Unable to Display Image from database at runtime

den 2005

Hi everyone,

Here is code working on..Trying to insert record with a column with Image
or VarBinary datatype in sql database from a existing jpeg image file, then
retrieve this image from database and display it in a Image web control
dynamically(at runtime).

The process after being displayed in the web control, user click insert/add
button, it converts the image(jpeg) file to bytes[] and store it the database
with Image or VarBinary Datatype. While in retrieval, I get the specific
record, store the Image or VarBinary Data in local byte[] variable, and then
convert this to a image (jpeg) file and used the filepath to display file.

The problem still unable to display image and create image file.

Anyone?

Thanks in Advanced...

den2005
--
MCP Year 2005, Philippines

Aug 2 '06 #1

Augustin Prasanna

Hi Den,

just wanted to know why you store the images as such in the database.

I believe, it is recommended to save images in file system and store the
filepaths in database rather than storing the entire image in database.

Augustin

"den 2005" wrote:

Hi everyone,

Here is code working on..Trying to insert record with a column with Image
or VarBinary datatype in sql database from a existing jpeg image file, then
retrieve this image from database and display it in a Image web control
dynamically(at runtime).

The process after being displayed in the web control, user click insert/add
button, it converts the image(jpeg) file to bytes[] and store it the database
with Image or VarBinary Datatype. While in retrieval, I get the specific
record, store the Image or VarBinary Data in local byte[] variable, and then
convert this to a image (jpeg) file and used the filepath to display file.

The problem still unable to display image and create image file.

