Array in Object passed by value?

Question

Maybe I am overlooking something, but I stumbled across this:

$employees = new stdClass();

$employee_id = 5;
$employee = array();
$employee["id"] = $employee_id;
$employee["name"] = "John;

$employees->$employee_id = $employee;

Now I want to change the employee Name:

$employee = $employees->$employee_id;
$employee["name"] = "Tom";

Now I have two problems:

The employee object seems not to be passed by reference, because the employee within the employees is still named John.
How would I retrieve the employee name? echo {$employee->$employee_id}["name"]; does not work

Thanks for the help,

Martin

Why are you using $employees->$employee_id and not $employees->employee_id ? — Ian
– Ian, Commented Jan 14, 2010 at 15:42
1.) see comment to answer from Emil. 2.) Why is it stupid code? Of course the employee name will not get changed immediately, but later on in a while loop when mysql data is read out. What would you suggest? — martin
– martin, Commented Jan 14, 2010 at 16:01

Emil Vikström · Accepted Answer · 2010-01-14 15:36:29Z

2

1) PHP does rarely pass by reference. If you want to force a pass by reference, use the =& operator:

$employees->$employee_id =& $employee;

Read more here: http://php.net/references

2) To use $employees->$employee_id in the first place is rather ugly, but here is a possible solution:

$current_employee =& $employee->$employee_id;
echo $current_employee['name'];

answered Jan 14, 2010 at 15:36

Emil Vikström

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1 Comment

martin Over a year ago

ad 1) ok. But actually Problem 1) is solved when I solve Problem 2, cause then I can assign it directly. ad 2) Why is this rather ugly? I can not use an associative array, cause when it is send to the Frontend (ZendAMF) the keys do not get maintained, so I need to use an object. Any other ideas?

Rik Heywood · Accepted Answer · 2010-01-14 15:41:55Z

1

You have an extra $ that should not be there.

//$employees->$employee_id = $employee;     Wrong
$employees->employee_id = $employee;

With the dollar, php uses the value of var. eg...

$name = "car";
$obj->car = 5;
$obj->$name = 10;

echo $obj->car;  // outputs 10

answered Jan 14, 2010 at 15:41

Rik Heywood

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3 Comments

Ian Over a year ago

That confused me as well.. he basically has a class member named "5". :P

Ian Over a year ago

It apparently is since he did it twice, but I'm pretty sure he doesn't realize what the extra dollar sign does..

martin Over a year ago

thx for the comments. please read the comment I made above. Yes I made it on purpose, for serializing the data properly, as ZendAMF does not treat Associative Arrays with numeric indices. They are not maintained after they are sent over the wire.

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