根据具有相同名称的条目数限制分页

Question

my issue is that I have a database with values like the ones below:\

|Entries|Name|
|12     |Bob |
|14     |Bob |
|15     |Bob |
|27     |Kevin|
|19     |Kevin|
|29     |Dale|
|18     |Dale|
|23     |Dale|
|13     |Dale|

I would like to have pagination where each page lists the entries for each name. So there would be a page for bob, then the next page would have kevins, and then the next would have dales. I know I must use limits to achieve this, but my question is how I would limit each page when the names(Bob,Kevin,Dale) all have different amount of entries.

Below is some php I used in the page where I just wanted 12 entries per page, This may or may not help, I just wanted to add it in here just in case.

$limit = 12;
$pages = ceil($count/$limit);
//check to see if pages is smaller than 
$page = min($pages, filter_input(INPUT_GET, 'page', FILTER_VALIDATE_INT, array(
    'options' => array(
        'default'   => 1,
        'min_range' => 1,
    ),
)));
$offset = ($page-1) * $limit;
$query = "SELECT `actual_quote`,`poster`,`formtype`,`id` FROM `data` WHERE `formtype` = 'Inspirational' LIMIT 13 OFFSET :offset";
$stmt = $db->prepare($query);
$stmt->bindParam(':offset', $offset, PDO::PARAM_INT); 
$stmt->execute();

Answer 1

To show the pages you can simply select the distinct names:

$query = "SELECT DISTINCT `Name` FROM `my_table` ORDER BY Name ASC";
$stmt = $db->prepare($query);
$stmt->execute();
$rows = $stmt->fetchAll();

for ($i = 1; $i <= count($rows); $i++) {
    $row = $rows[$i - 1];
    echo '<a href="?page=' . $i . '&name=' . $row['Name'] . '">Page ' . $i . ' (' . $row['Name'] . ')</a>';
}

On each page you can use the name parameter to get only the results for that name.