how to use constant from class as an argument definition in php function?

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遇见更好的自我
遇见更好的自我 2021-02-06 03:26

I have a class:

class FetchMode 
{
 const FetchAll = 0;
 const FetchOne = 1;
 const FetchRow = 2;}

and a function:

function get         


        
2条回答
  •  感动是毒
    2021-02-06 04:17

    You've hinted PHP to expect an instance of FetchMode (just like it says in the error message), but FetchMode::FETCH* passes the constant value. You'd have to use some sort of Enum instance (which we dont have natively in PHP. (Oh well, there is SplEnum but who uses that?)) or change the method signature to exclude the typehint.

    However, instead of a Switch/Case you could solve this more easily via Polymorphism and a Strategy pattern, e.g. instead of doing something like

    public function getRecordSet($mode)
    {
        switch ($mode) {
            case FetchMode::ALL:
                // code to do a fetchAll
                break;
            case FetchMode::ONE:
                // code to do a fetchOne
                break;
            default:
        }
    }
    

    which will increase the Cylcomatic Complexity of your class and forces changes to that class and FetchMode whenever you need to add additional FetchModes, you can do:

    public function getRecordSet(FetchMode $fetchModeStrategy)
    {
        return $fetchModeStrategy->fetch();
    }
    

    and then have an interface to protect the variation

    interface FetchMode
    {
        public function fetch();
    }
    

    and add concrete FetchMode classes for each supported FetchMode

    class FetchOne implements FetchMode
    {
        public function fetch()
        {
            // code to fetchOne
        }
    }
    class FetchAll …
    class FetchRow …
    

    This way, you'll never have to touch the class with that getRecordSet method again because it will work for any class implementing that FetchMode inteface. So whenever you have new FetchModes, you simply add a new class, which is much more maintainable in the long run.

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