behonest Posted March 9, 2009 Share Posted March 9, 2009 Dear Sir, i have been using xampp local server.There are availed mysql,phpmyadmin and more.i want to make a table to add information in it and to connect it with php.Do i need to download any mysql editor?please guide me. Link to comment Share on other sites More sharing options...
jlhaslip Posted March 9, 2009 Share Posted March 9, 2009 no, you can use phpmyadmin to add a database, a table, and even the data to the table rows. No need for a mysql editor (never heard of one)MySql also has its own MySqlAdmin software that is a lot like phpmyadmin. A graphical user interface is all it is, really. Link to comment Share on other sites More sharing options...
behonest Posted March 11, 2009 Author Share Posted March 11, 2009 yes there is a phpmyadmin server in XAMPP.So i suppose i dont need it anymore.Please make it clear for me.Do i need a mysql editor to make tables etc? or there is a mysql editor available in phpmyadmin? Thanks! Link to comment Share on other sites More sharing options...
behonest Posted March 18, 2009 Author Share Posted March 18, 2009 Dear Sir,recenlty,i am working on USER/LOGIN page of a web.i have made a database named as "login" with table name "form" and there r three fields such as "id","user" and "pass".Also there are two records.i am using the following php scripting to connect and activate user/login page.<?php session_start(); ?><html><body><form action='login.php?login=yes' method=post>Username: <input type=text name='user' /> <br/>Password: <input type=password name='pass' /> <br/><input type=submit value='Go' /></form> <?php$user=$_POST['user'];$pass=$_POST['pass'];$login=$_GET['login'];if ($login=='yes'){$con=mysql_connect('localhost','root', '');mysql_select_db('login');$get=mysql_query("SELECT count(id) FROM form WHERE user='$user' AND pass='$pass'"); $result=mysql_result($get, 0);mysql_close($con);if($result=1) echo "Login Failure";else {echo "login success";$_SESSION ['user']=$user;};};?></body></html>The problem is that when i enter the same user and password in input fields of web page it give error such as "LOGIN FAILURE.can u please guide me where i am wrong???? Link to comment Share on other sites More sharing options...
justsomeguy Posted March 18, 2009 Share Posted March 18, 2009 It looks like you're giving the error if the user was found. That should probably be the other way around. Link to comment Share on other sites More sharing options...
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