javascript - PHP 与 jQuery : how to do it in this case

Question

我有一个 jQuery 日历，一天有 3 个事件。如果任何客户预订了某个事件，我的日历会在我将其提交到 SQL 数据库后显示该事件已预订。

因此，如果我只在 SQL 中提交任何日期，我的日历就会显示它。这意味着通过 PHP

if($date > 0) { echo 'here is result';}

为了显示任何事件，我想为页面中的特定事件添加以下脚本: { "EventID": 1, "StartDateTime": new Date(2015,3, 1), "Title": "上午 10 点至下午 2 点", "URL": "用户名", "描述": "预订者 ", "CssClass": "Event_1"},

现在在我的 SQL 中，我为所有事件添加了 3 个事件 * 31 天 = 93 行。如果在sql中通过id在任何行的日期字段中找到任何数据，我的页面将显示上述脚本。

为了更清楚起见，我给出了这个错误的脚本，我想通过 jQuery 和 PHP 来正确执行该脚本。

<?php if($id = '1'){if($date > 0){ echo' ?>
    { "EventID": 1, "StartDateTime": new Date(2015,3,<? echo '$date';?>), "Title": "10am to 2pm", "URL": "#", "Description": "Booked by ", "CssClass": "Event_1" },
<?php ';}} ?>   

现在我不明白如何通过 jQuery 和 php 来做到这一点？在这里我读了很多文章，但我失败了。

这是我的代码示例:

<?php
include_once('db.php');
global $db;
$result = mysqli_query($db,"SELECT * FROM room ORDER BY id");
if(!$result) {
    die("Database query failed: " . mysqli_error());
}
    while($row = mysqli_fetch_assoc($result)) {
    $id=$row['id']; 
    $date=$row['date'];
    }
?>

<script type="text/javascript">
$().ready(function() {
// others script here

 var events = [
 { "EventID": 1, "StartDateTime": new Date(2015,3, 1), "Title": "10am to 2pm", "URL": "username1", "Description": "Booked by ", "CssClass": "Event_1" },}
 { "EventID": 2, "StartDateTime": new Date(2015,3, 1), "Title": "2pm to 6pm", "URL": "username2", "Description": "Booked by ", "CssClass": "Event_2" },
 { "EventID": 3, "StartDateTime": new Date(2015,3, 1), "Title": "6pm to 10pm", "URL": "username3", "Description": "Booked by ", "CssClass": "Event_3" }
// All events
 ];
});
</script>

Answer 1

根据您的代码，我做了一些调整。我还没有测试过，但应该可以工作，或者至少指向正确的方向:

<?php
include_once('db.php');
global $db;
$result = mysqli_query($db,"SELECT * FROM room ORDER BY id");
if(!$result) {
    die("Database query failed: " . mysqli_error());
}

$events = array();
    while($row = mysqli_fetch_assoc($result)) {
        if( !isset( $events[$row['date']]) ){ //create array of events for each day with results
            $events[$row['date']] = array();
        }

        $num_events = count($events[$row['date']]);
        $date = new DateTime($row['date']);

        $events[$row['date']][] = array(
            "EventID" => 1,
            "StartDateTime" => $date->format('Y-m-d'), //assuming real mysql datetime
            "Title" => "10am to 2pm",
            "URL" => "username1",
            "Description" => "Booked by ",
            "CssClass" => "Event_" . $num_events + 1 
        );
    }
?>

<script type="text/javascript">
$(function(){
    var events = '<?php echo json_encode($events);?>' 
    for (day in events){
        var day_events = events[day];
        for(var i= 0; i< day_events.length; ++i){
            day_events[i].StartDateTime = new Date(day_events[i].StartDateTime); //convert into proper date
            console.log(day_events[i]);
        }
    }
});
</script>