Display "Not Provided" when variable is null or empty in database - php

Question

i have a variable array that gets all what my function has retrieved.
$array = $funcs->searchCompany($bizName);
and then i used foreach to check if the value is null for varchar and 0 for int and then i replace its value to "Not Provided" so that everytime it is being called it will say "Not Provided"
foreach ($array as $var) {
if($var == " " || $var == 0) {
$var = "Not Provided";
}
}
$name = $var['name'];
$url = $var['url'];
$tagline = $var['tagline'];
$descrip = $var['descrip'];
$bemail = $var['bemail'];
$address = $var['address'];
$city = $var['city'];
but it seems wrong because it destroys the output instead.

Answer 1

You can use & here to pass the value of array to change inside foreach without actually worrying about which is the current array key, which is also sometimes called as passing a variable's value by reference.
Using foreach
foreach ($array as &$value) // note the &
{
if(empty($value)) $value = 'Not Provided';
// other values remain untouched
}
Using array_map()
$array = array_map(function($value){
if(empty($value))
return 'Not Provided';
return $value;
}, $array);
But i will suggest to go with foreach.

Answer 2

update your function so something like
foreach ($array as &$var) {
if($var == " " || $var == 0) {
$var = "Not Provided";
}
}
$name = $array['name'];
$url = $array['url'];
$tagline = $array['tagline'];
$descrip = $array['descrip'];
$bemail = $array['bemail'];
$address = $array['address'];
$city = $array['city'];

Answer 3

I would try something like this . I can not write comments so I am writing it as an answer.
What I understand from your code is your array has key value relation. That is most probably why it is not working with your single dimensional array iteration.
try this instead.
foreach ($array as $var => $value) {
if($value == " " || $value == 0 || $value == null) {
$array[$var] = "Not Provided";
}
}
echo "<pre>";
print_r($array);
echo "</pre>";
Give it a try.