On a foreach loop, it seems PHP reads the whole array at the beginning, so if you suddenly need to append new items to the array they won't get processed by the loop:
$a = array (1,2,3,4,5,6,7,8,9,10);
foreach ($a as $b)
{
echo " $b ";
if ($b ==5) $a[] = 11;
}
only prints out: 1 2 3 4 5 6 7 8 9 10
Just create a reference copy of the array you are looping
$a = array(1,2,3,4,5,6,7,8,9,10);
$t = &$a; //Copy
foreach ( $t as $b ) {
echo " $b ";
if ($b == 5)
$t[] = 11;
}
Or Just use ArrayIterator
$a = new ArrayIterator(array(1,2,3,4,5,6,7,8,9,10));
foreach ( $a as $b ) {
echo "$b ";
if ($b == 5)
$a->append(11);
}
Output
1 2 3 4 5 6 7 8 9 10 11
$a = [1,2,3,4,5,6,7,8,9,10]; while (($b = current($a)) !== false) { echo " $b "; if ($b === 5) $a[] = 11; next($a); }On the spirit of its-ok-to-ask-and-answer-your-own-questions, this is the best workaround I have found: convert it to a while loop.
$a = array (1,2,3,4,5,6,7,8,9,10);
$i = 0;
while ($i < count($a))
{
$b = $a[$i];
echo " $b ";
if ($b ==5) $a[] = 11;
$i++;
}
Not it properly gives out 1 2 3 4 5 6 7 8 9 10 11
for ($i = 0; $i < count($a); $i++) {
A simple way to do it is to create a second array which will keep the old array and while looping you can add new values there .
At the end of the loop the new array will have the old one + new inserted values
$arr = [1,2,3];
while ($el = current($arr)) {
var_dump($el);
$arr[] = 123; // infinite loop
next($arr);
}
