I'm trying to write some code that will retrieve a user's access level from my database and if it doesn't return any values then I want to set the access variable to 2.
I'm using...
$access_id = $useraccess["0"]["access_id"];
if (!$access_id) {$access_id = '2';}
Where $useraccess is the array I've got from the database. I set the variable $access_id to whatever the user's access_id field contains. Then I check the variable because if the access_id field is empty then I know the user hasn't had one specified and should be given the site default. The problem is one of the access levels is '0' which is returning an empty value and then setting it to two.
Has anyone any idea how to stop this please?
Giving the below syntax error
Dear Experts,
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