SELECT count_id, COUNT(count_id)
FROM select_table_name
where where_id > 1 and
GROUP BY count_id
HAVING COUNT(count_id) > 1
Learn Developing | Jobs portal project in ASP.NET, Free Download Online job portal system project in asp.net, Free Download Online Shopping Project in PHP
$start = 1; $end = 20; for ($i = $start; $i <= $end; $i++) { if ($i % 2 == 0) { echo $i . " is even.<br>"; ...
No comments:
Post a Comment