Ell20

Thanks for the help but it dosent seem to have solved the problem, still not menu or test message appearing.

Hey, Im attempting to make a special menu for the Admin of the site which only they can see. Im getting no errors but it dosent seem to be working. //Function function getUserInfo($user_id,$fields){ $sql="select $fields from `users` where user_id='$user_id' "; $query=mysql_query($sql); $rows=mysql_num_rows($query); if($rows==0){return;} $rs=mysql_fetch_assoc($query); return $rs; } //Code <?php require_once ('../mysql_connect.php'); include_once ('includes/functions.php'); $user_id = $_SESSION['user_id']; $userinfo=getUserInfo($user_id, '*'); $member_type = $userinfo['member_type']; $sql = "SELECT * FROM users WHERE member_type='$member_type'"; $result = @mysql_query($sql); if($result['member_type'] == "Admin"){ ?> Message Will Appear If Admin <a href="admin.php">Admin Test</a><br /> <? } ?> Any help would be appreciated Cheers

Thank you very much, seems to be working perfectly!!

This makes up my full query: $query = "INSERT INTO users (username, club_id, member_type, first_name, last_name, email, password, registration_date) VALUES ('$u', $_POST['clubdrop'], 'Member', '$fn', '$ln', '$e', PASSWORD('$p'), NOW() )"; $result = @mysql_query ($query); However by adding in the $_POST['clubdrop'] section has caused an error: Parse error: parse error, expecting `T_STRING' or `T_VARIABLE' or `T_NUM_STRING' in c:\program files\easyphp1-8\www\html\registermember.php on line 63 Appreciate your help

Thanks cmgmyr, your code works, however its the getting the club_id value to go into the database that I am having difficulties with? Cheers

Hey thanks for the reply, but im still very new to php so im sorry but I dont really understand how to do that. I know that I have already started a session however in another page called header.html which is included on the page. I have added some more code so you maybe able to see where I am going wrong. Cheers $query = "INSERT INTO users (username, member_type, first_name, last_name, email, password, registration_date) VALUES ('$u', 'Member', '$fn', '$ln', '$e', PASSWORD('$p'), NOW() )"; $result = @mysql_query ($query); if ($result) { echo '<h3>Thank you for registering!</h3>'; include ('includes/footer.html'); exit(); mysql_close(); } else { echo '<p><font color="red" size="+1">You could not be registered due to a system error. We apologize for any inconvenience.</font></p>'; } } else { echo '<p><font color="red" size="+1">That username is already taken.</font></p>'; } mysql_close(); } else { echo '<p><font color="red" size="+1">Please try again.</font></p>'; } } ?> <h1>Join Your Club</h1> <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"> <fieldset> <table border="0" align="center"> <tr> <td>First Name:</td> <td><input type="text" name="first_name" size="30" maxlength="15" value="<?php if (isset($_POST['first_name'])) echo $_POST['first_name']; ?>" /></td> </tr> <tr> <td>Last Name:</td> <td><input type="text" name="last_name" size="30" maxlength="30" value="<?php if (isset($_POST['last_name'])) echo $_POST['last_name']; ?>" /></td> </tr> <tr> <td>Email Address:</td> <td><input type="text" name="email" size="30" maxlength="40" value="<?php if (isset($_POST['email'])) echo $_POST['email']; ?>" /></td> </tr> <tr> <td>Select your club:</td> <td> <?php require_once ('../mysql_connect.php'); $sql="SELECT * FROM club"; $result=mysql_query($sql); $options=""; while ($row=mysql_fetch_array($result)) { $club_id=$row["club_id"]; $clubName=$row["clubn"]; $options.="<OPTION VALUE=\"$club_id\">".$clubName.'</option>'; } ?> <select name="clubdrop"> <option value=0>Select Your Club <?=$options?> </select>

Just read what I have written and its a bit misleading so ill try again. I want to store the $club_id value from the option selected into a table in the database. Here is my dynamic drop down box which works: <? require_once ('../mysql_connect.php'); $sql="SELECT * FROM club"; $result=mysql_query($sql); $options=""; while ($row=mysql_fetch_array($result)) { $club_id=$row["club_id"]; $clubName=$row["clubn"]; $options.="<OPTION VALUE=\"$club_id\">".$clubName.'</option>'; } ?> <select name="clubdrop"> <option value=0>Select Your Club <?=$options?> </select> Cheers

Hey, Im looking for some help in storing a value from a dynamic drop down box into a table. Here is my dynamic drop down box which works: <? require_once ('../mysql_connect.php'); $sql="SELECT * FROM club"; $result=mysql_query($sql); $options=""; while ($row=mysql_fetch_array($result)) { $club_id=$row["clubn"]; $clubName=$row["clubn"]; $options.="<OPTION VALUE=\"$club_id\">".$clubName.'</option>'; } ?> <select name="clubdrop"> <option value=0>Select Your Club <?=$options?> </select> I am already storing some values from other parts of the form but I am unsure on how to store the $club_id of the club selected into the table at the same time? Cheers

Missed out 1 coma thats it!

Im having some trouble with putting data from a form into 2 seperate tables. //Adds data into club table: club_id (auto increment), type of website and club name = working fine $query1 = "INSERT INTO club (webtype, clubn) VALUES ('$membertype', '$clubn')"; $result = @mysql_query ($query1); //I also need to put club_id into the users table so getting the data from the clubs table $sql = "SELECT * FROM club WHERE clubn='$clubn'"; $result = mysql_query($sql); $clubdata = mysql_fetch_array($result); $clubid=$clubdata['club_id']; //This query here dosent insert anything into the table, no errors, just no data $query2 = "INSERT INTO users (username, club_id, member_type, first_name, last_name, email, password) VALUES ('$username', '$clubid' 'Admin', '$fname', '$surname', '$email', '$pass1')"; $result = @mysql_query ($query2); } Appreciate any help Cheers

Im having some trouble with putting data from a form into 2 seperate tables. //Adds data into club table: club_id (auto increment), type of website and club name = working fine $query1 = "INSERT INTO club (webtype, clubn) VALUES ('$membertype', '$clubn')"; $result = @mysql_query ($query1); //I also need to put club_id into the users table so getting the data from the clubs table $sql = "SELECT * FROM club WHERE clubn='$clubn'"; $result = mysql_query($sql); $clubdata = mysql_fetch_array($result); $clubid=$clubdata['club_id']; //This query here dosent insert anything into the table, no errors, just no data $query2 = "INSERT INTO users (username, club_id, member_type, first_name, last_name, email, password) VALUES ('$username', '$clubid' 'Admin', '$fname', '$surname', '$email', '$pass1')"; $result = @mysql_query ($query2); } Appreciate any help Cheers