本文实例讲述了C#操作数据库中存取图片文件的方法。分享给大家供大家参考。具体如下:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
|
private string sqlconnstr = "Data Source=.;Database=db_test;User id=sa;PWD=123456" ;
/*功能:把一种图片插入到数据库中
*返回值:无
*/
void InsertImageToDB()
{
//将需要存储的图片读取为数据流
FileStream fs = new FileStream( @"D:/Bear.jpg" , FileMode.Open, FileAccess.Read);
Byte[] byte_fs = new byte [fs.Length];
fs.Read(byte_fs, 0, Convert.ToInt32(fs.Length));
fs.Close();
//建立数据库连接
SqlConnection conn = new SqlConnection(sqlconnstr);
conn.Open();
SqlCommand cmd = new SqlCommand();
cmd.Connection = conn;
cmd.CommandText = "insert into tb_test(image_id,image_file) values(@image_id,@image_file)" ;
SqlParameter[] param = new SqlParameter[2];
param[0] = new SqlParameter( "@image_id" , SqlDbType.Int);
param[0].Value = 1;
param[1] = new sqlParameter( "@image_file" , SqlDbType.Image);
param[1].Value = byte_fs;
for ( int index = 0; index < 2; index++)
{
cmd.Parameters.Add(param[i]);
}
//执行SQL语句
cmd.ExecuteNonQuery();
conn.Close();
}
/*功能:从数据库中读取图像文件,并显示在PictureBox控件中
*返回值:无
*/
void GetImageFromDB()
{
byte [] Data = new byte [0];
//建立数据库连接
SqlConnection conn = new SqlConnection(sqlconnstr);
conn.Open();
SqlCommand cmd = new SqlCommand();
cmd.Connection = conn;
cmd.CommandText = "select * from tb_parent" ;
SqlDataReader sdr = cmd.ExecuteReader();
sdr.Read();
Data = ( byte [])sdr[ "parent_image" ]; //读取第一个图片的位流
MemoryStream mystream = new MemoryStream(Data);
//用指定的数据流来创建一个image图片
System.Drawing.Image picbImage = System.Drawing.Image.FromStream(mystream, true );
mystream.Close();
picturebox1.Image = picbImage;
conn.Close();
}
|
希望本文所述对大家的C#程序设计有所帮助。