2012-08-06 48 views
0

我的數據庫中的事件不會顯示在我的完整日曆上。我使用了我在網上看到的示例代碼,因此請耐心等待我現在使用的代碼。我將非常感謝所有來自你們的幫助。謝謝!完整的日曆不會顯示我的活動從我的數據庫

<script> 
       $(document).ready(function() { 
       // page is now ready, initialize the calendar... 

       $('#calendar').fullCalendar({ 
        header: { 
         left: 'prev,next today', 
         center: 'title', 
         right: 'month,agendaWeek,agendaDay' 
        },editable: false, 

        events: "json_events.php", 

        eventDrop: function(event, delta) { 
          alert(event.title + ' was moved ' + delta + ' days\n' + 
          '(should probably update your database)'); 
        }, 

        loading: function(bool) { 
        if (bool) $('#loading').show(); 
        else $('#loading').hide(); 
       } 
       }); 
     }); 
    </script> 

json_events.php

<?php 
    include 'connect.php'; 
    session_start(); 
    $result = mysql_query("SELECT ID, title, startDate AS startDate FROM events"); 
    mysql_close(); 
    $events = array(); 
    while ($row=mysql_fetch_array($result)){ 
     $title = $row['title']; 
     $eventsArray['id'] = $row['ID']; 
     $eventsArray['title'] = $title; 
     $eventsArray['startDate'] = $row['startDate']; 
     $events[] = $eventsArray; 
    } 
    echo json_encode($events); 
?> 
+0

當您直接查看json_events.php時,能否粘貼json字符串的外觀? – ews2001 2012-08-14 03:12:25

回答

1
<?php 
    include 'connect.php'; 
    session_start(); 
    $result = mysql_query("SELECT ID, title, startDate AS startDate FROM events"); 
    mysql_close(); 
    $events = array(); 
    while ($row=mysql_fetch_array($result)){    
     $id = $row['ID']; 
     $title = $row['title']; 
     $start = $row['startDate']; 

     $events = array(
     'id' => "$id", 
     'title' => "$title", 
     'start' => "$start" 
     ); 

    } 
    echo json_encode($events); 
?> 

這應該爲你工作。如果你不想讓偶數成爲一天,只需在開始語句後添加'allDay'=>「」即可。

+0

如何將多個事件從多個來源顯示到fullcalendar頁面 – Keynes 2016-05-31 11:10:20

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