update database by text box based on the checked box's - php

Question

**Hi
i have been working on this php code .
I want to update the quantity that belongs to the checked box .
but the exact problem with my code is :if I have for example 3 checkboxs and when I enter the quantity for them all in the same time the updates are done correctly and if I only update the first quantity with out the other 2 the update is done correctly also ,But when I update the second or the third alone not in the same time it takes the quantity of the old value of the first :""(
so how can I change my code so I can update all the items or only the checked items.
here is my code to display the checkbox's with the quantity.
in the file manage_items.php :**
<?php
$DB_HOST ='localhost';
$DB_USER ='root';
$DB_PASSWORD='';
$DB_DATABASE='dks';
$con= mysql_connect($DB_HOST ,$DB_USER , $DB_PASSWORD);
if(!$con){
die('Failed to connect to server :'.mysql_error());
}
$db=mysql_select_db($DB_DATABASE);
if(!$db){
die("unable to select database");
}?>
$qry="SELECT * FROM catalog";
$result= mysql_query($qry);
if($result){
while($info = mysql_fetch_array($result))
{
print "<h3> cat:".$info['name']."</h3><div>";
$qryitem="SELECT * FROM item WHERE Id=". $info['Cid'];
$resultitem=mysql_query($qryitem);
if($resultitem){
?>
<form method="post" action="manage_item_action.php">
<?php
while($info=mysql_fetch_array($resultitem))
{
?>
<input type="checkbox" name="op[]" value="<?php echo $info['Id'];?>"/><?php echo $info['name'];?>
<label> Quantity <input type="text" name="Quantity[]" value="<?php echo $info['Quantity'];?>"/></label>
<br/>
<?php
}
}
else echo "There are no items.";
print "</div>";
}
}
?>
</div>
<input type="submit" value="update" name="submit"/>
</form>
and here is the excution of update
in the file manage_item_action.php
<?php
$DB_HOST ='localhost';
$DB_USER ='root';
$DB_PASSWORD='';
$DB_DATABASE='dks';
$con= mysql_connect($DB_HOST ,$DB_USER , $DB_PASSWORD);
if(!$con){
die('Failed to connect to server :'.mysql_error());
}
$db=mysql_select_db($DB_DATABASE);
if(!$db){
die("unable to select database");
}
$options=$_POST['op'];
$qun=$_POST['Quantity'];
$size =count($options);
for($i =0; $i<$size; $i++)
{
//$qryop="UPDATE item SET Quantity =".$qun."WHERE Id =".$options[$i]."';";
$resultop=mysql_query("update item set Quantity='".$qun[i]."'where Id='".$options[$i]."'");
}
if($resultop){
header("location: manage_items.php");}
else echo "there was an error"
?>
i dunno how to fix the problem !
but i think my main problem is that all the textbox's have the same name .
so when i send it to the other file for ecution it just take it as it is the last one .
help me please :$
*UPDATED*
*The Solution*
so I solved the problem of my code and its only needed the textbox to be disabled before checkin the checkbox so the array doesn't have any null index..only have the entered values without any nulls
:))

Answer 1

You need to give the 'Quantity' input field a unique name or cast it as an array for every checkbox listed, else when submitted the receiving form will take the last value for 'Quantity' as gospel.

Answer 2

That's because in your input form box, you have:
Quantity <input type="text" name="Quantity" ...
So, with each of the quantity inserted, the forwarded values are being updated. And the last value inserted into Quantity box will be taken in update. Instead, change this form input to this:
<!-- This will forward an array to php script -->
Quantity <input type="text" name="Quantity[]"...
and then, in your updation query, change to:
$resultop=mysql_query("UPDATE item SET Quantity='" . $qun[$i] . "'where Id='".$options[$i]."'");