php weird undefined offset error

Question

In my php applicaiton I have the following code:

try {
    $ordersIngredients[$userIngredient->getId()][$day] += $quantity;
} catch (\Exception $e) {
    ~r(array(
        $ordersIngredients[$userIngredient->getId()],
        $day,
        array_key_exists($day, $ordersIngredients[$userIngredient->getId()]),
        $e->getMessage()
    ));
}

The r() function prints the following:

array(4)
    0   =>  array(4)
        0   =>  0.9
        1   =>  null
        2   =>  null
        3   =>  1
    )
    1   =>  3
    2   =>  true
    3   =>  Notice: Undefined offset: 3
)

How can I have an error on un undefined offset when the offset actually exists given the dump of the array and array_key_exists ?

Side note: How do you have a function name that starts with ~? — phpisuber01
– phpisuber01, Commented Sep 3, 2015 at 18:23
@phpisuber01: Not part of the name php.net/manual/en/language.operators.bitwise.php — AbraCadaver
– AbraCadaver, Commented Sep 3, 2015 at 18:25
How is your catch catching this? It's a notice not an exception/error. — gen_Eric
– gen_Eric, Commented Sep 3, 2015 at 18:42
@RocketHazmat Maybe he's defined a custom error handler, as in stackoverflow.com/questions/5373780/… — Barmar
– Barmar, Commented Sep 3, 2015 at 18:50

gen_Eric · Accepted Answer · 2015-09-03 19:11:33Z

1

The issue here is that you were trying to append to a value that did not exist in the array.

$ordersIngredients[$userIngredient->getId()][$day] += $quantity;

This is the same as writing:

$ordersIngredients[$userIngredient->getId()][$day] = 
    $ordersIngredients[$userIngredient->getId()][$day] + $quantity;

So, what's happening is that PHP is trying to read the value at index 3, but it can't. That's what causes your notice. Since it's only a notice, PHP treats the undefined index as null and keeps going. It then adds null + $quantity to your array.

After finishing the line, it moves into your error handler and then to your catch block.

answered Sep 3, 2015 at 19:11

gen_Eric

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