我正在开发具有日历系统的项目。为了更有效地开发,我得到了日历代码from this site。
原始代码很好:我做了一些修改,但是我已经陷入了我的项目所需的修改之一。由于日历使用数据库对话框,我希望它在特定日期内显示数据库表的一些列。例如,如果我在2012/10/28录制了一个足球赛事(类似于原始数据库代码,来自我提到的网站),我希望其标题以相对日期显示。我努力尝试,但我找不到解决方案。请帮忙!
我用来显示日历的代码是:
$db = new PDO('mysql:host=localhost;dbname=calendar','root','');
$stmt = $db->prepare('SELECT time FROM events');
$stmt->execute();
$rawTimeStamps = $stmt->fetchAll(PDO::FETCH_ASSOC);
$cleanDateArray = array();
foreach ($rawTimeStamps as $t) {
$rawDate = $t['time'];
$rawDate = getdate($rawDate);
$cleanDate = mktime(0,0,0,$rawDate['mon'],$rawDate['mday'],$rawDate['year']);
$cleanDataArray[] = $cleanDate;
}
/* draw table */
$calendar = '<table cellpadding="0" cellspacing="0" class="calendar">';
/* table headings */
$headings = array('Sunday','Monday','Tuesday','Wednesday','Thursday','Friday','Saturday');
$calendar.= '<tr class="calendar-row"><td class="calendar-day-head">'.implode('</td><td class="calendar-day-head">',$headings).'</td></tr>';
/* days and weeks vars now ... */
$running_day = date('w',mktime(0,0,0,$month,1,$year));
$days_in_month = date('t',mktime(0,0,0,$month,1,$year));
$days_in_this_week = 1;
$day_counter = 0;
$dates_array = array();
/* row for week one */
$calendar.= '<tr class="calendar-row">';
/* print "blank" days until the first of the current week */
for($x = 0; $x < $running_day; $x++):
$calendar.= '<td class="calendar-day-np"> </td>';
$days_in_this_week++;
endfor;
/* keep going with days.... */
for($list_day = 1; $list_day <= $days_in_month; $list_day++):
$calendar.= '<td class="calendar-day">';
/*
* Assign a unique id to the day number. This unique id will be
* used by jQuery to locate events from the database that are
* on this day. The unique id is the timestamp for the first minute
* of $list_day of $month of $year.
*
* @author Josh Lockhart, http://www.newmediacampaigns.com, lines 40-41
*/
$timestamp = mktime(0,0,0,$month,$list_day,$year);
if (in_array($timestamp, $cleanDataArray)) {
$calendar.= '<div class="day-number day-number-event"><a id="'.$timestamp.'" href="#">'.$list_day.'</a></div>';
} else {
$calendar.= '<div class="day-number day-number-noevent">'.$list_day.'</div></div><div id="calendar-events"></div>';
}
$calendar.= '</td>';
if($running_day == 6):
$calendar.= '</tr>';
if(($day_counter+1) != $days_in_month):
$calendar.= '<tr class="calendar-row">';
endif;
$running_day = -1;
$days_in_this_week = 0;
endif;
$days_in_this_week++; $running_day++; $day_counter++;
endfor;
/* finish the rest of the days in the week */
if($days_in_this_week < 8):
for($x = 1; $x <= (8 - $days_in_this_week); $x++):
$calendar.= '<td class="calendar-day-np"> </td>';
endfor;
endif;
/* final row */
$calendar.= '</tr>';
/* end the table */
$calendar.= '</table>';
/* all done, return result */
return $calendar;
}
我已经在我的知识所允许的各种方式上进行了尝试,但我在PHP和MySQL方面受到限制。