CODEWITHSUNDEEP
phparraysreference
grom
grom

Reputation: 16116

PHP, Arrays, and References

Why does the following code not work as I was expecting?

<?php
$data = array(
    array('Area1', null, null),
    array(null, 'Section1', null),
    array(null, null, 'Location1'),
    array('Area2', null, null),
    array(null, 'Section2', null),
    array(null, null, 'Location2')
);
$root = array();
foreach ($data as $row) {
    if ($row[0]) {
        $area = array();
        $root[$row[0]] =& $area;
    } elseif ($row[1]) {
        $section = array();
        $area[$row[1]] =& $section;
    } elseif ($row[2]) {
        $section[] = $row[2];
    }
}
print_r($root);

Expected result:

Array(
    [Area1] => Array(                         
            [Section1] => Array(
                    [0] => Location1
                )                   
        )
    [Area2] => Array(           
            [Section2] => Array(              
                    [0] => Location2
                )                   
        )
)

Actual result:

Array(
    [Area1] => Array(                         
            [Section2] => Array(
                    [0] => Location2
                )                   
        )
    [Area2] => Array(           
            [Section2] => Array(              
                    [0] => Location2
                )                   
        )
)

Upvotes: 5

Views: 366

Answers (2)

grom
grom

Reputation: 16116

This will also works:

$root[$row[0]] = array();
$area =& $root[$row[0]];

Upvotes: 1

pix0r
pix0r

Reputation: 31280

If you modify your code on two lines as follows:

$area = array();

$section = array();

to this:

unset($area);
$area = array();

unset($section);
$section = array();

it will work as expected.

In the first version, $area and $section are acting as "pointers" to the value inside the $root array. If you reset the values first, those variables can then be used to create brand new arrays instead of overwriting the previous arrays.

Upvotes: 3

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