saran.tvmalai Posted September 1, 2010 Share Posted September 1, 2010 Dear All, Am new to PHP. now i want to online cargo tracking system in php. in that i create the reference id for each user to track the shipments. in that user must login and track the shipments. now i need the code for the specific user only view they shipments only. the other user can't be check the other shipments. so i need how to link the user and reference id. kindly advice me. Quote Link to comment Share on other sites More sharing options...
Rifts Posted September 1, 2010 Share Posted September 1, 2010 What? Can you explain more please and maybe give some examples of what you've done. Quote Link to comment Share on other sites More sharing options...
saran.tvmalai Posted September 1, 2010 Author Share Posted September 1, 2010 for example there are two users. user1 and user2. both have different reference id. user1 have the reference id is 001 user2 have the reference id is 002 if user1 login and track his shipment means they only possible to view his ref id 001. user1 never possible to check ref id 002. same as user2 never possible to check the ref id 001. Quote Link to comment Share on other sites More sharing options...
Rifts Posted September 1, 2010 Share Posted September 1, 2010 are you using a database? Do you have any coding you have started? Quote Link to comment Share on other sites More sharing options...
saran.tvmalai Posted September 1, 2010 Author Share Posted September 1, 2010 ya mysql is my database Quote Link to comment Share on other sites More sharing options...
Rifts Posted September 1, 2010 Share Posted September 1, 2010 you are going to need a login page and a member page for the login page use something like <html> <form method="post" action="memberpage.php"> Name:<input type="text" name="user"> Password:<input type="password" name="pass"> <input type="submit" value="submit" name="submit"> then in your member page you will have something like $con = mysql_connect("localhost","login","pass"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("my_db", $con); $user = $_POST['user']; $pass = $_POST['pass']; $result = mysql_query("SELECT * FROM yourtable WHERE user = $user, pass = $pass"); while($row = mysql_fetch_array($result)) { echo $row['refID']; } that should give you a good start Quote Link to comment Share on other sites More sharing options...
saran.tvmalai Posted September 1, 2010 Author Share Posted September 1, 2010 thanks alot Quote Link to comment Share on other sites More sharing options...
saran.tvmalai Posted September 1, 2010 Author Share Posted September 1, 2010 i got this error Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource Quote Link to comment Share on other sites More sharing options...
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