I have variable like this with the type of string
:
$price = "65000"
Now I tried setting it's type to integer
, so I tried these:
$finalprice = (int)$price; // returns 0
$finalprice = (intval)$price; // returns 0
So how to properly return the value 65000
as integer?
You wrong code, the first it's correct and the second is intval($variable)
$price = "65000";
$finalprice1 = (int) $price;
$finalprice2 = intval($price);
var_dump($finalprice1);
var_dump($finalprice2);
/*
int(65000)
int(65000)
*/
I hope it's just a typing error but in the variable you didn't close with ;
Another method you can use is settype
like:
$price = "65000";
settype($price,'int');
var_dump($price);
/*
int(65000)
*/
Reference: