[SOLVED] database storing issue... =(

[Derleek]

Ok, so I'm quite confused about this issue.

 

I set up a script to store 10 items that where selected by a user w/ drop down menu's.  This script is supposed to take each of the 10 selected values along with several user oriented variables into a database called 'benchracers'.

 

Everything seemed to be working fine, until i went to output the data stored.  The only thing that has an output value is "ID".

 

so can anyone figure out why the ONLY thing I am able to get an output value to is 'ID'.

 

here is the code:

if(isset($_POST))
		{
			dbconnect('thethrgu_moto');						
			foreach ($choices as $n => $choice) 
			{
				if ($choice == 'null') 
				{
					error('you did not completely fill out your top ten!!!');
				}
			}
			if (count(array_unique($choices)) != 10) {
				error('you had some duplicate riders, please make sure you do not enter a rider more than once!');
			} else {
				$race_id = 3;
				$paid = 1;
				$pick_id=1;
				$sql = "SELECT COUNT(*) FROM benchRacers WHERE id = $id && race_id = $race_id";
				$result = mysql_query($sql);
				if (!$result) {
				    echo mysql_error();
				}
				if (@mysql_result($result,0,0)>0) {
					echo "You have already entered your data for this race, would you like to re-do it?";
				}
				else
				{
					for($i=0;$i<10;$i++)
					{

					$sql = "INSERT INTO benchracers (id, pick_id, racer_id, race_id, paid)

						VALUES ($id,$pick_id,$choices[$i],$race_id,$paid)";
						echo $sql;
						mysql_query($sql) or die(mysql_error());
						$pick_id++;
					}
					$query = "SELECT * FROM benchracers"; 

					$result = mysql_query($query) or die(mysql_error());

					while($row = mysql_fetch_array($result, MYSQL_ASSOC)){
						echo "id: ".$row['id']." - Race_id: ".$row['race_id']." - Racer_id: ".$row['racer_id']." - Pick_id: ".$row['pick_id']."<br>";
					}
				}

			}
		}

 

and the table create:

$benchracers = "CREATE TABLE benchRacers(
id INT, 
pick_ID INT,
racer_ID INT, 
race_ID INT,
paid INT)";
mysql_query($query) or die(mysql_error());

 

It is entirely possible (i suppose) that the way i am retrieving the data is not valid, and is getting all funked up.

 

I don't know... please pick at my code if you so chose, i love figuring out new and better ways to get things done

[Derleek]

oh, i forgot to say that when i echo "$sql" all of the variables that are to be stored are set how they should be

[fenway]

oh, i forgot to say that when i echo "$sql" all of the variables that are to be stored are set how they should be

Let's see it.