Trying to access array offset on value of type null in

Posted: Fri May 05, 2023 4:04 am

Hello, running php 7.4 I got these errrors from file mainfileend.php

Notice: Trying to access array offset on value of type null in /www.bestbuildpc.org/modules/Your_Account/includes/mainfileend.php on line 24

Notice: Trying to access array offset on value of type null in /www.bestbuildpc.org/modules/Your_Account/includes/mainfileend.php on line 25

Notice: Trying to access array offset on value of type null in /www.bestbuildpc.org/modules/Your_Account/includes/mainfileend.php on line 29

Code:

if (isset($user)) {


   $uinfo = getusrinfo($user);


   $ulevel = $uinfo['user_level'];


   $uactive = $uinfo['user_active'];


   if (($ulevel < 1) OR ($uactive < 1)) {


      unset($user);


      unset($cookie);


      $sql = 'DELETE FROM `' . $prefix . '_bbsessions` WHERE `session_user_id`="' . $uinfo['user_id'] . '"';


      $db->sql_query($sql);

What should be the solution? This is happening when I try to deny a user or remove it.

Regular Joined: Oct 15, 2007 Posts: 87

That error is saying that null is being given with for the variable $uinfo, which must mean the function getusrinfo() is returning a null value instead of an array.

This may not be right, but, my guess would be to throw in a check to see if it's not null and is an array.

Let's take the full code:

php Code:

if (isset($user)) {


	$uinfo = getusrinfo($user);


	$ulevel = $uinfo['user_level'];


	$uactive = $uinfo['user_active'];


	if (($ulevel < 1) OR ($uactive < 1)) {


		unset($user);


		unset($cookie);


		$sql = 'DELETE FROM `' . $prefix . '_bbsessions` WHERE `session_user_id`="' . $uinfo['user_id'] . '"';


		$db->sql_query($sql);


	}


}

By changing that to something like this, should stop that error from happening.

php Code:

if (isset($user)) {


	$uinfo = getusrinfo($user);


	if ($uinfo !== null && is_array($uinfo)) {


		$ulevel = $uinfo['user_level'];


		$uactive = $uinfo['user_active'];


		if (($ulevel < 1) OR ($uactive < 1)) {


			unset($user);


			unset($cookie);


			$sql = 'DELETE FROM `' . $prefix . '_bbsessions` WHERE `session_user_id`="' . $uinfo['user_id'] . '"';


			$db->sql_query($sql);


		}


	}


}

I've set up Raven on my local running PHP 7.4.9, and I can't reproduce the error, so I don't know.

Posted: Fri May 05, 2023 4:18 pm

Thanks a lot. The errors is not coming back anymore. I wonder why u didn't get it on localhost running php 7.4.

New Member Joined: Apr 26, 2013 Posts: 1

You may have error reporting turned off on your localhost.

Posted: Fri May 05, 2023 8:03 pm

hicuxunicorniobestbuildpc wrote:

Thanks a lot. The errors is not coming back anymore. I wonder why u didn't get it on localhost running php 7.4.

n/p, glad it helped.

lonestar1983 wrote:

You may have error reporting turned off on your localhost.

Huh, I just looked and in the php.ini I have it turned on and even in the menu for WAMP. I will have to look into it because it's not writing it to a log file either.

Site Admin Joined: Aug 22, 2007 Posts: 1772

Thanks corpse! Just added the fix to the repo. [ Only registered users can see links on this board! Get registered or login! ]

I wouldn't recommend it to disable the error-reporting on a local environment, when it is used for testing. But I guess he meant it in another way like: why you are not using a installation on a local environment.

Posted: Mon May 08, 2023 11:58 am

neralex wrote:

Thanks corpse! Just added the fix to the repo. [ Only registered users can see links on this board! Get registered or login! ]

I wouldn't recommend it to disable the error-reporting on a local environment, when it is used for testing. But I guess he meant it in another way like: why you are not using a installation on a local environment.

As for disabling error-reporting, I am a bit confused. On my local, I still need to figure out what's going on, but as it is now, my local is not doing any error reporting, even though the quick look, it is turned on. I bet it's something in my WAMP settings, I just haven't gotten around to it to get it working. I also need to set it up to do it dump the error into an error.log file. I had done that on my old local that was on my old PC, but I guess I didn't do it on this newer PC yet.

Posted: Thu May 11, 2023 2:17 am

I've got the same issue with Survey module. file index.php

Warning: Trying to access array offset on value of type null in /modules/Surveys/index.php on line 211

Warning: Trying to access array offset on value of type null in /modules/Surveys/index.php on line 123

and when u vote error will come again.

Warning: Trying to access array offset on value of type null in /modules/Surveys/index.php on line 286

I took a look at the lines and I made the right modification I think. Just let me know

Line 123

Code:

$optionCount = $row2['optionCount'];

Replace with

Code:

$optionCount = isset($row2['optionCount']);

Line 211

Code:

$optionCount = $row2['optionCount'];

Replace with

Code:

$optionCount = isset($row2['optionCount']);

Line 286

Code:

$optionCount = $row2['optionCount'];

Replace with

Code:




$optionCount = isset($row2['optionCount']);

The error dissppear but I would like to know if it is correct.

Posted: Thu May 11, 2023 2:13 pm

When the array $row2 isn't exist, then the variable $optionCount has no value - just empty. I guess here it would be better to have "0" in this case.

I would do it in this way for example:

php Code:

$optionCount = is_array($row2) && isset($row2['optionCount']) ? $row2['optionCount'] : 0;

That means at first it checks if the variable $row2 is an array and if so, then it checks if the array-key 'optionCount' exists in the array $row2 and if so too, then the variable $optionCount gets the value of the array-key $row2['optionCount']. If these conditions are not true, the variable $optionCount get the value 0. So the variable always retains a value. Got it?

Posted: Fri May 12, 2023 2:31 am

Yes this line was much better. it fixed the error as well. Thanks