Fatal error: Call to a member function query() on a non-object in

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  • Call to a member function on a non-object [duplicate] 8 answers

Fatal error: Call to a member function query() on a non-object on line: $result = $conn->query($sql) or die(mysqli_error());

Who knows whats wrong and how to fix it?

<?php function dbConnect($usertype, $connectionType = 'mysqli') { $host = 'localhost'; $db = 'phpsols'; if ($usertype == 'read') { $user = 'psread'; $pwd = '123'; } elseif ($usertype == 'write') { $user = 'pswrite'; $pwd = '123'; } else { exit('Unrecognized connection type'); } if ($connectionType == 'mysqli') { return new mysqli($host, $user, $pwd, $db) or die ('Cannot open database'); } else { try { return new PDO("mysql:host=$host;dbname=$db", $user, $pwd); } catch (PDOException $e) { echo 'Cannot connect to database'; exit; } } } // connect to MySQL $conn = dbConnect('read'); // prepare the SQL query $sql = 'SELECT * FROM images'; // submit the query and capture the result **$result = $conn->query($sql) or die(mysqli_error());** // find out how many records were retrieved $numRows = $result->num_rows; ?> <!DOCTYPE HTML> <html> <head> <meta charset="utf-8"> <title>Connecting with MySQLi</title> </head> <body> <p>A total of <?php echo $numRows; ?> records were found.</p> </body> </html>

-------------Problems Reply------------

The culprit is most likely this line:

return new mysqli($host, $user, $pwd, $db) or die ('Cannot open database');

The do xyz or die() construct leads to funny behaviour in conjuction with the return statement (i.e. the whole thing is interpreted as an OR expression and because new mysqli will never be false, the "die" is never processed.). See a similar case here.

Do this instead:

$result = new mysqli($host, $user, $pwd, $db) ;
if (!$result) die (....);
return $result;

Also, slightly related, I think you will never catch a PDO connection error because this:

return new PDO("mysql:host=$host;dbname=$db", $user, $pwd);

will always exit the function, and never reach the catch block. As with your actual problem, the solution is to pass the object to a $result variable first.

I know this has been answered, but just wanted to add my own 2 cents worth....

I was googling "return new mysqli" to fix this very problem. The code is a snippet from an e-book (PHP Solutions. Dynamic Web Design Made Easy.) - 2nd Edition if I'm not too mistaken! Ironically the book is supposed to be teaching PHP basics (in this case DB connection methods) to poor souls like myself and easyrider. I have very limited understanding, as this book is my own introduction to PHP coding but I managed to suss out that creating or rather instantiating an actual object, passing it to a temporary variable and then returning that variable was the solution.

Annoyingly, this was demonstrated correctly in the first edition of said publication.

I dont know if Google will cache or index my post, but it might help others who are learning from the book, to discover that the code example is flawed, and as the time of writing, the error is not listed in the errata on the publishers website.

Category:mysqli Views:9 Time:2011-01-08
Tags: php mysqli

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