Establishing variable scope in PHP class?

Question

This is a very straightforward question that doesn't appear to be directly addressed at php.com - at least from looking through that section.

In any case, I have a class here with a specific function:

class CheckOut extends DB_MySQL{

public $fName;
public $lName;
public $numberOut;
public $p_id;

    /.../

    protected function publisherCheck($lName, $fName)
    {
        $this->lName = $lName;
        $this->fName = $fName;

        //Execute test
        $this->checkConnect();
        $stmt = $this->dbh->prepare("SELECT p_id FROM People WHERE lastName = :param1 AND firstName = :param2");
        $stmt->bindParam(':param1', $this->lName);
        $stmt->bindParam(':param2', $this->fName);
        $stmt->execute();

        //Determine value of test
        if($stmt == FALSE)
        {
            return FALSE;
        }
        else
        {
            $p_id = $stmt->fetch();
        }

    }

Just ignore the fact that there is no constructor posted with missing functions, etc. They're in this class - just not pertinent to my question.

Will setting $p_id in the last statement affect the variable declared initially in the header of the class? Essentially, will it be global within the class?

Any help is appreciated.

alex · Accepted Answer · 2012-09-18 03:42:02Z

3

Nope, it won't. You always need $this-> to tell PHP you're talking about the class properties, not local variables.

// Always assignment of a local variable.
$p_id = $stmt->fetch();

// Always assignment of a class property.
$this->p_id = $stmt->fetch();

answered Sep 18, 2012 at 3:42

alex

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1 Comment

Mlagma Over a year ago

Ha, didn't even catch that even though I did that for every other variable. Well, that answered it as well as fixed a future issue. Thanks.

user849137 · Accepted Answer · 2012-09-18 03:45:05Z

0

No. That's a local variable to your function. If you did $this->$p_id = 'blah'; then it would affect it. The variable you have defined in your class is a property, so it has to be accessed/altered with $this->...., whereas the variable you have in your function, is just a local variable (which you can play with by simply doing $p_id='....' ).

So,

$this->$p_id = '';//will alter the class property

and

$p_id = '';//will alter the local var defined/used in the function

answered Sep 18, 2012 at 3:45

user849137

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