WHY INSERT fail when I put an IF inside a FOR LOOP ?

vincej · May 10, 2012

Hi - I have spent 2 days trying to figure this out. I must be doing something stupid ...

I am trying to update the DB with a collection of variables off of a various POST array's. I have indexed the variables so that I can loop through them with a FOR.

However I am having all kinds of trouble:

1 - WEIGHT is giving 5 values even though the counter should only go 4 times.

2 - The WEIGHT variable is going into the wrong field in the DB even though I specify in my Query a specific place.

3 - I get undefined offset errors even though I have initialised the variables with their respective offsets.

4 - Adding the conditional statement adds even more offset errors even though they are initialised

I am starting to wonder if my entire approach to updating my DB is flawed.

As a student of PHP I think I need someone with more experience to look at what I am doing and tell me if something jumps out at them.

There is a lot of statements echoing out values - and they all seem to check out.

Man Many thanks for your advice

    function finishedorder(){  
$prodname = $_POST['prodname'];
$prodid =$_POST['prodid'];
$quantity = $_POST['quantity'];
$pricelb = $_POST['pricelb']; 
$customerid = $_POST['customerid']; 
$price = $_POST['price']; 
$orderid = $_POST['orderid']; 
$weight	= $_POST['weight'];

	echo "Varuable WEight"; print_r($weight);	 echo"<br/>";
echo "line 196 Weight"; print_r ($_POST['weight']); echo"<br/>";
echo "line 192 PriceLB"; print_r ($_POST['pricelb']); echo "<br/>";
echo "line 193 Price"; print_r ($_POST['price']); echo "<br/>";
echo "line 194 ProdID"; print_r ($_POST['prodid']); echo "<br/>";
    echo "line 195 OrderID"; print_r ($_POST['orderid']); echo "<br/><br/>";


$numloops = count($_POST['prodid']); 
for ($i = 0; $i < $numloops ; $i++) {										//  !!! Start of the for loop !!! 

if (!isset($orderid[$i])) $orderid[$i] ='';
if (!isset($prodid[$i])) $prodid[$i] ='';
if (!isset($weight[$i])) $weight[$i] ='';		
if (!isset($pricelb[$i])) $pricelb[$i] ='';									// This only initialises the Price & Weight variable to avoid an offset error message.	
if (!isset($weight[$i])) $weight[$i] ='';	


$ordervalue = $pricelb[$i] * $weight[$i];

echo " L. 216 The number is " . $i . "<br />";
echo "line 221 ProdID".$prodid[$i];echo "<br/>";
echo "line 217 Pricelb".$pricelb[$i];echo "<br/>";
echo 'Weight L. 219  weight[i]'. $weight[$i];echo "<br/>";
echo "line 220 weight".$weight; echo "<br/>";
echo "line 221 Numloops".$numloops; echo "<br/><br/>";
echo ' L. 225 $ordervalue  ' .$ordervalue;echo "<br/>";

}

//if ($pricelb[$i] == 0 ){													// conditonal statement  
//	$ordervalue =  $price[$i];
//	}



$sql ="
UPDATE confirmedorder
SET  weight='$weight[$i]', ordervalue='$ordervalue', confirmedorder.picking = 'finished', confirmedorder.sale = 'open'
WHERE prodID = '$prodid[$i]'
AND OrderID = '$orderid[$i]' ;
";

$this->db->query($sql); 

	}																	// !! End For Loop. !!

batwimp · May 10, 2012

Your array offset $i doesn't exist outside your for loop, so not sure what its value will be in your UPDATE statement.

xyph · May 10, 2012

"Your for loop. I do not think it ends where you think it ends."

batwimp · May 10, 2012

Inconceivable!

vincej · May 10, 2012

Thanks for the feedback :

I have checked the ending if the for loop and it does end where it ends. Dreamweaver checks it for you by balancing braces .

the Update statement is inside the For loop so it will know what the offset $i is

I'm not sure what is implied by 'inconceivable' ?

vincej · May 10, 2012

I should add that the For loop does Loop - it just is not giving me the results I expect.

and for the life of me I can not see what or what to do about it.

batwimp · May 10, 2012

According to my editing software, the for loop ends before the UPDATE is done.

