=sa;password=cos123");
        cn.Open();
        SqlCommand cmd = new SqlCommand(str, cn);
        cmd.Parameters.AddWithValue("@img", img);
        SqlDataReader dr = cmd.ExecuteReader();
        cn.Close();
    }
}

SqlConnection con = new SqlConnection(@"Data Source=LUVLY;Initial Catalog=practice;User ID=sa;Password=lenovo@345");
con.Open();
string cmdstr = "select * from testing";
SqlCommand cmd = new SqlCommand(cmdstr, con);
SqlDataAdapter da = new SqlDataAdapter(cmd);
DataTable dt = new DataTable();
da.Fill(dt);
GridView1.DataSource = dt;

GridView1.DataBind();

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