使用SQL值作为PHP变量和strtotime查找日期差异

Question

I am trying to determine how many days overdue a dvd is. My code is:

$query = "SELECT rental_date FROM rented_dvds WHERE dvd_id = '$dvd_id' ";
$result = mysqli_query($dbc, $query);
$newDate = DateTime::createFromFormat("l dS F Y", $result);
$startTimeStamp = strtotime("$newDate");
$endTimeStamp = strtotime("NOW");
$timeDiff = abs($endTimeStamp - $startTimeStamp);
$numberDays = $timeDiff/86400;  
$numberDays = intval($numberDays);

rented_dvds is captured when the DVD is rented using NOW(). When I run that code I get: "Warning: DateTime::createFromFormat() expects parameter 2 to be string, object given in .../.../return_dvd.php on line 34"

Line 34 is the line that starts with the $newDate variable.

This is all code that I've cobbled together from different sources, but I thought I had cobbled properly. I've been searching for an answer, and have tried a few different solutions, but I can't seem to make any of them work.

Also, this is my first post, so 'hi!' and thanks for any help!

Answer 1

Are you fetching in the right way ?because you are directly fetch $result which is give you in an pbject or array form ..you have to write

$row=mysqli_fetch_array($result);
$newDate = DateTime::createFromFormat("l dS F Y", $row['rental_date']);