"Inconceivable" is a quote from the same movie xyph was (sort of) quoting. Just for fun.

vincej · May 10, 2012

It's a mystery - I just pulled the code off phpfreaks nad put it back into Dreamweaver and it checks out. No matter - the key issue is trying to find a solution to my other problems,

Drummin · May 10, 2012

Assuming that the $_POST['prodid'] count is really WHAT you want to be using for loop ending value, moving the end loop tag so it surrounds the insert should fix the problem. Note: what you had commented as the closing loop was the end of the function loop.

  <?php
  function finishedorder(){  
$prodname = $_POST['prodname'];
$prodid =$_POST['prodid'];
$quantity = $_POST['quantity'];
$pricelb = $_POST['pricelb']; 
$customerid = $_POST['customerid']; 
$price = $_POST['price']; 
$orderid = $_POST['orderid']; 
$weight	= $_POST['weight'];

	echo "Varuable WEight"; print_r($weight);	 echo"<br/>";
echo "line 196 Weight"; print_r ($_POST['weight']); echo"<br/>";
echo "line 192 PriceLB"; print_r ($_POST['pricelb']); echo "<br/>";
echo "line 193 Price"; print_r ($_POST['price']); echo "<br/>";
echo "line 194 ProdID"; print_r ($_POST['prodid']); echo "<br/>";
    echo "line 195 OrderID"; print_r ($_POST['orderid']); echo "<br/><br/>";


$numloops = count($_POST['prodid']); 
for ($i = 0; $i < $numloops ; $i++) {										//  !!! Start of the for loop !!! 

if (!isset($orderid[$i])) $orderid[$i] ='';
if (!isset($prodid[$i])) $prodid[$i] ='';
if (!isset($weight[$i])) $weight[$i] ='';		
if (!isset($pricelb[$i])) $pricelb[$i] ='';									// This only initialises the Price & Weight variable to avoid an offset error message.	
if (!isset($weight[$i])) $weight[$i] ='';	


$ordervalue = $pricelb[$i] * $weight[$i];

echo " L. 216 The number is " . $i . "<br />";
echo "line 221 ProdID".$prodid[$i];echo "<br/>";
echo "line 217 Pricelb".$pricelb[$i];echo "<br/>";
echo 'Weight L. 219  weight[i]'. $weight[$i];echo "<br/>";
echo "line 220 weight".$weight; echo "<br/>";
echo "line 221 Numloops".$numloops; echo "<br/><br/>";
echo ' L. 225 $ordervalue  ' .$ordervalue;echo "<br/>";



//if ($pricelb[$i] == 0 ){													// conditonal statement  
//	$ordervalue =  $price[$i];
//	}



$sql ="
UPDATE confirmedorder
SET  weight='$weight[$i]', ordervalue='$ordervalue', confirmedorder.picking = 'finished', confirmedorder.sale = 'open'
WHERE prodID = '$prodid[$i]'
AND OrderID = '$orderid[$i]' ;
";

$this->db->query($sql); 

	}																	// !! End For Loop. !! 
  }
?>

xyph · May 10, 2012

If you tabulate your code properly, you will see your issue. I don't care what Dreamweaver says.

<?php

function finishedorder() {

$prodname = $_POST['prodname'];
$prodid = $_POST['prodid'];
$quantity = $_POST['quantity'];
$pricelb = $_POST['pricelb'];
$customerid = $_POST['customerid'];
$price = $_POST['price'];
$orderid = $_POST['orderid'];
$weight = $_POST['weight'];

echo "Varuable WEight";
print_r($weight);
echo"<br/>";

echo "line 196 Weight";
print_r($_POST['weight']);
echo"<br/>";

echo "line 192 PriceLB";
print_r($_POST['pricelb']);
echo "<br/>";

echo "line 193 Price";
print_r($_POST['price']);
echo "<br/>";

echo "line 194 ProdID";
print_r($_POST['prodid']);
echo "<br/>";

echo "line 195 OrderID";
print_r($_POST['orderid']);
echo "<br/><br/>";


$numloops = count($_POST['prodid']);
for ($i = 0; $i < $numloops; $i++) {
	if (!isset($orderid[$i]))
		$orderid[$i] = '';
	if (!isset($prodid[$i]))
		$prodid[$i] = '';
	if (!isset($weight[$i]))
		$weight[$i] = '';
	if (!isset($pricelb[$i]))
		$pricelb[$i] = '';
	if (!isset($weight[$i]))
		$weight[$i] = '';


	$ordervalue = $pricelb[$i] * $weight[$i];

	echo " L. 216 The number is " . $i . "<br />";
	echo "line 221 ProdID" . $prodid[$i];
	echo "<br/>";
	echo "line 217 Pricelb" . $pricelb[$i];
	echo "<br/>";
	echo 'Weight L. 219  weight[i]' . $weight[$i];
	echo "<br/>";
	echo "line 220 weight" . $weight;
	echo "<br/>";
	echo "line 221 Numloops" . $numloops;
	echo "<br/><br/>";
	echo ' L. 225 $ordervalue  ' . $ordervalue;
	echo "<br/>";
} // WOA THE FOR ENDS!