Expand|Select|Wrap|Line Numbers

1. //Get Image Record Stored and display them
2.     private void GetImageRecord()
3.     {
4.         if (CheckFields(2))
5.         {
6.             string sqlText = "Select * From PictureImages Where ImageName =
7. @ImgName";
8.             SqlCommand cmd = new SqlCommand(sqlText, conn);
9.             cmd.Parameters.Add("@ImgName", SqlDbType.VarChar, 20,
10. "ImageName").Value = txtImgName.Text.Trim();
11.             if (conn.State == ConnectionState.Closed)
12.                 conn.Open();
14.             SqlDataReader dr = cmd.ExecuteReader();
15.             string filePath = Request.PhysicalApplicationPath +
16. txtImgName.Text + ".jpg";
17.             FileStream fs = new FileStream(filePath, FileMode.Create,
18. FileAccess.Write);
20.             byte[] storedByte = null;
21.             //byte storedByte = null;
22.             //MemoryStream ms = new MemoryStream(storedByte);
23.             //StreamWriter sw = new StreamWriter(filePath);
25.             while (dr.Read())
26.             {
27.                 lblImgNo.Text = dr["ImageNo"].ToString();
28.                 txtImgDesc.Text = dr["Description"].ToString();
29.                 txtImgUrl.Text = dr["ImageUrl"].ToString();
30.                 storedByte = (byte[])dr["ImageData"];
31.                 //long lRes = dr.GetBytes(5,0,storedByte,0,100);
32.             }
33.             MemoryStream ms = new MemoryStream(storedByte, 0,
34. storedByte.Length);
35.             System.Drawing.Image img = System.Drawing.Image.FromStream(ms);
36. <<--- Error Occurred Here
37.             System.Drawing.Image otherImg = img.GetThumbnailImage(100, 100,
38. null, new IntPtr());
39.             //otherImg.Save(ms, img.RawFormat);
40.             otherImg.Save(filePath, System.Drawing.Imaging.ImageFormat.Jpeg);
42.             //byte[] byteImg = new byte[ms.Length];
43.             //ms.Position = 0;
44.             //ms.Read(byteImg, 0, byteImg.Length);
46.             //System.Drawing.Image img = System.Drawing.Image.FromStream(new
47. MemoryStream(storedByte, 0, storedByte.Length));
48.             //img.Save(filePath);
50.             //fs.Write(storedByte, 0, storedByte.Length);
51.             //fs.Flush();
52.             //fs.Close();
53.             if (File.Exists(filePath))
54.                 PicImg2.ImageUrl = filePath;
56.             if (conn.State == ConnectionState.Open)
57.                 conn.Close();
58.         }
59.     }
61. private void InsertProcess()
62.     {
63.         if (CheckFields(1))
64.         {
65.             FileStream fs = new
66. FileStream(txtImgFile.Text.Trim(),FileMode.OpenOrCreate,FileAccess.Read);
68.             byte[] imgByte = new byte[fs.Length];
69.             fs.Read(imgByte,0,(int)fs.Length);
70.             fs.Close();
71.             //Testing storing image as binary in database
73.             string sqlText = "Insert PictureImages
74. Values(@ImgName,@Desc,@Url,@Img)";
75.             SqlCommand cmd = new SqlCommand(sqlText,conn);
76.             cmd.Parameters.Add("@ImgName", SqlDbType.VarChar, 20,
77. "ImageName").Value = txtImgName.Text.Trim();
78.             cmd.Parameters.Add("@Desc", SqlDbType.VarChar, 100,
79. "Description").Value = txtImgDesc.Text.Trim();
80.             cmd.Parameters.Add("@Url", SqlDbType.VarChar, 50,
81. "ImageUrl").Value = txtImgUrl.Text.Trim();
82.             cmd.Parameters.Add("@Img", SqlDbType.Image, 16,
83. "ImageData").Value = imgByte;
84.             if (conn.State == ConnectionState.Closed)
85.                 conn.Open();
87.             int iRes = cmd.ExecuteNonQuery();
89.             if (conn.State == ConnectionState.Open)
90.                 conn.Close();
92.             if (iRes 0)
93.                 lblMessage.Text = "Successfully Added New Record.";
94.             else
95.                 lblMessage.Text = "Failed to add new record.";
96.         }
97.     }
99.     private void InsertProcess2()
100.     {
101.         if (CheckFields(0))
102.         {
103.             HttpPostedFile imgFile = txtFile1.PostedFile;
104.             int imgLen = imgFile.ContentLength;
105.             string imgType = imgFile.ContentType;
106.             byte[] imgByte = new byte[imgLen];
107.             imgFile.InputStream.Read(imgByte, 0, imgLen);
109.             //Testing storing image as binary in database
111.             string sqlText = "Insert PictureImages
112. Values(@ImgName,@Desc,@Url,@Img)";
113.             SqlCommand cmd = new SqlCommand(sqlText, conn);
114.             cmd.Parameters.Add("@ImgName", SqlDbType.VarChar, 20,
115. "ImageName").Value = txtImgName.Text.Trim();
116.             cmd.Parameters.Add("@Desc", SqlDbType.VarChar, 100,
117. "Description").Value = txtImgDesc.Text.Trim();
118.             cmd.Parameters.Add("@Url", SqlDbType.VarChar, 50,
119. "ImageUrl").Value = txtImgUrl.Text.Trim();
120.             cmd.Parameters.Add("@Img", SqlDbType.Image, 16,
121. "ImageData").Value = imgByte;
122.             if (conn.State == ConnectionState.Closed)
123.                 conn.Open();
125.             int iRes = cmd.ExecuteNonQuery();
127.             if (conn.State == ConnectionState.Open)
128.                 conn.Close();
130.             if (iRes 0)
131.                 lblMessage.Text = "Successfully Added New Record.";
132.             else
133.                 lblMessage.Text = "Failed to add new record.";
134.         }
135.     }
136.  

Anyone?

Thanks in Advanced...

den2005
--
MCP Year 2005, Philippines

Aug 2 '06 #2

Winista

This article should be of help..

http://www.netomatix.com/development...splayBlob.aspx

"den 2005" <de*****@discussions.microsoft.comwrote in message
news:F1**********************************@microsof t.com...

Hi everyone,

Here is code working on..Trying to insert record with a column with Image
or VarBinary datatype in sql database from a existing jpeg image file,
then
retrieve this image from database and display it in a Image web control
dynamically(at runtime).

The process after being displayed in the web control, user click
insert/add
button, it converts the image(jpeg) file to bytes[] and store it the
database
with Image or VarBinary Datatype. While in retrieval, I get the specific
record, store the Image or VarBinary Data in local byte[] variable, and
then
convert this to a image (jpeg) file and used the filepath to display file.

The problem still unable to display image and create image file.

