PHP 5.3.0 USE keyword -- how to backport in 5.2?

Question

I have some code that has been written in for php 5.3.0 using the USE function within PHP

can someone help me change this to work for 5.2.9?

$available  = array_filter($objects, function ($object) use ($week) { 
    return !in_array($object, $week);
});

thanks for the help

This may help you: http://stackoverflow.com/questions/1065188/in-php-5-3-0-what-is-the-function-use-identifier-should-a-sane-programmer-use — zaf, May 03 '11 at 12:01
You probably shouldn't call `use()` a function, as its a little confusing. — alex, May 03 '11 at 12:08

score 8 · Accepted Answer · edited May 03 '11 at 12:03

8

Not nice, but this would be an equivalent implementation.

class MyWeekFilter {
    protected $_week;

    public function __construct($week) {
        $this->_week = $week;
    }

    public function filter($object) {
        return !in_array($object, $this->_week);
    }
}

$filter    = new MyWeekFilter($week);
$available = array_filter($objects, array($filter, 'filter'));

edited May 03 '11 at 12:03

Shoe

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answered May 03 '11 at 12:01

Stefan Gehrig

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Well played, sir, well played. You just saved my life tonight. – dland Feb 13 '12 at 18:13

score 1 · Answer 2 · answered May 03 '11 at 12:57

1

Is there any difference between author's code

$available = array_filter($objects, function ($object) use ($week) { 
    return !in_array($object, $week);
});

and

$available = array_diff($objects, $week);

?

answered May 03 '11 at 12:57

binaryLV

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The author may have cut down the problem space to a simple example that is easier to follow. – dland Feb 14 '12 at 14:36

score 0 · Answer 3 · answered Jun 01 '12 at 06:00

0

$available  = array_filter($objects, create_function('$object', '
    $week = '.var_export($week,true).';
    return !in_array($object, $week);
'));

answered Jun 01 '12 at 06:00

newacct

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score -3 · Answer 4 · answered May 03 '11 at 12:03

-3

Try this:

$week = array(...); // defined and instantiated before...

function callback($object) {
    return !in_array($object, $week);
}
$available  = array_filter($objects, "callback");

answered May 03 '11 at 12:03

shadyyx

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This won't work. `$week` is a global variable, which must be *imported* into the function scope: `function callback($object) { global $week; return !in_array($object, $week); }` – Stefan Gehrig May 03 '11 at 12:04
1

`$week` is not available in callback unless pulled in using `global` – meouw May 03 '11 at 12:06
True. Sorry for missing this. – shadyyx May 03 '11 at 12:12

PHP 5.3.0 USE keyword -- how to backport in 5.2?

4 Answers4

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