Message: Attempt to increment/decrement property of non-object - php

Question

$order_temp = $this->mdl_admin->get_latest_order_id($_POST['parent_id']);
if ($order_temp) {
$order = (string)$order_temp->order++;
var_dump($order);
die();
}
This code above produce this error:
Message: Attempt to increment/decrement property of non-object
And the vardump is string '' (length=0)
If I do just something like this to vardump the variable:
$order_temp = $this->mdl_admin->get_latest_order_id($_POST['parent_id']);
if ($order_temp) {
var_dump($order_temp);
die();
}
the output is :
array (size=1)
0 =>
object(stdClass)[28]
public 'id' => string '16' (length=2)
public 'name' => string 'sssssssssssssss' (length=15)
public 'slug' => string 'aaaaaaaaa' (length=9)
public 'title' => string 'aaaaaa' (length=6)
public 'body' => string '<p>asdas asd asd </p>' (length=21)
public 'order' => string '1' (length=1)
public 'parent_id' => string '5' (length=1)
I just want to add 1 to the $order_temp->order so if e.g. the $order_temp->order = 2 the result $order = 3.
What am I doing wrong?

Answer 1

$order_temp is not an object, as the error message says. When you try to access (or in this case, increment) a property of something that isn't an object (and so does not have properties), you get this error.
Not knowing what $this->mdl_admin->get_latest_order_id does, I can only go so far, BUT, I can tell you to use explicit comparisons in if statements, and that your type coercion isn't doing what you expect.
Use explicit comparison:
if (is_object($order_temp) === true)
.. that might not be the cause of your bug, but that's good practice.
Your use of ((string)) in assignment coerces the presumed object variable to a string, which you then try to access property of. This is not the droid you are looking for. It appears, from PHP's perspective, that you want to do this:
"0"->order++;
...which doesn't make sense.
PHP uses type juggling[doc]. Whether your object's property order started out as a string or not, as soon as you use the increment operator ++, PHP juggles the type to become an integer. If you want a string, all you have to do is start treating it like a string:
$order_temp->order++; // <-- variable is an integer, $order_temp->order === 1
$order_temp->order .= " and bob's your uncle"; // <-- variable is a string, $order_temp->order === "1 and bob's your uncle"
For that reason, type coercion is seldom necessary in PHP.
Finally, you can do this exactly in the way that you're trying, but you'll have to understand the way the increment ++ operator works.
$number = 0;
$output = $number++;
echo "Output: ".$output; // Output: 0
$number = 0;
$output = ++$number;
echo "Output: ".$output; // Output: 1
... by placing the increment operator at the FRONT of the assignment, the return will be the incremented variable after the increment operation. If you put the operator at the end, then the return is the variable before the increment operation.
That means you could do this:
$order_number_as_string = (string) (++$one_order->order);
By enclosing the increment operator in parenthesis and type casting the outside, plus putting the increment operator as a prefix, you will get the expected results.
Documentation
Increment/Decrement operators - http://php.net/manual/en/language.operators.increment.php
Type juggling - http://php.net/manual/en/language.types.type-juggling.php
settype - http://php.net/manual/en/function.settype.php

Answer 2

<?php
class Test
{
public $order = 10;
}
$order_temp = new Test();
$order = (string)$order_temp->order++;
var_dump($order);
//string(2) "10"
$order = (string)$order_temp->order;
var_dump($order);
//string(2) "11"
?>
Your $order_temp is not an object. So that is what is causing the problems. Also note that you might not get the expected results, even if it works.
Your $order will contain the OLD value. See my test.
UPDATE
The var_dump of your $order_temp shows that its an array. You could use
$order_temp[0]->order++;
$order = $order_temp[0]->order
var_dump($order);