Cannot create PoolableConnectionFactory (Access de

2020-03-14 02:35发布

I have been trying to set up a DatabaseConnectionPool for a web app for the last couple of days with no success. I have read the relevant sections of the Tomcat docs and a great deal around the subject and think I'm doing everything right, but obviously not because I keep on getting the following error:

Cannot create PoolableConnectionFactory (Access denied for user ''@'localhost' (using password: YES))

I'm not getting the error when I start Tomcat running, but when I try to run the following servlet:

package twittersearch.web;
import javax.servlet.*;
import javax.servlet.http.*;
import java.io.*;
import java.util.*;
import java.sql.*;
import javax.sql.*;
import javax.naming.*;
import twittersearch.model.*;
public class ConPoolTest extends HttpServlet {

    public void doGet(HttpServletRequest request,
                  HttpServletResponse response) 
                  throws IOException, ServletException {
        Context ctx = null;
        DataSource ds = null;
        Connection conn = null;
        try {
            ctx = new InitialContext();
            ds = (DataSource)ctx.lookup("java:comp/env/jdbc/twittersearchdb");
            conn = ds.getConnection();
            if(conn != null) {
                System.out.println("have a connection from the pool");
            }
        } catch(SQLException e) {
            e.printStackTrace();
        } catch(NamingException e) {
            e.printStackTrace();
        } finally {
            try {
                if(conn!=null) {
                    conn.close();
                }
            } catch(SQLException e) {
                e.printStackTrace();
            }
        }
    }
}

The context for the webapp is:

<?xml version='1.0' encoding='utf-8'?>
<Context>

    <!-- Configure a JDBC DataSource for the user database -->
 <Resource name="jdbc/twittersearchdb" 
           type="javax.sql.DataSource" 
     auth="Container" 
     user="root" 
     password="mypwd" 
     driverClassName="com.mysql.jdbc.Driver" 
     url="jdbc:mysql://localhost:3306/twitter" 
     maxActive="8" 
     maxIdle="4"/>


</Context>

What I really don't understand is why the error isn't saying that the access is denied to 'root'@'localhost' when I've specified that that's the user.

What I've tried:

Do I have a duplicate context.xml? No. I deleted the default on in Tomcat 6.0/conf. I tried putting a context.xml in [mywebappname]/META-INF/context.xml but not only did this not work, but resulted in the creation of a file named TwitterSearch.xml which was autogenerated and put in the Tomcat 6.0/conf/Catalina/localhost directory. So now I'm just editing that one and thats the only one I have.

Is it the version of Tomcat? Well, I completely reinstalled the latest version of Tomcat 6.0 so I don't think it's that.

Are we missing some jars? I have the tomcat-dbcp.jar, jconnector.jar and all the other ones that I think I'm meant to have in the Tomcat 6.0/lib directory.

When I look at the passwords in the MySQL database they seem to have been coded for security purposes into a long alpha-numeric string. Is this normal? Should I have this in my context.xml or just the normal password?

I really don't know how I can sort this out and would really appreciate some expert advice.

Many thanks in advance.

Joe

2条回答
做个烂人
2楼-- · 2020-03-14 03:21

In your configuration context.xml for the webapp you have to change user="root" by username="root"

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狗以群分
3楼-- · 2020-03-14 03:34

Your setup looks fine. This looks to purly be a permissions problem.

You need to grant that user access in mysql. While Java will connect to localhost, it will do so using tcp/ip - however in mysql localhost and 127.0.0.1 have different meaning. Issuing this SQL should do the trick.

grant all on twitter.* to 'root'@'127.0.0.1' identified by 'mypwd';

That assumes Java resolves 'localhost' to 127.0.0.1, if things still doesn't work, you could try changing your connection string to "jdbc:mysql://127.0.0.1:3306/twitter"

Should I have this in my context.xml or just the normal password?

As you have it now, the plaintext password.

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