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With the effort you're putting into this, you could just learn the course material properly and pass the exam legitimately.
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Programming is work, it isn't finger painting. Luc Pattyn
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Hello everyone! I have been trying for three days on Eclipse 4.2 version.I have done all the work regarding Tomcat installation and configuration and JDBC 5.1 plugin installation.The problem is this;when I try to make a connection to the MySql server,it rejects although I provide it with proper info such as port no 3306,database "test".By the way I have no password for mysql server.Any help plz.thnx
Always believe in God and never let misfortunes make you give up
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Hi,
That is pretty vague; ou could Google for examples and find things such as this[^]; or you could show us the relevant parts of your code and the exact exception information you are getting, if you want specific help.
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countered the downvote 
V.
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Write a simple program that does NOTHING but a jdbc connection.
That means that it will NOT have any GUI components.
That means that it will NOT have any Tomcat/JEE specific code.
Set the connection string.
Run it.
Look at the exception.
Most likely causes are incorrect connection string, no access to the database and/or that the database is not running.
Only the first is solvable in java.
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Does the MySQL server run on the same machine? If no, configure it to accept connections from different computers.
Do you want to login as root from a different computer? Also that must be configured.
Have you tried a user with password?
Can you connect to your MySQL server with a MySQL tool, e.g. MySQLAdministrator.exe or MySQLQueryBrowser.exe?
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Yes the Sql server is in a same machine and I have also allowed remote connections and firewall access rules for both Tomcat and MySql.I can easily access MySql from Workbench as well as ping from connection builder tool from eclipse.The problem arises when I use com.mysql.jdbc.Driver in driver name in my code
Always believe in God and never let misfortunes make you give up
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which port do you use? is it already taken? what's the Exception saying?
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Dear fellows! I have tried everything.Made a program in notepad compiled it and when ran ame error then I used this command on elevated command prompt C:\>javap mysql.connector.java-5.1.20 and the result " Class ---- is unavailable".I run MySql over 3306 port it is working jhakass(means great) as i have 4GB RAM and CORE Processors 500 GB HD 32 Bit Win 7 Ultimate
Always believe in God and never let misfortunes make you give up
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adeel1981 wrote: javap
"javap" has nothing to do with compiling nor running a java application.
adeel1981 wrote: Class ---- is unavailable"
Class path problem or class naming problem. Which is a general problem that has nothing specifically to do with JDBC.
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I have tried everything.Made a program in notepad
yeah for sure. Why don't you use a IDE? Eclipse and Netbeans are free, there is no excuse for not using them.
You have NOT tried everything when you are not even using a proper IDE.
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Dude I hav alreasy tried Eclipse 4.2 installed all plugins and softwares,installed tomcat 7.
Always believe in God and never let misfortunes make you give up
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Why do you not address the actual problem you have and investigate why you cannot connect to MySQL? Either your parameters are wrong (you say they are correct) or the driver is not accessible on your system. You have already demonstrated that using javap your class does not exist in the local directory, so that should be a starting point to find out where it is.
Programming is work, it isn't finger painting. Luc Pattyn
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There has nothing to do with programming here. mysql.jdbc.driver is not being loaded.I am very first to java and apache and mysql but have done everything.I have personally installed it in java main directory and the newest version.When I installed it in Tomcat.Eclipse is easily making a connection to it
Always believe in God and never let misfortunes make you give up
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adeel1981 wrote: mysql.jdbc.driver is not being loaded.
OK, then you need to investigate why that is so. But from the information you have given here it is anyone's guess what is going wrong.
Programming is work, it isn't finger painting. Luc Pattyn
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In that case you should hope for heavenly help.
I can just tell you to use the tools that you are given.
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I made a list of string and store data inside it ,then i made a code on this list after that i have results....my question is i want to take this results and store it in list of string . how i do that????
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Create another list and add each result to it.
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Are you trying to keep the results on same List? Because I wonder, if you have already create a List and store data, you should not stuck with creating another List and store the results in that new List. Could you please clarify, so I can help you further if required. 
I appreciate your help all the time...
CodingLover
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Hi to all,
public class xyz {
public static int gUserID = 1;
}
When 1st user Login gUserID value is = 5
and second user login gUserID value is = 7
Problem is 1st user refresh or re-direct to another page gUserID is changing to 7.
How to set Global variable for session or any solution to above problem.
Thanks.
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How is the variable being updated?
The single variable is used by every instance of the class. If you want each instance to have a userId then you need something like this:
public class Test {
private static int nextUserId = 1;
private int userId;
public Test() {
this.userId = Test.nextUserId++;
}
public int getUserId() {
return this.userId;
}
}
Panic, Chaos, Destruction. My work here is done.
Drink. Get drunk. Fall over - P O'H
OK, I will win to day or my name isn't Ethel Crudacre! - DD Ethel Crudacre
I cannot live by bread alone. Bacon and ketchup are needed as well. - Trollslayer
Have a bit more patience with newbies. Of course some of them act dumb - they're often *students*, for heaven's sake - Terry Pratchett
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Hi Nagy Vilmos,
I m update gUserID from database. As soon as login to application. I m setting corresponding user id to that variable and I m utilizing that variable in complete project as global variable.
here I m facing problem. As soon as 2nd person login The Value is reset (2nd user id is updating).
I m trying your method now Thanks.
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