PHP error: Notice: Undefined offset

Question

i have created an api to fetch data from database to echo it as json and finally display it in html but while echoing the data,

i am getting below php errors

Notice: Undefined offset: 80 in /opt/lampp/htdocs/ReadExchange/api.php on line 16

Notice: Undefined offset: 79 in /opt/lampp/htdocs/ReadExchange/api.php on line 16

Notice: Undefined offset: 78 in /opt/lampp/htdocs/ReadExchange/api.php on line 16

and so on,

PHP Code

$db_result_set = $db->prepare('SELECT p.Id, p.FirstName, p.MiddleName, p.LastName, p.Gender, p.Location, p.Email, p.Mobile,b.BookTitle, b.BookGenre, b.BookWriter, b.BookDescription FROM personaldetails AS p LEFT JOIN bookdetails AS b ON b.UserId = p.Id ORDER BY p.Id DESC');;

$db_result_set->execute();

$final = [];
foreach ($db_result_set as $u) {

    if (!is_array($final[$u["Id"]])) {   //line 16 
        $final[$u["Id"]]= [
            "Id" => $u["Id"],
            "FirstName" => $u["FirstName"],
            "MiddleName" => $u["MiddleName"],
            "LastName" => $u["LastName"],
            "Gender" => $u["Gender"],
            "Location" => $u["Location"],
            "Email" => $u["Email"],
            "Mobile" => $u["Mobile"],
         ];
    }
    $final[$u["Id"]]["books"][] = [
        "BookTitle" => $u["BookTitle"],
        "BookGenre" => $u["BookGenre"],
        "BookWriter" => $u["BookWriter"],
        "BookDescription" => $u["BookDescription"],
    ];

}   
    
    header('Content-type: application/javascript');
    echo json_encode($final);

?>

i am not confirm but i think because of this i am not getting json data in html.

Possible duplicate of [Reference - What does this error mean in PHP?](http://stackoverflow.com/questions/12769982/reference-what-does-this-error-mean-in-php) — Henders, Jun 02 '16 at 08:20
`header('Content-type: application/javascript');` ? What are you trying to echo? Do you mean `header('Content-type: application/json');` ? — Raptor, Jun 02 '16 at 08:42

score 1 · Accepted Answer · answered Jun 02 '16 at 08:00

1

Not: if (!is_array($final[$u["Id"]]))

but: if (!isset($final[$u["Id"]]))

answered Jun 02 '16 at 08:00

nospor

4,190
1
16
25

score 0 · Answer 2 · answered Jun 02 '16 at 08:02

0

You are trying to check if an undefined offset exists in your array. You need to first need to check if the offset exists with isset.

So change your code to

if (isset($final[$u["Id"]]) && ! is_array($final[$u["Id"]))

answered Jun 02 '16 at 08:02

Marwelln

28,492
21
93
117

PHP error: Notice: Undefined offset

2 Answers2