saran.tvmalai Posted September 8, 2010 Share Posted September 8, 2010 Dear All, I have login script already. i have two database one is users and another one is reference in both user_name is common. In this if a user login and search the refid means that user datas onle possible to retrieve. for others datas they cant access. i wrote the below code but it display all user datas. kindly help me <?php session_start(); if (!$_SESSION["user_name"]) { // User not logged in, redirect to login page Header("Location: login.php"); / } // Member only content // ... $con = mysql_connect('localhost','root',''); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("login", $con); $user_name = $_POST['user_name']; $query = "select * from reference,users where reference.user_name=users.user_name and reference.refid='$refid'"; $result = mysql_query($query) or trigger_error('MySQL encountered a problem<br />Error: ' . mysql_error() . '<br />Query: ' . $query); while($row = mysql_fetch_array($result)) { echo $row['refid']; echo $row['origin']; echo $row['dest']; echo $row['date']; echo $row['exdate']; echo $row['user_name']; } echo "<p><a href=\"logout.php\">Click here to logout!</a></p>"; ?> <html> <form method="post" action="final.php"> Ref Id:<input type="text" name="refid"> <input type="submit" value="submit" name="submit"> </html> Quote Link to comment Share on other sites More sharing options...
HaLo2FrEeEk Posted September 8, 2010 Share Posted September 8, 2010 You call for a variable named $refid in your query, yet this variable doesn't exist before the query is executed, thus the variable's value will be empty, making your query look like this: SELECT * FROM reference.users WHERE reference.user_name = users.user_name AND reference.refid = '' I would also recommend that you capitilize the keywords in your query, like SELECT, FROM, WHERE, and AND. If makes t easier to read. Anyway, so right now you're selecting everything from the reference.users table where reference.user_name is equal to users.user_name and where the reference.refid is empty. Unless I'm missing something. I think the problem is here: WHERE reference.user_name = users.user_name You're telling it to select everything from the reference users table where the names are the same, which you said they were. You collect the POST variable user_name, why not use that in your query? $user_name = $_POST['user_name']; $refid = $_POST['refid']; $query = "SELECT * FROM reference.users WHERE reference.user_name = '".$user_name."' AND reference.refid = '".$refid."'"; Without more background into what exactly you're doing (I'm having difficulties understanding your description,) I can't help much more. Quote Link to comment Share on other sites More sharing options...
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