skiingguru1611

From my understanding you only need 1 table, then have some fields, such as Breed, Shelter, Gender, Friendly(for living with kids), Dogs (living with other dogs). As for the actual query to the DB, I'm not positive, I'll let someone else help you on that.

Try $queryi = "SELECT DISTINCT name, paid FROM `photoalbum` WHERE `userid` = '$id'"; I've never used DISTINCT, but I think that should work.

Could you please show us the "data.php" code?

The Error reporting thing helped me out alot. It made me realize there was no "=" at playerlevel = '$playerlevel'. Thank You. My Code now works. I appreciate the help, and again I am sorry for putting the NEEDED ASAP part in the title. It won't happen again.

Yes, they come from a form, and I use $_POST to retrieve them, and I have a bunch of functions running on them to secure the database. I'll check out the links.

Also, removing the @ didn't make the code execute. Any other suggestions?

Thank You. And I understand that mine is no more important than the others, but my neighbor needs to get it up so people can start registering. And, I've never been very good with die statements. Could you help me out there?

I am working on some code for my neighbor's camp, but for some reason the code that allows people to update their registration information will not update. <?php mysql_connect(localhost,$username,$password); @mysql_select_db($database) or die( "Unable to select database"); $query = "UPDATE registration SET playerfirst = '$playerfirst', playerlast = '$playerlast', playeraddress = '$playeraddress', playercity = '$playercity', playerstate = '$playerstate', playerzipcode = '$playerzipcode', guardianphone = '$homephone', guardiancell = '$cellphone', playeremail = '$playeremail', playerage = '$playerage', playerweight = '$playerweight', playerheight = '$playerheight', playergrade = '$playergrade', playerposition = '$playerposition', playerschool'$playerschool', playerlevel = '$playerlevel', playercoach = '$playercoach', playercountry = '$playercountry', roommate = '$roommate', guardianwork = '$workphone', guardianfirst = '$guardianfirst', guardianlast = '$guardianlast', guardianemail = '$guardianemail', emergencyfirst = '$emergencyfirst', emergencylast = '$emergencylast', emergencyhome = '$emergencyhomephone', emergencycell = '$emergencycellphone', emergencyemail = '$emergencyemail' WHERE user = '$user'"; mysql_query($query); mysql_close(); ?>

Something along the lines of: <?php $ip=USER IP ADDRESS HERE $sql="select * $table WHERE ip='$ip'"; $result=mysql_query($sql,$connection) or die(mysql_error()); $num = mysql_numrows($result); if($num != 0){ while($row=mysql_fetch_array($result)) { MESSAGE TELLING THE PERSON THEY ALREADY SUBMITTED THE INFO }else{ FORM YOU WANT THEM TO SUBMIT ?>

Andy-H what did you change that time. BECAUSE IT WORKS!!! THANK YOU SOO MUCH.

I won't be on for like 6 hours. So I will see your replies in the morning. Thanks for all your help. Keep posting anything you think may help.

Gives me this error with or without an entry in the DB.

Well the output depends on the user that is logged in.

After if(!isset($row['user'])){ ?> I close the php code block and then put the form in html. I do not echo it or anything. Could that be the problem?

This sql query selects what I need when the code runs correctly. select user, first, last, birth, address, city, state, zip, email, grade, team, coach, height, weight, gpa, sat, phone, other from $table WHERE user='$user'