Why is 'the array pushed element' not echoed?
function dofoo1() {
$array = array("1aaa", "2bbbb", "3cccc");
$count = '######';
$array1 = array_push($array, $count);
return $array1;
}
$foo1 = dofoo1();
echo $foo1[3];
Add a comment
|
4 Answers
No need to assign array_push to a variable.
function dofoo1() {
$array = array("1aaa", "2bbbb", "3cccc");
$count = '######';
array_push($array, $count);
return $array;
}
$foo1 = dofoo1();
echo $foo1[3];
2 Comments
Latestarter
It's different from one in javascript. Gracias!
Muhammet
@Latestarter De nada :)
As described in the php docs array_push() alters the given array and returns only the number of elements.
Therefore you have to return $array instead of $array1.
If you just want to add one element, it is even better to avoid array_push() and use $array[] = $count; instead.
This usage is recommended in the docs of array_push().
So your code should look like this:
function dofoo1() {
$array = array("1aaa", "2bbbb", "3cccc");
$count = '######';
$array[] = $count;
return $array;
}
$foo1 = dofoo1();
echo $foo1[3];
Comments
You can simply use array merge
function dofoo1() {
$array = array("1aaa", "2bbbb", "3cccc");
$count = '######';
return array_merge($array, array($count));
}
$foo1 = dofoo1();
echo $foo1[3];
Comments
array_push() Returns the new number of elements in the array.
So, You should return the array itself in which you have pushed,
Change,
return $array1;
to
return $array;
