I have the following (summarised) code
define pauls_code($a, $b)
{
$c = $a + $b;
echo $a;
echo $b;
}
pauls_code(1,2);
echo $c; // how do I get this to print outside of the function? I have tried everything. I am hoping to see 123
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Greg
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Paul Dowrick
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1why don't you return the value from the function and call the function from outside , how do you expect to access a variable out of its scope ? – Fisherman Oct 01 '14 at 08:02
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Use `global`. Or make it a return. Or pass a variable by reference. – Aneri Oct 01 '14 at 08:02
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There are so many syntax errors. You should use return in the function – TBI Oct 01 '14 at 08:03
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don't make a function for the sake of it, give it a clear purpose and a meaningful name... – SirDarius Oct 01 '14 at 08:04
2 Answers
3
make your function like this and return the $c
function pauls_code($a, $b){
$c = $a + $b;
return $c;
}
echo pauls_code(1,2);
Ram Sharma
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0
To declare a function in php you simply put function pauls_code($param1, $param2) not define
You can not access $c outside of the function, read more about scope: http://php.net/manual/en/language.variables.scope.php