Array check undefined offset php

Question

I'll try to explain it.

i have a array:

 $arrayTime = array(0=>"07",1=>"09", 3=>"13", 4=>"15", 5=>"17", 6=>"19");

Here you can see that is not defined offset 2 and now i need for my array and on offset 2 push number 0(for example) I tried use this:

if($arrayTime[$i]==""){
   $arrayTime[$i]=0;
}

Yes it works but 50 to 50 array looks like this:

$arrayTime = array(0=>"07",1=>"09", 3=>"13", 4=>"15", 5=>"17", 6=>"19",2=>"0");

but on the line where is the if it throws an error:

Notice: Undefined offset: 2 in C:\wamp\www\xxx.php on line 10

So i need same result, but without error. Thanks for your help all :)

It doesn't throw an error, it is noticing you are trying to access an undefined index, warning you about a possible bug in your code. Usually this comes in a package, you trying to script something that is 'bad practice'. The fact you're trying to resolve it is good. Instead of $arrayTime[$i]=="" do empty($arrayTime[$i]) — Xorifelse
– Xorifelse, Commented Mar 27, 2017 at 0:34
Ou nice thank you, if you want point, answer my question. I will mark it as correct. — Dolis
– Dolis, Commented Mar 27, 2017 at 0:38

Xorifelse · Accepted Answer · 2021-03-05 14:24:41Z

31

First of all, it doesn't throw an error. It gives you a warning about a possible bug in your code.

if($arrayTime[$i]==""){}

This attempts to access $arrayTime[$i] to retrieve a value to compare against your empty string.

The attempt to read and use a non-existing array index to get a value for comparison is the reason why it throws the warning as this is usually unexpected. When the key does not exist null is used instead and the code continues executing.

if(null == ""){} // this evaluates to true.

Because you are comparing against an empty string "", your answer would be empty():

if(empty($arrayTime[$i])){}

It means you are expecting a key not to exist and at the same time you are checking the value for emptyness. See the type comparison table to see what is and what is not considered 'empty'.

The same rules apply to isset() and is_null(), it wont throw the notice if the key does not exist. So choose the function that best serves your needs.

Keep in mind that by using any of these functions you are checking the value and not if the key exists in the array. You can use array_key_exists() for that.

if(array_key_exists($i, $arrayTime)){}

edited Mar 5, 2021 at 14:24

answered Mar 27, 2017 at 0:42

Xorifelse

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1 Comment

Dolis Over a year ago

Oukey, thank you so much for your help and for the valuable information! :)

Pabhoz · Accepted Answer · 2017-03-27 00:52:03Z

5

to add zeroes to your non-defined indexes without getting a Notice you should evaluate if the desired index to compare exists, so instead of comparing directly try checking the existence of the index first by using isset method, checking if the variable is defined and isn't NULL.

So your code to validate should look like this:

    //check for the index before tryin' to acces it
    if( !isset($arrayTime[$i]) ){
       $arrayTime[$i]=0;
    }

Hope it works for you.

edited Mar 27, 2017 at 0:52

answered Mar 27, 2017 at 0:43

Pabhoz

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6 Comments

RiggsFolly Over a year ago

Except now when an occurance does not exist, it wont run the code to add the occurance!

Xorifelse Over a year ago

isset() would be valid for undefined or null. Since the OP is comparing against "", isset() will return true, while the valid answer would be !isset() or empty(), where empty() is more dynamic on what is empty.

Pabhoz Over a year ago

@Xorifelse isset check if the index is not defined, and in your examples index 2 wasn't so isset should return true and fill index 2 with zero value

Pabhoz Over a year ago

@Xorifelse well, your problem need to check for not defined indexes to fill them all with 0, not precisely to check indexes filled with "", so =="" validation doesn't got to go, but if the problem is now by checking not defined index and "" filled ones it should.

Pabhoz Over a year ago

@Xorifelse if( empty($arrayTime[$i]) ){ $arrayTime[$i]=0; } will replace a defined index with a false value too not only the not defined as !isset()

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