//if ($pricelb[$i] == 0 ){
	// $ordervalue =  $price[$i];
//}



$sql = "
UPDATE confirmedorder
SET  weight='$weight[$i]', ordervalue='$ordervalue', confirmedorder.picking = 'finished', confirmedorder.sale = 'open'
WHERE prodID = '$prodid[$i]'
AND OrderID = '$orderid[$i]' ;
";

$this->db->query($sql);
}

?>

As a student of PHP, you should uninstall Dreamweaver.

vincej · May 10, 2012

Bugger me - I take it all back. You were right, I was indeed missing a brace at the end !!!! Thank you !

However, it's almost working but not quite. One of the WEIGHT variables is not in the correct field even though I specify a PRODID AND ORDERID. look at Prodid 101. It should say WEIGHT:7 instead the 7 is in the next row. To illustrate the problem here is a print of the table:

ID OrderID CustomerID ProdID Prodname Price Quantity Pricelb Weight OrderValue Picking Sale

151 JAC9506 5 17 Ribs 60.00 1 5.00 6.00 30.00 finished open

152 JAC9506 5 28 25 Piece Bag 15.00 1 0.00 0.00 15.00 finished open

153 JAC9506 5 101 breasts 10.25 1 2.00 0.00 0.00 finished open

154 JAC6193 5 27 Sirloin Steak 24.00 1 0.00 7.00 24.00 finished open

Drummin · May 10, 2012

That print must be an html table not the DB table. What does the DB table look like when doing a clean update,i.e. closing browser and reloading fresh copy of the page and posting your form again. Does new DB values match?

vincej · May 10, 2012

It is a cut and paste from MYSQL print preview.

I tried closing FF and reopening twice - didn't help. Bummer .

Drummin · May 11, 2012

Interesting, so print preview does First Caps on field names. They don't match your insert names.

vincej · May 11, 2012

If they need to match that will be me .. I always do first caps on table names. So far I have not seen MySQL ever be case dependant .... mistake ?

Drummin · May 11, 2012

What do you see when adding echo sql. Does product 101 have weight of 7?

$sql ="
UPDATE confirmedorder
SET  weight='$weight[$i]', ordervalue='$ordervalue', confirmedorder.picking = 'finished', confirmedorder.sale = 'open'
WHERE prodID = '$prodid[$i]'
AND OrderID = '$orderid[$i]' ;
";
echo "$sql<br />";

vincej · May 11, 2012

Ok - I ran the test, and the results are consistent with the MYSQL print:

UPDATE confirmedorder SET  weight='0.00', ordervalue='0', confirmedorder.picking = 'finished', confirmedorder.sale = 'open' WHERE prodID = '101' AND OrderID = 'JAC9506' ;

Thanks for sticking with me on this !

btw - how do you get colored code on this site .. mine comes out in bw.

Drummin · May 11, 2012

Are all your form tags arrays? name='prodname[]' etc. Seems like a simple foreach loop should handle this fine. Btw. code with <?php ?> will have color.

ALSO is product id ever duplicated in the form? Just wondering why you have count($_POST['prodid']). OK, I assume this is an array... Never mind.

Drummin · May 11, 2012

Sorry if I'm not seeing the big picture and how this code fits into what you have going, but it would seem to me that a simple foreach loop would grab the the "post array key" and using this you identify each item and make this update. Again sorry if this is way off.

