PHP else if methods isn't working?

Question

Hello I am trying to implement after they pay it checks how much they paid and inputs different variables in sql database with if methods.

The problem: The problem is all the if methods return to this first option(inserts paid=1, and 30days) not sure what the problem is, is my if method broken?what's wrong please explain! thanks!

if(number_format($amount, 2) == 8.00) 
{
    $mysqli = new mysqli(******);
    $stmt = $mysqli->prepare("UPDATE `as_users` SET paid='1', reg_date=CURRENT_TIMESTAMP, end_date=DATE_ADD(CURRENT_TIMESTAMP(), INTERVAL 30 DAY) WHERE username = ?");
    $stmt->bind_param('s', $username);
    $stmt->execute();
} elseif (number_format($amount, 2) == 10.00) 
{
    $mysqli = new mysqli(******);
    $stmt = $mysqli->prepare("UPDATE `as_users` SET paid='2', reg_date=CURRENT_TIMESTAMP, end_date=DATE_ADD(CURRENT_TIMESTAMP(), INTERVAL 30 DAY) WHERE username = ?");
    $stmt->bind_param('s', $username);
    $stmt->execute();
} elseif (number_format($amount, 2) == 100.00) 
{
    $mysqli = new mysqli(******);
    $stmt = $mysqli->prepare("UPDATE `as_users` SET paid='2', reg_date=CURRENT_TIMESTAMP, end_date=DATE_ADD(CURRENT_TIMESTAMP(), INTERVAL 365 DAY) WHERE username = ?");
    $stmt->bind_param('s', $username);
    $stmt->execute();
}

Edit: I tried with $amount and also tried using the names of the array from dropdown menu I have

array("Basic Package-Monthly", "8.00", "Month", "1", "0", "0"),
array("Premium Package-Monthly", "10.00", "Month", "1", "0", "0"),
array("Premium Package-Annually", "100.00", "Year", "1", "0", "0"),

You only need the $mysqli = new mysqli(******); once, above the if statement. — Robert Harvey
– Robert Harvey, Commented Jan 23, 2014 at 2:00
oh okay thanks robert, do u have any idea to fix the problem tho? — Ghandour
– Ghandour, Commented Jan 23, 2014 at 2:02
Same with number_format($amount, 2). If you put the result of that in a variable, you can just reference that variable in your if statements. It has the added bonus of putting the numeric value in one place so that you can check it. — Robert Harvey
– Robert Harvey, Commented Jan 23, 2014 at 2:03
try to echo out / console log the $amount value before if condition to check what actual value is — Digital Alchemist
– Digital Alchemist, Commented Jan 23, 2014 at 2:03
You need to clean up the code first, then we can figure out what is wrong with it (assuming the problem doesn't resolve itself naturally when you clean up the code). — Robert Harvey
– Robert Harvey, Commented Jan 23, 2014 at 2:05

Ohgodwhy · Accepted Answer · 2014-01-23 02:15:51Z

6

Why worry about the conditional statement, let the switch handle it. Leverage your prepared statement fully.

$mysqli = new mysqli('******');
$paid = false;
$recurring = false;

switch(number_format($amount, 2)):
    case '8.00':
        $paid = 1;
        $recurring = 30;
        break;
    case '10.00':
        $paid = 2;
        $recurring = 30;
        break;
    case '100.00':
        $paid = 2;
        $recurring  = 365;
        break;
endswitch;

$stmt = $mysqli->prepare("UPDATE `as_users` SET paid=?, reg_date=CURRENT_TIMESTAMP, end_date=DATE_ADD(CURRENT_TIMESTAMP(), INTERVAL ? DAY) WHERE username = ?");
$stmt->bind_param('dis', $paid, $recurring, $username);
$stmt->execute();

edited Jan 23, 2014 at 2:15

answered Jan 23, 2014 at 2:09

Ohgodwhy

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7 Comments

Ghandour Over a year ago

smart method, but 2 are monthly and the 100$ version is yearly

Ghandour Over a year ago

I recieved Fatal error: Call to a member function prepare() on a non-object in(the line that starts the prepare statement)

Ohgodwhy Over a year ago

@user3158844 Then your mysqli connection string failed.

Ghandour Over a year ago

oh wow accidently wrote my mysqli password incorrectly hold on fixing

Ghandour Over a year ago

wow it still only fired first one and I payed for the 10$ version hmm

|

Digital Alchemist · Accepted Answer · 2014-01-23 02:09:58Z

1

try to use

number_format($amount, 2, '.', '') instead of number_format($amount, 2) for returning english notation without thousands separator.

i.e

english notation without thousands separator

$number = 1234.56;
$english_format_number = number_format($number, 2, '.', '');

1234.56

and to compare two float values use BC math function

$b = 10.00;
$a = number_format($amount, 2, '.', '');
    bccomp($a, $b)==0  // returns true if both values are 10.00

answered Jan 23, 2014 at 2:09

Digital Alchemist

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