Updating mysql database with php using variables

Question

I am having problems updating mysql with php using varables.

    mysqli_query($connection,  "UPDATE passwords SET used=1, time_used='{$time}'
                                WHERE password='{$key}'
                               ");

I was given the error:

Warning: mysqli_query() expects parameter 1 to be mysqli, resource given in C:\wamp\www\key_check.php on line 47

any ideas why?

Thanks!

EDIT: Whole Code: http://pastebin.com/raw.php?i=W5cx8pBP

The "new mysqli" solution seems to be giving problems when trying to

$result = mysql_query("SELECT * FROM passwords", $connection);

Thanks :)

post your mysqli connection code.. $connection

Jhonathan H.
– Jhonathan H.

2013-03-27 01:49:32 +00:00
Commented Mar 27, 2013 at 1:49 — Jhonathan H.
– Jhonathan H., Commented Mar 27, 2013 at 1:49

Jhonathan H. · Accepted Answer · 2013-03-27 02:00:45Z

2

Your connection setting must look like

$connection = new mysqli($host,$username,$pass,$db);

Then execute the query using your way or by this way also

$query="UPDATE passwords SET used=1, time_used='{$time}'
                            WHERE password='{$key}'
                           ";
$stmt = $connection->query($sql);

note: using prepared statements for mysqli can also possible and great. By somehow you also needed to bind parameters in there..

edited Mar 27, 2013 at 2:00

answered Mar 27, 2013 at 1:50

Jhonathan H.

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Comments

Saturnix · Accepted Answer · 2013-03-27 01:53:19Z

1

You have to declare $connection by creating a new mysqli object. If you fail to do so, you can check the documentation for mysqli constructor

Here's the code from the documentation.

$connection = new mysqli('localhost', 'my_user', 'my_password', 'my_db');

if ($connection->connect_error) {
    die('Connect Error (' . $connection->connect_errno . ') '
            . $connection->connect_error);

answered Mar 27, 2013 at 1:53

Saturnix

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