Filtering array values using integers in PHP

Question

I am trying to filter a php array by comparing a field containing a unix date string ("start-unix-date") with the current unix time (simplified in this example as "15"). I want to keep only items where the unix date is in the future (>= than present).

Here is a simplified version of my code - based on this filtering method:

$my_array = array(
    array( 'start-date-unix' => '20' ), 
    array( 'start-date-unix'  => '10')
);

$unix_now = 15;

function filter_future($var) {
    return (is_array($var) && $var['start-date-unix'] >= $unix_now)
}

$filtered_array = array_filter($my_array, "filter_future");

var_dump($filtered_array);

The expected result:

The $filtered_array should keep only the item with a "start-date-unix" value of 20.

Actual result:

That filter doesn't do anything, every value is accepted. The $filtered_array still contains both items.

What works:

In the filter_future, if I enter the number (15) instead of the $unix_now variable, it works as expected: only the "20" item is kept.

So I suspect that there's an issue with $unix_now being a string vs an integer. I tried many methods of enforcing it being an integer but nothing seems to work :(

Mark Baker · Accepted Answer · 2013-09-11 11:34:33Z

1

$unix_now is out of scope within the callback function:

Use a closure with the use keyword to pass the $unix_now value to the callback

$unix_now = 15;
$filtered_array = array_filter(
    $my_array, 
    function ($var) use ($unix_now) {
        return (is_array($var) && $var['start-date-unix'] >= $unix_now)
    }
);

answered Sep 11, 2013 at 11:34

Mark Baker

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1 Comment

Ben Over a year ago

This seems like the more elegant option out of the 4 (so far) answers (IMHO)

iiro · Accepted Answer · 2013-09-11 11:32:15Z

0

$unix_now should be a global variable inside filter_future-function.

$unix_now = 15;

function filter_future($var) {
    global $unix_now;
    return (is_array($var) && (int) $var['start-date-unix'] >= $unix_now)
}

$filtered_array = array_filter($my_array, "filter_future");

var_dump($filtered_array);

The result:

array(1) { [0]=> array(1) { ["start-date-unix"]=> string(2) "20" } }

answered Sep 11, 2013 at 11:32

iiro

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1 Comment

Manu Over a year ago

Sure, it's just that stackoverflow imposes a 10 min delay before I could approve, and by then I was back in my workflow :)

Your Common Sense · Accepted Answer · 2013-09-11 11:33:20Z

0

$my_array = array(
    array( 'start-date-unix' => '20' ), 
    array( 'start-date-unix'  => '10')
);

$unix_now = 15;

foreach ($my_array as $i => $row)
{
    if ($row['start-date-unix'] < $unix_now)
    {
        unset($my_array[$i]);
    }
}

answered Sep 11, 2013 at 11:33

Your Common Sense

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Comments

rakeshjain · Accepted Answer · 2013-09-11 11:34:23Z

0

Change your function like below:

function filter_future($var) {
    $unix_now = 15;
    return (is_array($var) && $var['start-date-unix'] >= $unix_now);
}

answered Sep 11, 2013 at 11:34

rakeshjain

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Comments

david · Accepted Answer · 2013-09-11 11:43:09Z

0

move $unix_now = 15; inside function filter_future

function filter_future($var) {
    $unix_now = 15;
    return (is_array($var) && $var['start-date-unix'] >= $unix_now);
}

$unix_now inside your function is not recognize outside variable $unix_now. So put it inside your function is better

answered Sep 11, 2013 at 11:43

david

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1 Comment

Manu Over a year ago

Actually that function, $unix_now, is being defined at a totally different place in my code. My example here was a very simplified version.

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