Getting the wrong result in mysql? - php

Question

Hello i have a html form in side php everything is working fine and i have just been showed how to do hidden fields from this website.
I have a submit button on each and every result letting the user pick what they want. When they press the submit button i want the info to be submitted and added to the database. But for some reason when the user click submit on item 1 it adds the last item into the database e.g item 6 ?? So there is 6 results and a submit button for each one so 6 buttons. When the user presses submit on number 1 item it submits number 6 for some resson.
<form method="post" action="buydo.php">
<label><br />
<br />
</label>
<p>
<?php
$sql = "SELECT * FROM sell
ORDER BY Pokemon_level ASC";
$res = mysql_query($sql) or die(mysql_error());
while ($v = mysql_fetch_array($res)) {
echo '
<div class="auction_box">
<img src="http://myrpg.net/new_rpg/'.$v['Pokemon_pic'].'" width="100" height="100"><br/>
£'.$v['price'].'<br/>
<label id="pokemonName'.$v['id'].'">'.$v['pokemon_name'].'</label><br/>
<label>Level '.$v['Pokemon_level'].'</level><br/>
<label>Exp '.$v['exp'].'</level><br/>
<label>Time Left:';
echo '</label>
<br/>
<input type="hidden" name="Name" value="'.$v['pokemon_name'].'">
<input type="hidden" name="level" value="'.$v['Pokemon_level'].'">
<input type="hidden" name="vid" value="'.$v['id'].'">
<input type="hidden" name="price" value="'.$v['price'].'">
<input type="hidden" name="exp" value="'.$v['exp'].'">
<input type="submit" id="'.$v['id'].'" class="buy_submit" value="Buy Now" /> </div>';
}
?>
</p>
<p> </p>
</form>
That is the select with the submit buttons for each result.
Then i insert the info they have chosen.
include 'config.php';
session_start() ;
$name = mysql_real_escape_string($_POST['Name']);
$Pokemon_level = mysql_real_escape_string($_POST['level']);
$idofpokemonsell = mysql_real_escape_string($_POST['vid']);
$price = mysql_real_escape_string($_POST['price']);
$exp = mysql_real_escape_string($_POST['exp']);
$sql = "SELECT * FROM `users` WHERE `username` = '" . $_SESSION['username'] . "'";
$result = mysql_query($sql) or die(mysql_error());
$values = mysql_fetch_array($result);
echo $values['money'] ;
if ( $values['money'] == $price ) {
echo "Give them the pokemon yay";
}
if ( $values['money'] > $price ) {
mysql_query("INSERT INTO `user_pokemon` (`pokemon`, `belongsto`, `exp`, `slot`, `level`) VALUES ('$name','" . $_SESSION['username'] . "','$exp','0','$Pokemon_level')") or die(mysql_error());
echo "Your money is over";
}
Like i say the insert only inserts item 6 no matter if u press item 1 - 5 it inserts 6 the level , exp, name everything of item 6

Answer 1

Since you display it all in one form, the latest values are overwriting the earlier ones. Try making each item its own form by moving your <form> and </form> tags inside your while loop

Answer 2

or try to track with jquery clicked button id and fill a hidden field with the clicked id

Answer 3

This is because your html is wrong. The values get overwritten
Basically you have to options:
render a form for each item
do it with a dropdown or radio buttons
With the second option you need to rewrite your code more, but it is more elegant, if you ask me. So what do you need to do.
Move the submit button out of the loop
Only generate code with the id and labels like this:
<option value="'.$v['id'].'"/>YOUR LABEL</option>
When the form is submitted, you have to retrieve the values corresponding to your the submitted id