array_walk not doing the walk for me

Question

When I pass an array as an extra argument to array walk, it does not get interpreted as I would expect.

function boom($item, $z )
{
    print_r("\n".$item);
    print_r("\n".$z);
}

$z=[ "alpha", "bravo" ];
$x=[ "one", "two" ];
array_walk( $x, 'boom', $z );

The output:

one
0 
two 
1

Why is zero and one printed? Why don't I get alpha and bravo printed? How can I get alpha and bravo included in the output?

Thanks!

Read the manual, the second argument is the key and the third one will be your extra data — Rizier123, Mar 18 '16 at 17:12
Ah! Well spotted... I have my php man page open prior to raising the question but now I see what you suggest... let me try... — , Mar 18 '16 at 17:16
Yes! Thanks for that... Do you want to put that suggestion into an answer so I can give you the credit/points for it? — , Mar 18 '16 at 17:17

Daniel Dudas · Accepted Answer · 2016-03-18T17:27:41.910

1

You have to change the code to pass $z as 3rd parameter like this:

function boom($item, $i, $z )
{
    print_r("\n".$item);
    print_r("\n".$z[$i]);
}

$z=[ "alpha", "bravo" ];
$x=[ "one", "two" ];
array_walk( $x, 'boom', $z );

You can find more here: http://php.net/manual/en/function.array-walk.php

If the optional userdata parameter is supplied, it will be passed as the third parameter to the callback.

edited Mar 18 '16 at 17:27

answered Mar 18 '16 at 17:23

Daniel Dudas

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1

You have to use $z[$i]. The third parameter passed to array_walk can be anything and does not need to be an array. So you do get the full parameter passed to the callback function. – NineBerry Mar 18 '16 at 17:26
@NineBerry fixed that. Thanks. – Daniel Dudas Mar 18 '16 at 17:27

array_walk not doing the walk for me

1 Answers1