php itertation loop in hidden input not iterating

Question

I've got one nagging little bug in this script. I'm going through my cart items and passing them into hidden inputs. The cart_id ($obj->id) is working fine into the value="" but my iteration loop that gives each value a unique name="" (cart_id_1, cart_id_2 etc) is NOT iterating.

<?php         

           $pass_cart_q = "SELECT c.id FROM carts AS c WHERE c.user_session_id='$sid'";
           $result = $mysqli->query($pass_cart_q);

                    $i = 1; 
                while ($obj = $result->fetch_object()) {

                    echo "<input type=\"hidden\" name=\"cart_id_".$i."\" value=\"  .$obj->id.  \"><br>";  
                    $i = $i++;
                }
                mysqli_close();?>

Each name field is coming through as cart_id_1

I've just found the thread that states you can cast a hidden field into an array using square brackets - name="name[]" ... I was operating on the assumption that I would be overwriting the values, but I will see if this works. — shotdsherrif
– shotdsherrif, Commented Aug 19, 2011 at 10:20

Sleeperson · Accepted Answer · 2011-08-19 10:04:53Z

2

$i=$i++;

That's the problem just do:

 $i++

answered Aug 19, 2011 at 10:04

Sleeperson

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Kaken Bok · Accepted Answer · 2011-08-19 10:02:07Z

2

Please replace $i = $i++; with just $i++.

$i = 1;
$i = $i++;

echo $i, "\n"; // 1

$i = 1;
$i = ++$i;

echo $i, "\n"; // 2

$i = 1;
$i++;

echo $i, "\n"; // 2

$i = 1;
++$i;

echo $i, "\n"; // 2

answered Aug 19, 2011 at 10:02

Kaken Bok

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2 Comments

feeela Over a year ago

The increment operator VAR++ returns the value of VAR and afterwards increasing the value of VAR by one. You may simply use VAR++ as suggested or VAR = ++VAR, to get the already increased value.

Kaken Bok Over a year ago

Yes! The interesting thing is with $i = $i++ that after assigning the value of the right $i to the left $i the incrementor does not operate on our left $i. There are at that time apparently two different references for the same variable name.

Community · Accepted Answer · 2020-06-20 09:12:55Z

0

What $i = $i++ will cause it literally this: "make $i equal to $i and then increase it by one", but the $i will still remain the same. To solve this, simply replace $i = $i++; with $i++.

Manual Entry

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Aug 19, 2011 at 10:04

Madara's Ghost

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1 Comment

Kaken Bok Over a year ago

Why the $i will still remain the same?

Ibrahim Azhar Armar · Accepted Answer · 2011-08-19 10:06:04Z

0

you are assigning the incremented value to $i variable. and hence it is not able to iterate. instead you should remove that assignment variable $i and it should only be $i++

answered Aug 19, 2011 at 10:06

Ibrahim Azhar Armar

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