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使用asp.net将图片上传并存入SqlServer中,然后从SqlServer中读取并显示出来

2006-09-08 15:42 1286 查看
一,上传并存入SqlServer
数据库结构
create table test
{
id identity(1,1),
FImage image
}
相关的存储过程
Create proc UpdateImage
(
@UpdateImage Image
)
As
Insert Into test(FImage) values(@UpdateImage)
GO

在UpPhoto.aspx文件中添加如下:
<input id="UpPhoto" name="UpPhoto" runat="server" type="file">
<asp:Button id="btnAdd" name="btnAdd" runat="server" Text="上传"></asp:Button>

然后在后置代码文件UpPhoto.aspx.cs添加btnAdd按钮的单击事件处理代码:
private void btnAdd_Click(object sender, System.EventArgs e)
{
//获得图象并把图象转换为byte[]
HttpPostedFile upPhoto=UpPhoto.PostedFile;
int upPhotoLength=upPhoto.ContentLength;
byte[] PhotoArray=new Byte[upPhotoLength];
Stream PhotoStream=upPhoto.InputStream;
PhotoStream.Read(PhotoArray,0,upPhotoLength);

//连接数据库
SqlConnection conn=new SqlConnection();
conn.ConnectionString="Data Source=localhost;Database=test;User Id=sa;Pwd=sa";

SqlCommand cmd=new SqlCommand("UpdateImage",conn);
cmd.CommandType=CommandType.StoredProcedure;

cmd.Parameters.Add("@UpdateImage",SqlDbType.Image);
cmd.Parameters["@UpdateImage"].Value=PhotoArray;

//如果你希望不使用存储过程来添加图片把上面四句代码改为:
//string strSql="Insert into test(FImage) values(@FImage)";
//SqlCommand cmd=new SqlCommand(strSql,conn);
//cmd.Parameters.Add("@FImage",SqlDbType.Image);
//cmd.Parameters["@FImage"].Value=PhotoArray;

conn.Open();
cmd.ExecuteNonQuery();
conn.Close();
}

二,从SqlServer中读取并显示出来
在需要显示图片的地方添加如下代码:
<asp:image id="imgPhoto" runat="server" ImageUrl="ShowPhoto.aspx"></asp:image>

ShowPhoto.aspx主体代码:
private void Page_Load(object sender, System.EventArgs e)
{
if(!Page.IsPostBack)
{
SqlConnection conn=new SqlConnection()
conn.ConnectionString="Data Source=localhost;Database=test;User Id=sa;Pwd=sa";

string strSql="select * from test where id=2";//这里假设获取id为2的图片
SqlCommand cmd=new SqlCommand()
reader.Read();
Response.ContentType="application/octet-stream";
Response.BinaryWrite((Byte[])reader["FImage"]);
Response.End();
reader.Close();
}
}

3,在winform中将图片存入sqlserver,并从sqlserver中读取并显示在picturebox中

1,存入sqlserver
数据库结构和使用的存储过过程,同上面的一样
1.1,在窗体中加一个OpenFileDialog控件,命名为ofdSelectPic
1.2,在窗体上添加一个打开文件按钮,添加如下单击事件代码:
Stream ms;
byte[] picbyte;
//ofdSelectPic.ShowDialog();
if (ofdSelectPic.ShowDialog()==DialogResult.OK)
{
if ((ms=ofdSelectPic.OpenFile())!=null)
{
//MessageBox.Show("ok");
picbyte=new byte[ms.Length];
ms.Position=0;
ms.Read(picbyte,0,Convert.ToInt32(ms.Length));
//MessageBox.Show("读取完毕!");

//连接数据库
SqlConnection conn=new SqlConnection();
conn.ConnectionString="Data Source=localhost;Database=test;User Id=sa;Pwd=sa";

SqlCommand cmd=new SqlCommand("UpdateImage",conn);
cmd.CommandType=CommandType.StoredProcedure;

cmd.Parameters.Add("@UpdateImage",SqlDbType.Image);
cmd.Parameters["@UpdateImage"].Value=picbyte;

conn.Open();
cmd.ExecuteNonQuery();
conn.Close();

ms.Close();
}
}

2,读取并显示在picturebox中
2.1 添加一个picturebox,名为ptbShow
2.2 添加一个按钮,添加如下响应事件:
SqlConnection conn=new SqlConnection();
conn.ConnectionString="Data Source=localhost;Database=test;User Id=sa;Pwd=sa";

string strSql="select FImage from test where id=1";

SqlCommand cmd=new SqlCommand(strSql,conn);

conn.Open();
SqlDataReader reader=cmd.ExecuteReader();
reader.Read();

