PHP assigning variables

Question

$start  = $this->getWeekRange($date);
$end    = $start->modify("+6 days");

echo $start->format("Y-m-d");
echo $end->format("Y-m-d"); 
exit();

OUTPUT:

2013-12-08
2013-12-08

it should be

2013-12-02
2013-12-08

why is that both $start and $end has the same value? even though I've already assigned the value on the $start variable before modifying it then assigning it to $end.

var_dump($start, $end);

u_mulder
– u_mulder

2013-12-07 12:07:47 +00:00
Commented Dec 7, 2013 at 12:07 — u_mulder
– u_mulder, Commented Dec 7, 2013 at 12:07

DAG · Accepted Answer · 2013-12-07 12:29:59Z

2

Objects are assigned by reference in PHP (and many other languages).

This means that $end and $start are pointing to the same object. In order to get a clone of that object you have to use clone:

$end = clone $start;

Now you have a seperate object in $end which has the same properties as $start; until you call methods or modify one of them.

For your example you should put the above line in the second line of your example and modify this line:

$end = $start->modify("+6 days");

to:

$end->modify("+6 days");

edited Dec 7, 2013 at 12:29

answered Dec 7, 2013 at 12:15

DAG

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1 Comment

DAG Over a year ago

Yeah, I haven' t seen a diffrent way of doing this so far. Additionally using pure language approaches is always good.

Mehdi · Accepted Answer · 2013-12-07 12:05:06Z

0

try this:

$start  = $this->getWeekRange($date);
$end    = $start;
$end->modify("+6 days");

echo $start->format("Y-m-d");
echo $end->format("Y-m-d"); 
exit();

answered Dec 7, 2013 at 12:05

Mehdi

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