How to retrieve and display images from a database in a JSP page?

Question

How can I retrieve and display images from a database in a JSP page?

Answer 1

I used SQL SERVER database and so the answer's code is in accordance. All you have to do is include an <img> tag in your jsp page and call a servlet from its src attribute like this

<img width="200" height="180" src="DisplayImage?ID=1">

Here 1 is unique id of image in database and ID is a variable. We receive value of this variable in servlet. In servlet code we take the binary stream input from correct column in table. That is your image is stored in which column. In my code I used third column because my images are stored as binary data in third column. After retrieving input stream data from table we read its content in an output stream so it can be written on screen. Here is it

import java.io.*;  
import java.sql.Connection;
import java.sql.ResultSet;
import java.sql.SQLException;
import java.sql.Statement;
import java.util.logging.Level;
import java.util.logging.Logger;
import javax.servlet.*;  
import javax.servlet.http.*;  
import model.ConnectionManager;
public class DisplayImage extends HttpServlet { 
    public void doGet(HttpServletRequest request,HttpServletResponse response)  
             throws IOException  
    { 
    Statement stmt=null;
    String sql=null;
    BufferedInputStream bin=null;
    BufferedOutputStream bout=null;
    InputStream in =null;

    response.setContentType("image/jpeg");  
    ServletOutputStream out;  
    out = response.getOutputStream();  
    Connection conn = ConnectionManager.getConnection();

    int ID = Integer.parseInt(request.getParameter("ID"));
        try {
            stmt = conn.createStatement();
            sql = "SELECT * FROM IMAGETABLE WHERE ID="+ID+"";
            ResultSet result = stmt.executeQuery(sql);
            if(result.next()){
                in=result.getBinaryStream(3);//Since my data was in third column of table.
            }
            bin = new BufferedInputStream(in);  
            bout = new BufferedOutputStream(out);  
            int ch=0;   
            while((ch=bin.read())!=-1)  
                {  
                bout.write(ch);  
            }  

        } catch (SQLException ex) {
            Logger.getLogger(DisplayImage.class.getName()).log(Level.SEVERE, null, ex);
        }finally{
        try{
            if(bin!=null)bin.close();  
            if(in!=null)in.close();  
            if(bout!=null)bout.close();  
            if(out!=null)out.close();
            if(conn!=null)conn.close();
        }catch(IOException | SQLException ex){
            System.out.println("Error : "+ex.getMessage());
        }
    }


    }  
}  

After the execution of your jsp or html file you will see the image on screen.

Answer 2

You can also create custom tag for displaying image.

1) create custom tag java class and tld file.

2) write logic to display image like conversion of byte[] to string by Base64.

so it is used for every image whether you are displaying only one image or multiple images in single jsp page.

Answer 3

Let's see in steps what should happen:

• JSP is basically a view technology which is supposed to generate HTML output.
• To display an image in HTML, you need the HTML <img> element.
• To let it locate an image, you need to specify its src attribute.
• The src attribute needs to point to a valid http:// URL and thus not a local disk file system path file:// as that would never work when the server and client run at physically different machines.
• The image URL needs to have the image identifier in either the request path (e.g. http://example.com/context/images/foo.png) or as request parameter (e.g. http://example.com/context/images?id=1).
• In JSP/Servlet world, you can let a Servlet listen on a certain URL pattern like /images/*, so that you can just execute some Java code on specific URL's.
• Images are binary data and are to be obtained as either a byte[] or InputStream from the DB, the JDBC API offers the ResultSet#getBytes() and ResultSet#getBinaryStream() for this, and JPA API offers @Lob for this.
• In the Servlet you can just write this byte[] or InputStream to the OutputStream of the response the usual Java IO way.
• The client side needs to be instructed that the data should be handled as an image, thus at least the Content-Type response header needs to be set as well. You can obtain the right one via ServletContext#getMimeType() based on image file extension which you can extend and/or override via <mime-mapping> in web.xml.

That should be it. It almost writes code itself. Let's start with HTML (in JSP):

<img src="${pageContext.request.contextPath}/images/foo.png">
<img src="${pageContext.request.contextPath}/images/bar.png">
<img src="${pageContext.request.contextPath}/images/baz.png">

You can if necessary also dynamically set src with EL while iterating using JSTL:

<c:forEach items="${imagenames}" var="imagename">
    <img src="${pageContext.request.contextPath}/images/${imagename}">
</c:forEach>

Then define/create a servlet which listens on GET requests on URL pattern of /images/*, the below example uses plain vanilla JDBC for the job:

@WebServlet("/images/*")
public class ImageServlet extends HttpServlet {

    // content=blob, name=varchar(255) UNIQUE.
    private static final String SQL_FIND = "SELECT content FROM Image WHERE name = ?";

    @Resource(name="jdbc/yourDB") // For Tomcat, define as <Resource> in context.xml and declare as <resource-ref> in web.xml.
    private DataSource dataSource;
    
    @Override
    protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        String imageName = request.getPathInfo().substring(1); // Returns "foo.png".

        try (Connection connection = dataSource.getConnection(); PreparedStatement statement = connection.prepareStatement(SQL_FIND)) {
            statement.setString(1, imageName);
            
            try (ResultSet resultSet = statement.executeQuery()) {
                if (resultSet.next()) {
                    byte[] content = resultSet.getBytes("content");
                    response.setContentType(getServletContext().getMimeType(imageName));
                    response.setContentLength(content.length);
                    response.getOutputStream().write(content);
                } else {
                    response.sendError(HttpServletResponse.SC_NOT_FOUND); // 404.
                }
            }
        } catch (SQLException e) {
            throw new ServletException("Something failed at SQL/DB level.", e);
        }
    }

}

That's it. In case you worry about HEAD and caching headers and properly responding on those requests, use this abstract template for static resource servlet.

See also:

• How should I connect to JDBC database / datasource in a servlet based application?
• How to upload an image and save it in database?
• Simplest way to serve static data from outside the application server in a Java web application

Answer 4

I suggest you address that as two problems. There are several questions and answer related to both.

1. How to load blob from MySQL

   See for instance Retrieve image stored as blob

2. How to display image dynamically

   See for instance Show thumbnail dynamically

Answer 5

Try to flush and close the output stream if it does not display. Blob image = rs.getBlob(ImageColName); InputStream in = image.getBinaryStream(); // Output the blob to the HttpServletResponse response.setContentType("image/jpeg"); BufferedOutputStream o = new BufferedOutputStream(response.getOutputStream());

    byte by[] = new byte[32768];
    int index = in.read(by, 0, 32768);
    while (index != -1) {
        o.write(by, 0, index);
        index = in.read(by, 0, 32768);
    }
    o.flush();
    o.close();