Date not converting in the array

Question

I var_dumped an array that shows these elements (there are 266, but I am showing you two).

array(266) {
  [1]=>
  array(5) {
    ["date_created"]=>
    string(10) "1381816800"
    ["project_number"]=>
    string(5) "02783"
    ["name"]=>
    string(9) "sdfsdfdfd"
    ["description"]=>
    string(13) "dsfsfdsfdssdf"
    ["manager"]=>
    string(11) "Kevin Allen"
  }
  [21]=>
  array(5) {
    ["date_created"]=>
    string(10) "1381816800"
    ["project_number"]=>
    string(5) "02783"
    ["name"]=>
    string(9) "sdfsdfdfd"
    ["description"]=>
    string(13) "dsfsfdsfdssdf"
    ["manager"]=>
    string(16) "Carter Hilkewich"
  }
}

I needed to convert the date in this array too: m-d-Y so I wrote:

private function dateConverter($array){
    foreach($array as $key=>$value){
        if(isset($value['date_created'])){
            $value['date_created'] = date("m-d-Y", strtotime($value['date_created']));
        }
    }

    return $array;
}

Which you pass the array into, walks through, converts the date and returns the array. simple. 'Cept what it's returning is the exact same array. So I am wondering do I need to save the "new" array into a separate array? I have a similar function that does this with objects and I never had to save a"new" object.

thoughts?

The strtotime() is unnecessary since the time is already a timestamp. — John Conde
– John Conde, Commented Oct 16, 2013 at 17:51

keithhatfield · Accepted Answer · 2013-10-16 17:51:44Z

1

Two things:

The foreach does not pass the value by reference, so modifying it directly will not store it back to the original array
The date_created field is already a timestamp, no need for strtotime

Try this:

private function dateConverter($array){
    foreach($array as $key => &$value){
        if(isset($value['date_created'])){
            $value['date_created'] = date("m-d-Y", $value['date_created']);
        }
    }

    return $array;
}

answered Oct 16, 2013 at 17:51

keithhatfield

3,2831 gold badge20 silver badges25 bronze badges

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

4 Comments

keithhatfield Over a year ago

The value in date_created appears to be a unix timestamp, the same kind that is returned by strtotime.

keithhatfield Over a year ago

The biggest problem with your code is the fact that you're modifying a copy of the array value and not the value in the original array. Notice the ampersand before $value in the foreach of the function I posted; that causes it to be passed into the array by reference and stored back to the original array when you change the value.

LogicLooking Over a year ago

I see that and I am still lost on "pass by reference" I have read the manual a couple of times. So this will pass the value I intend instead of just, as you say, a copy? so If I passed apples, I pass the original apples and NOT the copy of apples I would pass with out &?

keithhatfield Over a year ago

Passing by reference is simply telling where to find the value rather than passing a copy of the value. So, if you needed to pass an apple to your friend and passed it as $apple, you would copy your apple and give it to him. He could eat it, and you'd still have your apple. If you passed it by reference as &$apple you would tell him, "There's an apple on the counter in the kitchen next to the sink." If he takes a bite of the apple, your apple has a bite taken out of it because you just told him where the apple was, rather than giving him his own copy. Hard to explain pointers in 500 chars.

Collectives™ on Stack Overflow

Date not converting in the array

1 Answer 1

4 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

4 Comments

Your Answer

Sign up or log in

Post as a guest

Related