Why "foreach" don't echo? - php

Question

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<?php
$a = 3;
if ($a>1){
$arr = array (1,2,3);
}
foreach ($arr as $b) {
echo $b[0];
echo $b[1];
echo $b[2];
}
var_dump($arr);
?>
I don't know why it can not echo in foreach?
But var_dump($arr) still run with result:
array(3) { [0]=> int(1) [1]=> int(2) [2]=> int(3) }
But when I wrote: $arr[] it can run.
<?php
$a = 3;
if ($a>1){
$arr[]= array (1,2,3);
}
foreach ($arr as $b) {
echo $b[0];
echo $b[1];
echo $b[2];
}
var_dump($arr);
?>
Result:
123
array(1) { [0]=> array(3) { [0]=> int(1) [1]=> int(2) [2]=> int(3) } }
Both of them are same result with var_dump. So what different between $arr and $arr[] ?

Answer 1

These two lines:
$arr = array (1,2,3);
and
$arr[]= array (1,2,3);
are not equivalent. Look more closely at the two var_dump outputs, and you'll see the difference.
In the former, you're creating a one dimensional array - when you try to loop over it, you'll get an iteration with $b set to each of the three values (1, 2 and 3) in turn. Each time, $b is an integer. Any "index" of it will return null, since you can't deference scalar values (other than strings). This is defined in the manual here:
Array dereferencing a scalar value which is not a string silently yields NULL, i.e. without issuing an error message.
And when you echo null, nothing happens. It's the equivalent of an empty string, and so no output is produced.
In the second case, you're creating a two dimensional array. Writing
$arr[]= array (1,2,3);
when $arr is empty is the same as writing
$arr = array(array (1,2,3));
This time, when you loop over it, you get a single iteration, with $b set to the inner array. Now, echo-ing $b[0], $b[1] and $b[2] refers to the integers in the array, so you get your expected output of
123

Answer 2

Use this code instead of you written above:
<?php
$a = 3;
if ($a>1){
$arr = array (1,2,3);
}
foreach ($arr as $b) {
echo $b;
echo '<br>';
}
var_dump($arr);
?>