Expand|Select|Wrap|Line Numbers

1. //Get Image Record Stored and display them
2.    private void GetImageRecord()
3.    {
4.        if (CheckFields(2))
5.        {
6.            string sqlText = "Select * From PictureImages Where ImageName =
7. @ImgName";
8.            SqlCommand cmd = new SqlCommand(sqlText, conn);
9.            cmd.Parameters.Add("@ImgName", SqlDbType.VarChar, 20,
10. "ImageName").Value = txtImgName.Text.Trim();
11.            if (conn.State == ConnectionState.Closed)
12.                conn.Open();
14.            SqlDataReader dr = cmd.ExecuteReader();
15.            string filePath = Request.PhysicalApplicationPath +
16. txtImgName.Text + ".jpg";
17.            FileStream fs = new FileStream(filePath, FileMode.Create,
18. FileAccess.Write);
20.            byte[] storedByte = null;
21.            //byte storedByte = null;
22.            //MemoryStream ms = new MemoryStream(storedByte);
23.            //StreamWriter sw = new StreamWriter(filePath);
25.            while (dr.Read())
26.            {
27.                lblImgNo.Text = dr["ImageNo"].ToString();
28.                txtImgDesc.Text = dr["Description"].ToString();
29.                txtImgUrl.Text = dr["ImageUrl"].ToString();
30.                storedByte = (byte[])dr["ImageData"];
31.                //long lRes = dr.GetBytes(5,0,storedByte,0,100);
32.            }
33.            MemoryStream ms = new MemoryStream(storedByte, 0,
34. storedByte.Length);
35.            System.Drawing.Image img = System.Drawing.Image.FromStream(ms);
36. <<--- Error Occurred Here
37.            System.Drawing.Image otherImg = img.GetThumbnailImage(100, 100,
38. null, new IntPtr());
39.            //otherImg.Save(ms, img.RawFormat);
40.            otherImg.Save(filePath,
41. System.Drawing.Imaging.ImageFormat.Jpeg);
43.            //byte[] byteImg = new byte[ms.Length];
44.            //ms.Position = 0;
45.            //ms.Read(byteImg, 0, byteImg.Length);
47.            //System.Drawing.Image img =
48. System.Drawing.Image.FromStream(new
49. MemoryStream(storedByte, 0, storedByte.Length));
50.            //img.Save(filePath);
52.            //fs.Write(storedByte, 0, storedByte.Length);
53.            //fs.Flush();
54.            //fs.Close();
55.            if (File.Exists(filePath))
56.                PicImg2.ImageUrl = filePath;
58.            if (conn.State == ConnectionState.Open)
59.                conn.Close();
60.        }
61.    }
63. private void InsertProcess()
64.    {
65.        if (CheckFields(1))
66.        {
67.            FileStream fs = new
68. FileStream(txtImgFile.Text.Trim(),FileMode.OpenOrCreate,FileAccess.Read);
70.            byte[] imgByte = new byte[fs.Length];
71.            fs.Read(imgByte,0,(int)fs.Length);
72.            fs.Close();
73.            //Testing storing image as binary in database
75.            string sqlText = "Insert PictureImages
76. Values(@ImgName,@Desc,@Url,@Img)";
77.            SqlCommand cmd = new SqlCommand(sqlText,conn);
78.            cmd.Parameters.Add("@ImgName", SqlDbType.VarChar, 20,
79. "ImageName").Value = txtImgName.Text.Trim();
80.            cmd.Parameters.Add("@Desc", SqlDbType.VarChar, 100,
81. "Description").Value = txtImgDesc.Text.Trim();
82.            cmd.Parameters.Add("@Url", SqlDbType.VarChar, 50,
83. "ImageUrl").Value = txtImgUrl.Text.Trim();
84.            cmd.Parameters.Add("@Img", SqlDbType.Image, 16,
85. "ImageData").Value = imgByte;
86.            if (conn.State == ConnectionState.Closed)
87.                conn.Open();
89.            int iRes = cmd.ExecuteNonQuery();
91.            if (conn.State == ConnectionState.Open)
92.                conn.Close();
94.            if (iRes 0)
95.                lblMessage.Text = "Successfully Added New Record.";
96.            else
97.                lblMessage.Text = "Failed to add new record.";
98.        }
99.    }
101.    private void InsertProcess2()
102.    {
103.        if (CheckFields(0))
104.        {
105.            HttpPostedFile imgFile = txtFile1.PostedFile;
106.            int imgLen = imgFile.ContentLength;
107.            string imgType = imgFile.ContentType;
108.            byte[] imgByte = new byte[imgLen];
109.            imgFile.InputStream.Read(imgByte, 0, imgLen);
111.            //Testing storing image as binary in database
113.            string sqlText = "Insert PictureImages
114. Values(@ImgName,@Desc,@Url,@Img)";
115.            SqlCommand cmd = new SqlCommand(sqlText, conn);
116.            cmd.Parameters.Add("@ImgName", SqlDbType.VarChar, 20,
117. "ImageName").Value = txtImgName.Text.Trim();
118.            cmd.Parameters.Add("@Desc", SqlDbType.VarChar, 100,
119. "Description").Value = txtImgDesc.Text.Trim();
120.            cmd.Parameters.Add("@Url", SqlDbType.VarChar, 50,
121. "ImageUrl").Value = txtImgUrl.Text.Trim();
122.            cmd.Parameters.Add("@Img", SqlDbType.Image, 16,
123. "ImageData").Value = imgByte;
124.            if (conn.State == ConnectionState.Closed)
125.                conn.Open();
127.            int iRes = cmd.ExecuteNonQuery();
129.            if (conn.State == ConnectionState.Open)
130.                conn.Close();
132.            if (iRes 0)
133.                lblMessage.Text = "Successfully Added New Record.";
134.            else
135.                lblMessage.Text = "Failed to add new record.";
136.        }
137.    }
138.  

