John Davidson

Message:

I am trying to show echo 2 but its not working

$zipcode1 = 07300-011;

  $zipcode= str_replace("-","",$zipcode1);

$zipcode = 07300011;

   if ($zipcode >= 20000000 && $zipcode <= 26600999) { 

     echo '1';

}

   elseif ($zipcode >= 07000001 && $zipcode <= 07399999) { 



     echo '2';

}

else {

    echo 'no value';

    }

    

Please let me know where i am doing wrong. Thank you for the help
Result = 2

---

My Answer:
The issue in your code is with the way you are assigning the value to the $zipcode variable. You are using a leading zero in the value (07300011), which is causing PHP to interpret it as an octal number. Octal numbers in PHP are represented with a leading zero, and in this case, 07300011 is equivalent to 2500009 in decimal.

To fix this issue, you should remove the leading zero from the value assigned to $zipcode. Here is the corrected code:

php
$zipcode1 = '07300-011';
$zipcode = str_replace('-', '', $zipcode1);

if ($zipcode >= 20000000 && $zipcode <= 26600999) {
echo '1';
} elseif ($zipcode >= 7000001 && $zipcode <= 7399999) {
echo '2';
} else {
echo 'no value';
}


With this correction, the code should now correctly output '2' when the $zipcode value is within the specified range.