PHP undefined offset notice

Question

I have the following if condition:

if ( 
    $items[ $count + 1 ]->menu_item_parent != $parent_id && 
    $submenu && 
    $items[ $count + 1 ]->menu_item_parent != $item->ID) {

    //something

}

I get the notice:

Undefined offset: 13

So I assume the 13 does not exist in the $items array and it can be avoided by checking with in_array. However I'm unsure how I can use that function in the above condition.

You will want to have a look at the output of what you're testing on. What does that reveal? — SaeX
– SaeX, Commented Apr 30, 2015 at 8:15
Also instead of $count+1 can you use ++$count or $count++ whichever goes with you rather than that? — Preeti Maurya
– Preeti Maurya, Commented Apr 30, 2015 at 8:18
@PreetiMaurya In this case, $count+1 is necessary, because $count++ would actually alter the value of $count, whereas $count+1 does not (which lets him use that more than once). — Seer
– Seer, Commented Apr 30, 2015 at 8:29

Sanjay Kumar N S · Accepted Answer · 2015-04-30 08:16:55Z

7

if ( isset($items[ $count + 1 ]) && 
    $items[ $count + 1 ]->menu_item_parent != $parent_id && 
    $submenu && 
    $items[ $count + 1 ]->menu_item_parent != $item->ID) {

    //something

}

answered Apr 30, 2015 at 8:16

Sanjay Kumar N S

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1 Comment

Sanjay Kumar N S Over a year ago

The idea is that, first need to check whether the items array's index is set or not before accesing that index element. So if that is set only it checks for the succeeding condtions otherwise it will return false and exit from the if condition.

Bob Nocraz · Accepted Answer · 2015-04-30 10:27:13Z

3

You could first check if $items[$count + 1] exists. Like

if( 
   //check first if $items[$count + 1] exists, else it breaks here
   isset($items[ $count + 1 ] )

   //now you know $item[$count + 1] exists, so you can continue
   &&
     ( 
        $items[$count + 1]->menu_item_parent != $parent_id && 
        $submenu && 
        $items[ $count + 1 ]->menu_item_parent != $item->ID
     ) 
   ){

    //something

  }

edited Apr 30, 2015 at 10:27

Bob Nocraz

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answered Apr 30, 2015 at 8:19

Michel

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Comments

Seer · Accepted Answer · 2015-04-30 08:27:10Z

1

You should use a prefix test using isset method.

The call would look like that:

if (isset($items[ $count + 1 ])) { //do stuffs}

edited Apr 30, 2015 at 8:27

Seer

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answered Apr 30, 2015 at 8:15

Answers_Seeker

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Comments

Loko · Accepted Answer · 2015-04-30 08:28:53Z

Ok so the in_array docs say:

in_array ( mixed $needle , array $haystack)

Where

needle = the searched value

haystack = the array

And you are not looking for the value, you're looking for the key

When you look at the bottom under see also, it says:

array_key_exists() - Checks if the given key or index exists in the array

which is what you are looking for:

array_key_exists() returns TRUE if the given key is set in the array. key can be any value possible for an array index.

Perfect! Exactly what you want!

array_key_exists ( mixed $key , array $array )

where

key = value to check

array = array (duh!)

if ( array_key_exists($count + 1, $items))
    $items[ $count + 1 ]->menu_item_parent != $parent_id && 
    $submenu && 
    $items[ $count + 1 ]->menu_item_parent != $item->ID) {

    //something

}

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