quick php question - GoFuckYourself.com

mkx · 11-25-2013, 05:44 PM

I suck with php. Don't post unless i spent over an hour trying to figure out a simple equation.

What is the correct way to say this?

If the item is part1, part2, or the ebay fees are under 25, sig is 0, otherwise sig is 1.

I tried putting the brackets in different spots with no luck, sig keeps being declared as 0 even though the item is for part1 and the ebay fees are over 25.

Below is the most recent code I tried.

Code:

<?php if ($item->SKU != 'part1' || $item->SKU != 'part2' || $recipient->ebay_fees < '25'){ $sig='0';} else { $sig='1';} print $sig ?>

lezinterracial · 11-25-2013, 05:52 PM

!= means not equal right? and == means equal?

I'm not good at PHP either.

alcstrategy · 11-25-2013, 05:56 PM

Quote:

Originally Posted by mkx

I suck with php. Don't post unless i spent over an hour trying to figure out a simple equation.

What is the correct way to say this?

If the item is part1, part2, or the ebay fees are under 25, sig is 0, otherwise sig is 1.

I tried putting the brackets in different spots with no luck, sig keeps being declared as 0 even though the item is for part1 and the ebay fees are over 25.

Below is the most recent code I tried.

Code:

<?php if ($item->SKU != 'part1' || $item->SKU != 'part2' || $recipient->ebay_fees < '25'){ $sig='0';} else { $sig='1';} print $sig ?>

why are you using strings for ints like $recipient->ebay_fees < '25' dont use quotations because it is an integer not string. Same thing with the assignments.

php isnt stricly typed but just for purposes of practice

if (($item->SKU != 'part1' || $item->SKU != 'part2') || $recipient->ebay_fees < '25'){ $sig='0';} else { $sig='1';} should work

alcstrategy · 11-25-2013, 06:02 PM

also actually the other poster is right != is not equal and == is equals so not sure if it was a typo in your explanation or just wrong coding

mkx · 11-25-2013, 06:15 PM

Quote:

Originally Posted by alcstrategy

why are you using strings for ints like $recipient->ebay_fees < '25' dont use quotations because it is an integer not string. Same thing with the assignments.

php isnt stricly typed but just for purposes of practice

if (($item->SKU != 'part1' || $item->SKU != 'part2') || $recipient->ebay_fees < '25'){ $sig='0';} else { $sig='1';} should work

Doesn't seem to work, still makes sig 0.

if I echo $recipient->ebay_fees it shows a value of 27, thats the way I am pulling it.

mkx · 11-25-2013, 06:17 PM

I intentionally put does not equal !=

mkx · 11-25-2013, 06:18 PM

Basically sig should be 1 if any of those 3 criteria are met which 2 of 3 are.

lezinterracial · 11-25-2013, 06:33 PM

Now, I am no PHP expert but maybe it will make more sense, If you say what the criteria is as opposed to what it isn't

If I am hearing your right, it should be completely backwards.
if the item is 1 or the item is 2 or the fee < 25 sig 1, is that correct? If so.

if (($item->SKU == 'part1' || $item->SKU == 'part2') || $recipient->ebay_fees > '25'){ $sig='1';} else { $sig='0';}

EddyTheDog · 11-25-2013, 06:33 PM

Have you tried using 'Switch-Case' instead?..

http://php.net/manual/en/control-structures.switch.php

Example:

PHP Code:


			
<?php

switch($beer)

{

    case 'tuborg';

    case 'carlsberg';

    case 'heineken';

        echo 'Good choice';

    break;

    default;

        echo 'Please make a new selection...';

    break;

}

?>

dunhill · 11-25-2013, 06:35 PM

PHP Code:


			
<?php $sig = ($item->SKU == 'part1' or $item->SKU == 'part2' or $recipient->ebay_fees < 25?0:1)

mkx · 11-25-2013, 06:43 PM

Quote:

Originally Posted by dunhill

PHP Code:


			
<?php $sig = ($item->SKU == 'part1' or $item->SKU == 'part2' or $recipient->ebay_fees < 25?0:1)

Thanks for all the suggestions, this one worked in declaring sig as 1 but if sig is not 1 then it does not declare sig as anything, it should be declared as 0 if not 1, i think I can do this myself though

I did the not equals because sometimes there are multiple items and this was my way of troubleshooting it

mkx · 11-25-2013, 06:50 PM

Got it all fixed, thanks again!

