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up vote 2 down vote favorite

I need to be able to echo 1 and 2 with an array_push in the same foreach loop of that array.

<?php
$arr = array('1');

foreach ($arr as &$arr_number) {
   echo "$arr_number\n"; //this print 1
   array_push($arr, '2');
}

foreach ($arr as &$arr_number) {
   echo "$arr_number\n"; //this print 1 and then 2
}
?>

EDIT: Solved it myself by not using a foreach but a while loop

    <?php
    $arr = array('1');

 while ( list($key, $value) = each($arr) ) {
    echo "$value\n";

    if ( !in_array('2', $arr) ) {
       array_push($arr, '2');
    }
 }
 ?>
share|improve this question
1  
unable to understand question. can you please elaborate? –  عثمان غني Mar 4 '14 at 7:28
1  
What you're trying to do? –  hindmost Mar 4 '14 at 7:29
1  
What is the aim for using references in your code? –  zerkms Mar 4 '14 at 7:30
    
@echo: echo 1 and 2 –  Zurd Mar 4 '14 at 7:36
    
@zerkms: yeah, probably no point in using references... doesn't matter, code still doesn't work with foreach without references –  Zurd Mar 4 '14 at 7:42

1 Answer 1

up vote 4 down vote

I need to be able to echo 1 and 2 with an array_push in the same foreach loop of that array.

Implode it.

<?php
$arr = array('1');

foreach ($arr as &$arr_number) {

   array_push($arr, '2');
   echo implode(' ',$arr); //"prints" 1 2
}
share|improve this answer
    
would that not be an infinite loop ? –  KarelG Mar 4 '14 at 7:33
2  
How would it ? FYI, Demo –  Shankar Damodaran Mar 4 '14 at 7:34
    
you're pushing an variable into an array which you're looping through. I thought that the iterator would detect the newly added item and thus continuing the loop. But it's an instance at that moment. My excuses for my confusion. –  KarelG Mar 4 '14 at 7:37
    
@KarelG The implode function above simply concatenates all the array values together with a space. It won't affect the loop. Also the iterator does not detect the newly added value which is why the OP is receiving only 1 as output. –  I Can Has Kittenz Mar 4 '14 at 7:37
    
@KarelG No, it's not. see the first answer of stackoverflow.com/questions/2348077/… –  Hereblur Mar 4 '14 at 7:39

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