Using php pointers on function?

Question

I'm creating a website and somewhere in the code I need to query for a user attribute (ex:account state) and in the same row I have the reason, case is account state is "suspended".

I'm trying to minimize the requests to the database, so I created a function to verify account state.

function getAccountState($userid,$reason){}

What I am trying to do is if account state is "suspended" I would change the $reason to "the database reason".

I've already done that but if I change the $reason inside the function, outside the function it will not change. I searched for "php pointers" on google but I think there is not such thing.

Is there a way to do this? Other way I'll just make another database request...

PHP doesn't have pointers or dynamic memory. Even references aren't exactly like references you'd see in C++ (although they're used pretty much the same way). — Major Productions
– Major Productions, Commented Feb 19, 2013 at 17:12

jeroen · Accepted Answer · 2013-02-19 17:07:21Z

2

You could of course pass the variable by reference but as you don't seem to need it, I would just return it from the function:

function getAccountState($userid){
  // your code
  return $reason;
}

and call it like:

$reason = getAccountState($userid);

If you want to stay your code as it is now, you could pass the variable by reference:

function getAccountState($userid,&$reason){}
                                 ^ like so

answered Feb 19, 2013 at 17:07

jeroen

91.8k22 gold badges117 silver badges132 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

Ricardo Neves Over a year ago

Well, I was returning the "state" attribute, so the first way is not the one i'm looking. The second one was perfect. Thanks for the fast response !

jeroen Over a year ago

@Ricardo Neves You could return an array or just use the second solution :-)

juco · Accepted Answer · 2013-02-19 17:08:31Z

0

You could consider passing it in by reference. Or perhaps just changing the function to return the correct information.

answered Feb 19, 2013 at 17:08

juco

6,3423 gold badges27 silver badges42 bronze badges

Comments

Karisters · Accepted Answer · 2013-02-19 17:10:00Z

0

In the definition of functions you can tell that $reason argument is passed by reference, not value. To do so, use & in front of variable:

function getAccountState($userid,& $reason){}

answered Feb 19, 2013 at 17:10

Karisters

1865 bronze badges

Comments

Fabien · Accepted Answer · 2013-02-19 17:13:03Z

0

You are looking for references, in PHP terminology, not pointers.

They work this way :

function getAccountState($userid, &$reason){  // Notice the &
    $reason = "database locked";  // Use it as a regular variable
}


getAccountState(12345, $reason);  // Here, it is written as a regular variable, but it is a ref.
echo $reason;   // echoes "database locked"

answered Feb 19, 2013 at 17:13

Fabien

13.6k10 gold badges48 silver badges66 bronze badges

Collectives™ on Stack Overflow

Using php pointers on function?

4 Answers 4

2 Comments

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

2 Comments

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related