MemoryStream ms=new MemoryStream((byte[])reader["FImage"]);

Image image=Image.FromStream(ms,true);

reader.Close();
conn.Close();

ptbShow.Image=image;

在UpPhoto.aspx文件中添加如下:
<input id="UpPhoto" name="UpPhoto" runat="server" type="file">
<asp:Button id="btnAdd" name="btnAdd" runat="server" Text="上传"></asp:Button>

然后在后置代码文件UpPhoto.aspx.cs添加btnAdd按钮的单击事件处理代码:
private void btnAdd_Click(object sender, System.EventArgs e)
{
//获得图象并把图象转换为byte[]
HttpPostedFile upPhoto=UpPhoto.PostedFile;
int upPhotoLength=upPhoto.ContentLength;
byte[] PhotoArray=new Byte[upPhotoLength];
Stream PhotoStream=upPhoto.InputStream;
PhotoStream.Read(PhotoArray,0,upPhotoLength);

//连接数据库
SqlConnection conn=new SqlConnection();
conn.ConnectionString="Data Source=localhost;Database=test;User Id=sa;Pwd=sa";

SqlCommand cmd=new SqlCommand("UpdateImage",conn);
cmd.CommandType=CommandType.StoredProcedure;

cmd.Parameters.Add("@UpdateImage",SqlDbType.Image);
cmd.Parameters["@UpdateImage"].Value=PhotoArray;

//如果你希望不使用存储过程来添加图片把上面四句代码改为:
//string strSql="Insert into test(FImage) values(@FImage)";
//SqlCommand cmd=new SqlCommand(strSql,conn);
//cmd.Parameters.Add("@FImage",SqlDbType.Image);
//cmd.Parameters["@FImage"].Value=PhotoArray;

conn.Open();
cmd.ExecuteNonQuery();
conn.Close();
}

二,从SqlServer中读取并显示出来
在需要显示图片的地方添加如下代码:
<asp:image id="imgPhoto" runat="server" ImageUrl="ShowPhoto.aspx"></asp:image>

ShowPhoto.aspx主体代码:
private void Page_Load(object sender, System.EventArgs e)
{
if(!Page.IsPostBack)
{
SqlConnection conn=new SqlConnection()
conn.ConnectionString="Data Source=localhost;Database=test;User Id=sa;Pwd=sa";

string strSql="select * from test where id=2";//这里假设获取id为2的图片
SqlCommand cmd=new SqlCommand()
reader.Read();
Response.ContentType="application/octet-stream";
Response.BinaryWrite((Byte[])reader["FImage"]);
Response.End();
reader.Close();
}
}

3,在winform中将图片存入sqlserver,并从sqlserver中读取并显示在picturebox中

1,存入sqlserver
数据库结构和使用的存储过过程,同上面的一样
1.1,在窗体中加一个OpenFileDialog控件,命名为ofdSelectPic
1.2,在窗体上添加一个打开文件按钮,添加如下单击事件代码:
Stream ms;
byte[] picbyte;
//ofdSelectPic.ShowDialog();
if (ofdSelectPic.ShowDialog()==DialogResult.OK)
{
if ((ms=ofdSelectPic.OpenFile())!=null)
{
//MessageBox.Show("ok");
picbyte=new byte[ms.Length];
ms.Position=0;
ms.Read(picbyte,0,Convert.ToInt32(ms.Length));
//MessageBox.Show("读取完毕!");

//连接数据库
SqlConnection conn=new SqlConnection();
conn.ConnectionString="Data Source=localhost;Database=test;User Id=sa;Pwd=sa";

SqlCommand cmd=new SqlCommand("UpdateImage",conn);
cmd.CommandType=CommandType.StoredProcedure;

cmd.Parameters.Add("@UpdateImage",SqlDbType.Image);
cmd.Parameters["@UpdateImage"].Value=picbyte;

conn.Open();
cmd.ExecuteNonQuery();
conn.Close();

ms.Close();
}
}

2,读取并显示在picturebox中
2.1 添加一个picturebox,名为ptbShow
2.2 添加一个按钮,添加如下响应事件:
SqlConnection conn=new SqlConnection();
conn.ConnectionString="Data Source=localhost;Database=test;User Id=sa;Pwd=sa";

string strSql="select FImage from test where id=1";

SqlCommand cmd=new SqlCommand(strSql,conn);

conn.Open();
SqlDataReader reader=cmd.ExecuteReader();
reader.Read();

MemoryStream ms=new MemoryStream((byte[])reader["FImage"]);

Image image=Image.FromStream(ms,true);

reader.Close();
conn.Close();

ptbShow.Image=image;
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