user_shifts: 15 freie noch kommende schichten anzeigen (und nicht bereits vergangene)

main
root 13 years ago
parent c3a9e66c32
commit 28b50acb88

@ -89,7 +89,7 @@ function user_shifts() {
$day_timestamp = DateTime :: createFromFormat("Y-m-d-Hi", $day . "-0000")->getTimestamp();
if ($id == 0)
$shifts = sql_select("SELECT * FROM `Shifts` JOIN `Room` ON (`Shifts`.`RID` = `Room`.`RID`) ORDER BY `start`");
$shifts = sql_select("SELECT * FROM `Shifts` JOIN `Room` ON (`Shifts`.`RID` = `Room`.`RID`) WHERE `start` > " . sql_escape(time()) . " ORDER BY `start`");
else
$shifts = sql_select("SELECT * FROM `Shifts` WHERE `RID`=" . sql_escape($id) . " AND `start` >= " . sql_escape($day_timestamp) . " AND `start` < " . sql_escape($day_timestamp +24 * 60 * 60) . " ORDER BY `start`");
@ -160,4 +160,4 @@ function make_room_select($rooms, $id, $day) {
$html[] = '<a href="' . page_link_to('user_shifts') . '&room_id=0">Next 15 free shifts</a>';
return join(' | ', $html);
}
?>
?>

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