11-25-2013, 05:44 PM	#1
mkx Confirmed User Industry Role: Join Date: Nov 2003 Location: Toronto Posts: 4,001	quick php question I suck with php. Don't post unless i spent over an hour trying to figure out a simple equation. What is the correct way to say this? If the item is part1, part2, or the ebay fees are under 25, sig is 0, otherwise sig is 1. I tried putting the brackets in different spots with no luck, sig keeps being declared as 0 even though the item is for part1 and the ebay fees are over 25. Below is the most recent code I tried. Code: <?php if ($item->SKU != 'part1' \|\| $item->SKU != 'part2' \|\| $recipient->ebay_fees < '25'){ $sig='0';} else { $sig='1';} print $sig ?>

11-25-2013, 05:52 PM	#2
lezinterracial Confirmed User Industry Role: Join Date: Jul 2012 Posts: 3,073	!= means not equal right? and == means equal? I'm not good at PHP either. __________________ Live Sex Shows

11-25-2013, 06:33 PM	#8
lezinterracial Confirmed User Industry Role: Join Date: Jul 2012 Posts: 3,073	Now, I am no PHP expert but maybe it will make more sense, If you say what the criteria is as opposed to what it isn't If I am hearing your right, it should be completely backwards. if the item is 1 or the item is 2 or the fee < 25 sig 1, is that correct? If so. if (($item->SKU == 'part1' \|\| $item->SKU == 'part2') \|\| $recipient->ebay_fees > '25'){ $sig='1';} else { $sig='0';} __________________ Live Sex Shows Last edited by lezinterracial; 11-25-2013 at 06:35 PM..

11-25-2013, 06:33 PM	#9
EddyTheDog Just Doing My Own Thing Industry Role: Join Date: Jan 2011 Location: London, Spain, New Zealand, GFY - Not Croydon... Posts: 25,044	Have you tried using 'Switch-Case' instead?.. http://php.net/manual/en/control-structures.switch.php Example: PHP Code: `<?php switch($beer) { case 'tuborg'; case 'carlsberg'; case 'heineken'; echo 'Good choice'; break; default; echo 'Please make a new selection...'; break; } ?>` Last edited by EddyTheDog; 11-25-2013 at 06:36 PM..

11-25-2013, 06:35 PM	#10
dunhill Confirmed User Industry Role: Join Date: Jul 2013 Posts: 89	PHP Code: `<?php $sig = ($item->SKU == 'part1' or $item->SKU == 'part2' or $recipient->ebay_fees < 25?0:1)`

11-25-2013, 06:02 PM	#4
alcstrategy Confirmed User Industry Role: Join Date: May 2012 Posts: 124	also actually the other poster is right != is not equal and == is equals so not sure if it was a typo in your explanation or just wrong coding

11-25-2013, 06:17 PM	#6
mkx Confirmed User Industry Role: Join Date: Nov 2003 Location: Toronto Posts: 4,001	I intentionally put does not equal !=

11-25-2013, 06:18 PM	#7
mkx Confirmed User Industry Role: Join Date: Nov 2003 Location: Toronto Posts: 4,001	Basically sig should be 1 if any of those 3 criteria are met which 2 of 3 are.

11-25-2013, 06:50 PM	#12
mkx Confirmed User Industry Role: Join Date: Nov 2003 Location: Toronto Posts: 4,001	Got it all fixed, thanks again!