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Witam mam problem próbuję id wybranego charname wyświetlić w formularzu aby potem te id dodać do bazy danych
id próbuje wyświetlić za pomocą tego

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<?php echo $charname[1]=$characterID[1]	?>
[PHP] pobierz, plaintext 

lecz to wyświetla id innego charname patrzałem za pomocą $_POST ale też nie wyświetla

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<?php if ($_POST['buy']==false): ?>
            <?php
				$id=$_SESSION['username'];
				$query ="SELECT * FROM accounts,characters WHERE accounts.accountID=characters.ownerID AND username='$id'";
 
				$result = mysql_query($query) or die(mysql_error());
				// Print out the contents of each row into a table 
				echo "<select size=1 name=char_receiver>";
				while($row = mysql_fetch_array($result)){
					$charname[1] = $row['charname'];
					$characterID[1] = $row['CharacterID'];
					echo '<option>'.$charname[1].'</option>';
				}
					echo "</select>";
 
				mysql_free_result($result);
 
				?>
 
 
            <form method='POST' action='2.php'>
				<table cellspacing="10">
				<tr>
            		<td>ID</td><td>Stack ID</td><td>Nazwa</td><td>Cena</td><td></td>
            		</tr>
            <tr>
            <td><?php echo $_POST['itemid']?><input type="hidden" name="item1" value="<?php echo $_POST['itemid']?>"/></td>
            <td><?php echo $_POST['itemstack_id']?><input type="hidden" name="stack1" value="<?php echo $_POST['itemstack_id']?>"/></td>
				<td><?php echo $_POST['name']?><input type="hidden" value="<?php echo $name ?>"/></td>
				<td><?php echo $_POST['cost']?><input type="hidden" value="<?php echo $cost ?>"/></td>
                <td><input type="hidden" name="char_receiver" value="<?php echo $charname[1]=$characterID[1]	?>"/></td>
				<td><input type='submit' name="buy" value='Kup'></td>
                </tr>
            </table>
 
            </form> 
            <?php endif; ?>
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Ten post edytował pajurpl 1.10.2014, 19:54:59