Php:undefined variable derived from array

Question

I have an array say $packages. I have the code like this to prevent the undefined variable error.

$hospital_charge=0;
    if(!empty($packages)){
    foreach($packages as $package){
        $hospital_charge=$package['hospital_charge'];

Then in subsequent code I use

$health_card_discount = ((($a*5)/100)+($hospital_charge*.05));

That is if $hospital_charge have a null value, I get the error undefined variable hospital_charge,

so to prevent that I defined $hospital_charge=0

is this the proper way of doing it or there is better way of achieving this?

Note:I am using PHP 5.4.16

taxicala · Accepted Answer · 2015-04-29 18:53:43Z

1

You can always use isset (it will check that the variable is defined and not null:

$health_card_discount = isset($hospital_charge) ? ((($a*5)/100)+($hospital_charge*.05)) : (($a*5)/100);

But the way you are doing it is also valid.

answered Apr 29, 2015 at 18:53

taxicala

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2 Comments

Thanks taxicala - Assuming that you want to check both $a and $hospital_charge with isset how you can achieve that?

Same way, use a ternary comparison but i recomend moving the code to a function with an if statement.