Anyone?

Thanks in Advanced...

den2005
--
MCP Year 2005, Philippines

Aug 2 '06 #3

den 2005

Augustin,
That's the requirement what I am doing right now.

Winista,
Thanks for the link, but I am seen this approach, I am looking for
approach that converts stored sql binary/image datatype data back to an image
(jpeg) file.

dennis
--
MCP Year 2005, Philippines
"Winista" wrote:

This article should be of help..

http://www.netomatix.com/development...splayBlob.aspx

"den 2005" <de*****@discussions.microsoft.comwrote in message
news:F1**********************************@microsof t.com...

Hi everyone,

Here is code working on..Trying to insert record with a column with Image
or VarBinary datatype in sql database from a existing jpeg image file,
then
retrieve this image from database and display it in a Image web control
dynamically(at runtime).

The process after being displayed in the web control, user click
insert/add
button, it converts the image(jpeg) file to bytes[] and store it the
database
with Image or VarBinary Datatype. While in retrieval, I get the specific
record, store the Image or VarBinary Data in local byte[] variable, and
then
convert this to a image (jpeg) file and used the filepath to display file.

The problem still unable to display image and create image file.

Expand|Select|Wrap|Line Numbers

1.  //Get Image Record Stored and display them
2.     private void GetImageRecord()
3.     {
4.         if (CheckFields(2))
5.         {
6.             string sqlText = "Select * From PictureImages Where ImageName =
7.  @ImgName";
8.             SqlCommand cmd = new SqlCommand(sqlText, conn);
9.             cmd.Parameters.Add("@ImgName", SqlDbType.VarChar, 20,
10.  "ImageName").Value = txtImgName.Text.Trim();
11.             if (conn.State == ConnectionState.Closed)
12.                 conn.Open();
13.  
14.             SqlDataReader dr = cmd.ExecuteReader();
15.             string filePath = Request.PhysicalApplicationPath +
16.  txtImgName.Text + ".jpg";
17.             FileStream fs = new FileStream(filePath, FileMode.Create,
18.  FileAccess.Write);
19.  
20.             byte[] storedByte = null;
21.             //byte storedByte = null;
22.             //MemoryStream ms = new MemoryStream(storedByte);
23.             //StreamWriter sw = new StreamWriter(filePath);
24.  
25.             while (dr.Read())
26.             {
27.                 lblImgNo.Text = dr["ImageNo"].ToString();
28.                 txtImgDesc.Text = dr["Description"].ToString();
29.                 txtImgUrl.Text = dr["ImageUrl"].ToString();
30.                 storedByte = (byte[])dr["ImageData"];
31.                 //long lRes = dr.GetBytes(5,0,storedByte,0,100);
32.             }
33.             MemoryStream ms = new MemoryStream(storedByte, 0,
34.  storedByte.Length);
35.             System.Drawing.Image img = System.Drawing.Image.FromStream(ms);
36.  <<--- Error Occurred Here
37.             System.Drawing.Image otherImg = img.GetThumbnailImage(100, 100,
38.  null, new IntPtr());
39.             //otherImg.Save(ms, img.RawFormat);
40.             otherImg.Save(filePath,
41.  System.Drawing.Imaging.ImageFormat.Jpeg);
42.  
43.             //byte[] byteImg = new byte[ms.Length];
44.             //ms.Position = 0;
45.             //ms.Read(byteImg, 0, byteImg.Length);
46.  
47.             //System.Drawing.Image img =
48.  System.Drawing.Image.FromStream(new
49.  MemoryStream(storedByte, 0, storedByte.Length));
50.             //img.Save(filePath);
