从mysql数据库获取事件并放在日历上

Question

我有一个php日历，可以准确显示我想要的方式。 现在，我需要获取存储在mysql数据库中的事件并将其放在日历上。 我以为我已经这样做了，但是我无法弄清楚出了什么问题。

日历代码：

    <?php

    $db = mysql_connect("***", "***", "***") or die("Could not connect.");
    $link = mysql_select_db("***",$db);


  /* draws a calendar */
  function draw_calendar($month,$year,$events = array()){

/***********************************************CONTROLS************************************************/
    /* date settings */
    $month = (int) ($_GET['month'] ? $_GET['month'] : date('m'));
    $year = (int)  ($_GET['year'] ? $_GET['year'] : date('Y'));

    if($month < 10){
            $month = str_pad($month, 2, '0', STR_PAD_LEFT);
        }
    /* select month control */
    $select_month_control = '<select name="month" id="month">';
    for($x = 1; $x <= 12; $x++) {
      $select_month_control.= '<option value="'.$x.'"'.($x != $month ? '' : ' selected="selected"').'>'.date('F',mktime(0,0,0,$x,1,$year)).'</option>';
    }
    $select_month_control.= '</select>';
    /* select year control */
    $year_range = 7;
    $select_year_control = '<select name="year" id="year">';
    for($x = ($year-floor($year_range/2)); $x <= ($year+floor($year_range/2)); $x++) {
      $select_year_control.= '<option value="'.$x.'"'.($x != $year ? '' : ' selected="selected"').'>'.$x.'</option>';
    }
    $select_year_control.= '</select>';
    /* "next month" control */
    $next_month_link = '<a href="?month='.($month != 12 ? $month + 1 : 1).'&year='.($month != 12 ? $year : $year + 1).'" class="control">Next Month >></a>';
    /* "previous month" control */
    $previous_month_link = '<a href="?month='.($month != 1 ? $month - 1 : 12).'&year='.($month != 1 ? $year : $year - 1).'" class="control"><<   Previous Month</a>';
    /* bringing the controls together */
    $controls = '<form method="get">'.$select_month_control.$select_year_control.'&nbsp;<input type="submit" name="submit" value="Go" />&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;'.$previous_month_link.'&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;'.$next_month_link.' </form>';
    echo $controls;

/*******************************************START CALENDAR*******************************************/
    /* 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 */
    $running_day = date('w',mktime(0,0,0,$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">&nbsp;</td>';
        $days_in_this_week++;
    endfor;
    /* keep going with days */
    for($list_day = 1; $list_day <= $days_in_month; $list_day++):
        if($list_day == date("d") && $month == date("n") && $year == date("Y")) {
            $calendar.= '<td class="calendar-day current">';
        }
        else {
            $calendar.= '<td class="calendar-day">';
        }
        /* add in the day number */
        if($list_day < 10){
            $list_day = str_pad($list_day, 2, '0', STR_PAD_LEFT);
        }
        $calendar.= '<div class="day-number">'.$list_day.'</div>';

        /******************************************CHECK FOR 

EVENTS*******************************/
            $event_day = $year.'-'.$month.'-'.$list_day;
            if(isset($events[$event_day])) {
                foreach($events[$event_day] as $event) {
                  $calendar.= '<div class="event">'.$event['title'].'</div>';
                }
            }
            else {
                $calendar.= str_repeat('<p>&nbsp;</p>',2);
            }
            /*****************************************************************************************/

            $calendar.= '</div></td>';  
            if($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">&nbsp;</td>';
            endfor;
        endif;
        /* final row */
        $calendar.= '</tr>';
        /* end the table */
        $calendar.= '</table>';
        /** DEBUG **/
      $calendar = str_replace('</td>','</td>'."\n",$calendar);
      $calendar = str_replace('</tr>','</tr>'."\n",$calendar);
        /* all done, return result */
        return $calendar;
      }

    /* get all events for the given month */
    $events = array();
    $query = "SELECT Name, DATE_FORMAT(Date,'%Y-%m-%d') AS Date FROM events WHERE Date LIKE '$year-$month'";
    $result = mysql_query($query) or die('cannot get results!');
    while($row = mysql_fetch_assoc($result)) {
      $events[$row['Date']][] = $row;
    }

    echo '<h2 style="float:left; padding-right:30px;">'.date('F',mktime(0,0,0,$month,1,$year)).' '.$year.'</h2>';
    echo '<div style="float:left;">'.$controls.'</div>';
    echo '<div style="clear:both;"></div>';
    echo draw_calendar($month,$year,$events);
    echo '<br /><br />';

?>

我的数据库表称为事件，其字段为id，名称，描述，日期，时间和地点。

为了使事件显示在日历上，我需要解决什么问题？

编辑：这是回显的查询：选择名称，DATE_FORMAT（Date，'％Y-％m-％d'）AS日期从事件WHERE日期类似'2013-02'

Answer 1

尝试使用以下查询修改您的select语句：

SELECT Name, DATE_FORMAT(Date,'%Y-%m-%d') AS Date 
   FROM events 
 WHERE Date LIKE '$year-$month%'

另外，您正在使用不推荐使用的MySQL php函数。 看一下以下链接：

为什么我不应该在PHP中使用mysql_ *函数？

我可以使用什么代替mysql_函数？

前置陈述

用MySQLi准备的语句

Answer 2

请在查询前添加以下代码。 同时查看查询中的更改。 它的工作...经过测试

$year=$_REQUEST['year'];
$month=$_REQUEST['month'];
$month = sprintf("%02s", $month);
//echo $year;
$query = "SELECT Name, DATE_FORMAT(Date,'%Y-%m-%d') AS Date FROM events WHERE Date LIKE '$year-$month-%'";