PHP警告：遇到非数字值

Question

I'm getting the following error "PHP Warning: A non-numeric value encountered on line 11". The following code is for a pagination. I have tried everything I know, not sure what else to do. Any help is appreciated.

8 $rowsPerPage = 20;
9 if(isset($_GET['page'])){$pageNum = $_GET['page'];}
10 if(empty($pageNum)) {$pageNum = 1;}
11 $offset = ($pageNum - 1) * $rowsPerPage;

Answer 1

Something is not numeric (as obvious as that sounds, it is the error) and given you manually set all things to integer other than the GET value, try this:

$rowsPerPage = 20;
if (isset($_GET['page'])) {
    $pageNum = (int) $_GET['page'];
}
if(empty($pageNum)) {
    $pageNum = 1;
}
$offset = ($pageNum - 1) * $rowsPerPage;

Notice the cast to int for the GET param.

You could also just cast to int and PHP will default to integer 0 if it's not already int:

$pageNum = (int) $_GET['page'];

---

Couldn't help myself - you can also use a ternary to make your setting of the $pageNum cleaner (IMO):

$rowsPerPage = 20;

$pageNum = isset($_GET['page'])
    ? (int) $_GET['page']
    : 1;

$offset = ($pageNum - 1) * $rowsPerPage;