51.  
52.             //fs.Write(storedByte, 0, storedByte.Length);
53.             //fs.Flush();
54.             //fs.Close();
55.             if (File.Exists(filePath))
56.                 PicImg2.ImageUrl = filePath;
57.  
58.             if (conn.State == ConnectionState.Open)
59.                 conn.Close();
60.         }
61.     }
62.  
63.  private void InsertProcess()
64.     {
65.         if (CheckFields(1))
66.         {
67.             FileStream fs = new
68.  FileStream(txtImgFile.Text.Trim(),FileMode.OpenOrCreate,FileAccess.Read);
69.  
70.             byte[] imgByte = new byte[fs.Length];
71.             fs.Read(imgByte,0,(int)fs.Length);
72.             fs.Close();
73.             //Testing storing image as binary in database
74.  
75.             string sqlText = "Insert PictureImages
76.  Values(@ImgName,@Desc,@Url,@Img)";
77.             SqlCommand cmd = new SqlCommand(sqlText,conn);
78.             cmd.Parameters.Add("@ImgName", SqlDbType.VarChar, 20,
79.  "ImageName").Value = txtImgName.Text.Trim();
80.             cmd.Parameters.Add("@Desc", SqlDbType.VarChar, 100,
81.  "Description").Value = txtImgDesc.Text.Trim();
82.             cmd.Parameters.Add("@Url", SqlDbType.VarChar, 50,
83.  "ImageUrl").Value = txtImgUrl.Text.Trim();
84.             cmd.Parameters.Add("@Img", SqlDbType.Image, 16,
85.  "ImageData").Value = imgByte;
86.             if (conn.State == ConnectionState.Closed)
87.                 conn.Open();
88.  
89.             int iRes = cmd.ExecuteNonQuery();
90.  
91.             if (conn.State == ConnectionState.Open)
92.                 conn.Close();
93.  
94.             if (iRes 0)
95.                 lblMessage.Text = "Successfully Added New Record.";
96.             else
97.                 lblMessage.Text = "Failed to add new record.";
98.         }
99.     }
100.  
101.     private void InsertProcess2()
102.     {
103.         if (CheckFields(0))
104.         {
105.             HttpPostedFile imgFile = txtFile1.PostedFile;
106.             int imgLen = imgFile.ContentLength;
107.             string imgType = imgFile.ContentType;
108.             byte[] imgByte = new byte[imgLen];
109.             imgFile.InputStream.Read(imgByte, 0, imgLen);
110.  
111.             //Testing storing image as binary in database
112.  
113.             string sqlText = "Insert PictureImages
114.  Values(@ImgName,@Desc,@Url,@Img)";
115.             SqlCommand cmd = new SqlCommand(sqlText, conn);
116.             cmd.Parameters.Add("@ImgName", SqlDbType.VarChar, 20,
117.  "ImageName").Value = txtImgName.Text.Trim();
118.             cmd.Parameters.Add("@Desc", SqlDbType.VarChar, 100,
119.  "Description").Value = txtImgDesc.Text.Trim();
120.             cmd.Parameters.Add("@Url", SqlDbType.VarChar, 50,
121.  "ImageUrl").Value = txtImgUrl.Text.Trim();
122.             cmd.Parameters.Add("@Img", SqlDbType.Image, 16,
123.  "ImageData").Value = imgByte;
124.             if (conn.State == ConnectionState.Closed)
125.                 conn.Open();
126.  
127.             int iRes = cmd.ExecuteNonQuery();
128.  
129.             if (conn.State == ConnectionState.Open)
130.                 conn.Close();
131.  
132.             if (iRes 0)
133.                 lblMessage.Text = "Successfully Added New Record.";
134.             else
135.                 lblMessage.Text = "Failed to add new record.";
136.         }
137.     }
138.  

Anyone?

Thanks in Advanced...

den2005
--
MCP Year 2005, Philippines

Aug 3 '06 #4