PHP variable is not shown as an integer

Question

if($koltukk%4 == 2){ if($koltukk > 1000){ $koltukH = "B"; (int)$koltukR = ($koltukk / 4) + 1;//Doesnt work (int) } else{ $koltukH = "E"; (int)$koltukR = ($koltukk / 4) + 1;//Doesnt work (int) } } $koltukR = ($koltukk / 4) + 1;

I want to get the $koltukR variable as an integer but i couldn't do it (int) did not work

Get rid of the (int)'s. They aren't valid in PHP the way you are using them. — John Conde
– John Conde, Commented Jan 7, 2014 at 21:31
I am curious, @John Conde, why do you put your answer in the comments instead of writing an answer? I have seen that before but I don't understand... — Fabien Warniez
– Fabien Warniez, Commented Jan 7, 2014 at 21:33
@DamienPirsy he is not, you are getting confused with the variable names. It is $koltukH and $koltukk — Junior
– Junior, Commented Jan 7, 2014 at 21:34
@FabienWarniez Because John is posting a comment and not an answer. — The Blue Dog
– The Blue Dog, Commented Jan 7, 2014 at 21:36

Mike Brant · Accepted Answer · 2014-01-07 21:32:58Z

You need to use the (int) casting on the other side of the assignment operator:

$koltukR = (int)(($koltukk / 4) + 1);

Or, use intval() like this:

$kolturR = intval(($koltukk / 4) + 1);

Jessica · Accepted Answer · 2014-01-07 21:33:05Z

$koltukR = intval(($koltukk / 4) + 1);

You should use better variable names, move the math out of the if/else since both are the same, and you shouldn't even need to cast this as an int manually.

Community · Accepted Answer · 2023-11-17 20:24:57Z

0

PHP has an intval() method that turns a variable into an integer. Pass in your variable as a parameter.

intval()

edited Nov 17, 2023 at 20:24

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answered Jan 7, 2014 at 21:34

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Comments

user8675 · Accepted Answer · 2014-01-07 21:37:01Z

<?php if($koltukk%4 == 2) { if($koltukk > 1000) { $koltukH = "B"; $koltukR = (int)(($koltukk / 4) + 1); } else{ $koltukH = "E"; $koltukR = (int)(($koltukk / 4) + 1); } } echo $koltukR; ?>

Community · Accepted Answer · 2023-11-20 22:45:59Z

One important note here: intval() is NOT round(). intval() is similar to floor().

I think what you really want is round():

Here's the real answer: K.I.S.S.

if($k%4 == 2){ if($k > 1000){ $H = "B" } else{ $H = "E"; } } $R = round($k / 4) + 1;

Stealing an example from https://www.php.net/intval to illustrate:

echo number_format(8.20*100, 20), "<br />"; echo intval(8.20*100), "<br />"; echo floor(8.20*100), "<br />"; echo round(8.20*100), "<br />"; 819.99999999999988631316 819 819 820

Good to point out that this can definitely be refactored, though not actually answering the original question about integer type casting.

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