Pass variables when calling method

Question

I am trying to pass variables when calling a method. I am trying to do something like this:

 $c->house_num.$id($name);

The name of the method is house_num_(number). I then want to pass another variable $name.

I am getting an error, what is the right way to do this? Thankyou

you should be getting an undefined constant house_num notice here. — Funk Forty Niner, Oct 21 '15 at 18:55
Why is the ID part of the method name, instead of another argument? — Barmar, Oct 21 '15 at 18:56
@Fred-ii- No, it's parsed as `$c->house_num . $id($name)`, so it's looking for a property named `house_num` — Barmar, Oct 21 '15 at 18:57
I am trying to make it dynamic so that the id can be passed in, as there are multiple house_num_(number) methods — Ryan, Oct 21 '15 at 18:57
@Ryan My point is that you shouldn't have multiple `house_num_` methods. You should have one `house_num($number, $name)` method. — Barmar, Oct 21 '15 at 18:58
@Barmar thanks. I questioned even myself after posting my comment about it. I should've deleted it when my little voice spoke to me. — Funk Forty Niner, Oct 21 '15 at 18:58
@Ryan Any time you find yourself creating variables or methods in a numeric sequence, you should think about using an array instead. — Barmar, Oct 21 '15 at 18:59
Possible duplicate of [How to call PHP function from string stored in a Variable](http://stackoverflow.com/questions/1005857/how-to-call-php-function-from-string-stored-in-a-variable) — Djizeus, Oct 21 '15 at 19:57

score 1 · Answer 1 · answered Oct 21 '15 at 18:59

1

$c->house_num($id,$name);

And then use somthing like

function house_num($id,$name){
switch($id){
 //call functions...
}

answered Oct 21 '15 at 18:59

itay

score 0 · Answer 2 · answered Oct 21 '15 at 19:02

0

you should be use that :

$func = 'house_num_'.$id;

$c->$func($name);

answered Oct 21 '15 at 19:02

Lucas Burg

2 Answers2