<?php
function finishedorder(){
foreach ($_POST['prodid'] as $arrkey => $prodid){
	$pricelb = $_POST['pricelb'][$arrkey];
	$price = $_POST['price'][$arrkey]; 
	$orderid = $_POST['orderid'][$arrkey]; 
	$weight	= $_POST['weight'][$arrkey];

	$ordervalue = $pricelb * $weight;

	$sql ="	UPDATE confirmedorder SET weight='$weight', ordervalue='$ordervalue', picking = 'finished', sale = 'open' WHERE prodID = '$prodid' AND OrderID = '$orderid'";
$this->db->query($sql);
}
}
?>

vincej · May 11, 2012

Hi Drummin - thanks for this idea, also thank you very much for the generosity of your time and spirit. I am always amazed how some people get a kick out of patronizing posters whilst people like you are so generous. Where are you ? I'm in Canada.

I would love to change the approach. The reason why I chose a for loop is because my POST arrays come in looking like this:

Array ProdID:

 
Array
( [0] => 17 [1] => 28 [2] => 101 [3] => 27 )

a For loop allows me to increment the index like this: $prodid[$i]. If there is a way of incrementing the index, I'm not sure how to do that .. any ideas ?

Many thanks !

Drummin · May 11, 2012

You probably could sort() the list first then do the foreach loop.

<?php
function finishedorder(){
sort($_POST['prodid']);
foreach ($_POST['prodid'] as $arrkey => $prodid){
	$pricelb = $_POST['pricelb'][$arrkey];
	$price = $_POST['price'][$arrkey]; 
	$orderid = $_POST['orderid'][$arrkey]; 
	$weight	= $_POST['weight'][$arrkey];

	$ordervalue = $pricelb * $weight;

	$sql ="	UPDATE confirmedorder SET weight='$weight', ordervalue='$ordervalue', picking = 'finished', sale = 'open' WHERE prodID = '$prodid' AND OrderID = '$orderid'";
$this->db->query($sql);
}
}
?>

vincej · May 11, 2012

Ok - then presumably $arrkey takes care of keeping all the different POST arrays synchronized ? so for example

Prodid array([2] => 101) is correctly synchronized with Pricelb array([2] => 101) etc etc etc

correct ?

vincej · May 11, 2012

Oops - that should have read:

Prodid array([2] => 101) is correctly synchronized with Weight array([2] => 7 )[/size] etc etc etc

Drummin · May 11, 2012

Yes you are correct.

vincej · May 11, 2012

heah Drummin - Thanks again for your code, however I'm not making it work properly.

The SQL Query works on the first pass - then it appears to fail without any PHP errors

One thing I have noticed is that for some reason the $arrkey and $prodid value is getting messed up when I compare it to the POST ARRAY.

Take for example the second UPDATE. $prodid 27 has an $arrkey of 2 yet when you look into the POST it has an array key of [3]

This might be causing the SQL query to fail ? NB - I have attached a PNG off the table so you can see how it is failing.

MANY THANKS !



line 196 WeightArray ( [0] => 6 [1] => 0.00 [2] => 0.00 [3] => 7 [4] => 0.00 [5] => 0.00 ) 
line 192 PriceLBArray ( [0] => 5.00 [1] => 0.00 [2] => 2.00 [3] => 0.00 ) 
line 193 PriceArray ( [0] => 60.00 [1] => 15.00 [2] => 10.25 [3] => 24.00 ) 
line 194 ProdIDArray ( [0] => 17 [1] => 28 [2] => 101 [3] => 27 ) 
line 195 OrderIDArray ( [0] => JAC9506 [1] => JAC9506 [2] => JAC9506 [3] => JAC6193 ) 

UPDATE confirmedorder SET weight='6', ordervalue='30', picking = 'finished', sale = 'open' WHERE prodID = '17' AND OrderID = 'JAC9506'
arrkey0
prodid 17
Pricelb5.00

UPDATE confirmedorder SET weight='0.00', ordervalue='0', picking = 'finished', sale = 'open' WHERE prodID = '27' AND OrderID = 'JAC9506'
arrkey1
prodid 27
Pricelb0.00

UPDATE confirmedorder SET weight='0.00', ordervalue='0', picking = 'finished', sale = 'open' WHERE prodID = '28' AND OrderID = 'JAC9506'
arrkey2
prodid 28
Pricelb2.00

UPDATE confirmedorder SET weight='7', ordervalue='0', picking = 'finished', sale = 'open' WHERE prodID = '101' AND OrderID = 'JAC6193'
arrkey3
prodid 101
Pricelb0.00

$C:\Users\Vince\Desktop\Capture$

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WHY INSERT fail when I put an IF inside a FOR LOOP ?

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