Parameter is not valid.

Question

i have stored jpeg images in database table. now i want to retrieve them. here is the code
MemoryStream stream = new MemoryStream();
            try
            {
                  cnn.Open();
                  cmd = new
                     SqlCommand("select pics from images", cnn);
                  byte[] image = (byte[])cmd.ExecuteScalar();

                  stream.Write(image, 0, image.Length);
                  Bitmap bitmap = new Bitmap(stream);
                  Response.ContentType = "image/gif";
                  bitmap.Save(Response.OutputStream,System.Drawing.Imaging.ImageFormat.Bmp);
            }
            finally
            {
                  cnn.Close();
                  stream.Close();
            }
but when i run the application it gives me the following error
Parameter is not valid.
Bitmap bitmap = new Bitmap(stream);
can somebody tells me what i need to correct.

Vulpes · Answer

Something I've just noticed is that you said in your opening post that the image type is jpeg whereas you're using a content type of image/gif. If that's the case  I'd replace: Response.ContentType = image/gif; with: Response.ContentType = image/JPEG;

saifullah khan · Answer

YOU R rite sir. it workd now. thank you soooo much. you are the great.

saifullah khan · Answer

now it worked but i shows the following out put instead of image its the output

Javeed M Shaikh · Answer

can you post the sample data from the pics column from the database just a few 50 or 60 characters from the begining of the data. Also did you try to add this line in the upload as suggested by Vulpes: stream.Seek(0 SeekOrigin.Begin); it should be like this in your code: FileUpload1. FileContent .Seek(0 SeekOrigin.Begin); FileUpload1.PostedFile.InputStream.Read(pic 0 len);

saifullah khan · Answer

i changed the query it uploaded the image. i checked it in database table in the pics colm i written instead of pics name. and after the exution it gives the same error

Vulpes · Answer

Try changing this line: cmd = new SqlCommand(insert into images values(' + pic + ') cnn); to these: cmd = new SqlCommand(insert into images values(@picture) cnn); cmd.Parameters.Add(@picture pic);

saifullah khan · Answer

here is the code for uploading image: protected void Button1_Click(object sender EventArgs e) { int len = FileUpload1.PostedFile.ContentLength; byte[] pic = new byte[len]; FileUpload1.PostedFile.InputStream.Read(pic 0 len); cnn.Open(); cmd = new SqlCommand(insert into images values(' + pic + ') cnn); cmd.ExecuteNonQuery(); } here is the code for retriving from db. protected void Page_Load(object sender EventArgs e) { MemoryStream stream = new MemoryStream(); try { cnn.Open(); cmd = new SqlCommand(select pics from images cnn); byte[] image = (byte[])cmd.ExecuteScalar(); stream.Write(image 0 image.Length); Image imag =Image.FromStream(stream); Response.ContentType = image/gif; imag.Save(Response.OutputStream System.Drawing.Imaging.ImageFormat.Bmp); } finally { cnn.Close(); stream.Close(); }

Javeed M Shaikh · Answer

Hi Saifullah what Vulpes is trying to say in his previous post is for you to check if you are doing the following when you are inserting the image in the database: stream.Seek(0 SeekOrigin.Begin); Please check the above in your image upload code.

saifullah khan · Answer

still the same error Parameter is not valid. Image imag =Image.FromStream(stream);

Vulpes · Answer

I'd see if it makes any difference if you use Image.FromStream rather than the Bitmap constructor: MemoryStream stream = new MemoryStream(); try { cnn.Open(); cmd = new SqlCommand(select pics from images cnn); byte[] image = (byte[])cmd.ExecuteScalar(); stream.Write(image 0 image.Length); Image image = Image.FromStream(stream); Response.ContentType = image/gif; image.Save(Response.OutputStreamSystem.Drawing.Imaging.ImageFormat.Bmp); } finally { cnn.Close(); stream.Close(); }

saifullah khan · Answer

i have used your given line but it still gives an error

Vulpes · Answer

I assume you're confident that the images in the database are valid. Have you done anything with the MemoryStream before entering the try block? If you have you'll need to set the stream back to the beginning before writing the image to it: stream.Seek(0 SeekOrigin.Begin);