[PHP]Arrays and Notice: Undefined offset

Question

Can you please tell what is wrong here, because i get an annoying message in my if statement:

Notice: Undefined offset: 1

The code:

$maxlvls[14] = array(0, 30, 25, 20, 15, 1, 20, 25, 30, 30, 30, 30, 30, 10, 20);

$lvle[14] = array(0, (int)$_GET['ratuszlvl'], (int)$_GET['koszarylvl'], (int)$_GET['stajnialvl'], (int)$_GET['warsztatlvl'], (int)$_GET['palaclvl'], (int)$_GET['kuznialvl'], (int)$_GET['ryneklvl'], (int)$_GET['tartaklvl'], (int)$_GET['cegielnialvl'], (int)$_GET['hutalvl'], (int)$_GET['zagrodalvl'], (int)$_GET['spichlerzlvl'], (int)$_GET['schoweklvl'], (int)$_GET['murlvl']);

if($lvle[1] <= $maxlvls[1] && $lvle[1] >= 0)
{
    echo "smth";
}

Where do you expect $lvle[1] to come from, if you only assigned a value to $lvle[14] …? — C3roe
– C3roe, Commented May 11, 2014 at 13:40
I'm really disapointed from the questions under this tag recently — Ivan Yonkov
– Ivan Yonkov, Commented May 11, 2014 at 13:41
@user3576397 does C++ allow you to set only 14th element in array, but to use its 1st element instead? — Ivan Yonkov
– Ivan Yonkov, Commented May 11, 2014 at 13:41

souslov · Accepted Answer · 2014-05-11 13:46:33Z

4

Instead of

$maxlvls[14] =
$lvle[14] =

Use just

$maxlvls = 
$lvle =

You don't have to declare (actually you can't) array's length in php(I guess you are doing this). However,assigning like :

$maxlvls[14] = array(...);

Means that you are assigning an array to the element of array $maxlvls with key 14.

edited May 11, 2014 at 13:46

answered May 11, 2014 at 13:40

souslov

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1 Comment

Althain Over a year ago

So simple explanation... It works and you are right that i was trying to set an array length. I'm new in PHP - previously i was learning C++.

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