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Call to a member function prepare() on a non-object PHP Help

Asked 2010-12-16T17:09:22.593
12

I am trying to write a PHP function. It is very simple. It is just a prepared statement that queries the database, but I can not get this to work. I keep recieving the error Call to a member function prepare() on a non-object. here is the code:

$DBH = new mysqli("host", "test", "123456", "dbname");
function selectInfo($limit, $offset){
    $stmt = $DBH->prepare("SELECT * FROM information LIMIT ?,?");
    $stmt->bind_param("ii", $limit, $offset);
    $stmt->execute();
    }
selectInfo();

Any time I call the function i get that error. Can someone please help?

php mysqli prepared-statement
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2 Answers

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selectInfo($DBH);

function selectInfo($DBH,$limit, $offset){
    $stmt = $DBH->prepare("SELECT * FROM information LIMIT ?,?");
    $stmt->bind_param("ii", $limit, $offset);
    $stmt->execute();
    }
answered 2010-12-16T17:12:08.687
3

That's simply. $DBH doesn't exist within selectInfo() function. Variable defined in global scope won't be visible within function and vice-versa. Read more about variables scope on manual pages.

How to solve it? Pass that variable as a argument of the function:

$dbh = new MySQLi(...);

function selectInfo(MySQLi $dbh, $limit, $offset) {
    $stmt = $dbh->prepare(...);
    ...
}
answered 2010-12-16T17:14:21.913

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