Build an array dynamically in php [duplicate]

Question

I need that this array become dynamic: (initial output)

$data[0] = array('var0');

I have a for cycle that returns (after a few conditions) something as a VAR:

for($i = 0; $i < N; $i++) 
    {
    (...) END AS var1;
    (...) END AS var2;
    (...) END AS var3;
    (...) END AS varN;
    (...) END AS (...)
    }

Expected output after the first cycle:

 $data[0] = array('var0','var1');

Expected output after the second cycle:

 $data[0] = array('var0','var1','var2');

And go on (expected final output):

 $data[0] = array('var0','var1','var2','var3','varN');

How could it be done?

psx · Accepted Answer · 2013-01-21 15:41:04Z

3

To "push" something onto the end of the array, use the [] syntax:

$data[0][] = 'something to add to array';

E.g.

$data[0] = array();
$data[0][] = 'first';
$data[0][] = 'second';

$data[0] will contain:

array('first','second');

edited Jan 21, 2013 at 15:41

answered Jan 21, 2013 at 15:38

psx

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Shomz · Accepted Answer · 2013-01-21 15:42:24Z

3

for($i = 0; $i < N; $i++) {
    $data[0][] = 'var' . $i;
}

answered Jan 21, 2013 at 15:42

Shomz

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Comments

hakre · Accepted Answer · 2013-01-21 15:51:37Z

Why not prefix the numbers range, for example via array_map:

$data[$i] = array_map(function($v) {return 'var' . $v;}, range(0, $i));

And if you've got a range inside the range, you can do that, too:

$prefix = function($prefix) {
    return function($v) use ($prefix) {
        return $prefix . $v;
    };
};

$map = function($v) use ($prefix) {
    return array_map($prefix('var'), range(0, $v));
}

$data = array_map($map, range(0, $n - 1));

Or all at once (compacted):

$data = array_map(function($v) {
    return array_map(function($v) { return 'var' . $v;}, range(0, $v));       
}, range(0, $n - 1);

webnoob · Accepted Answer · 2013-01-21 15:39:09Z

0

Maybe you want: array_push(). You can add your vars to your array in your loop then.

answered Jan 21, 2013 at 15:39

webnoob

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