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我目前有一个看起来像这样的函数：

function getEvents($weekNumStart, $weekNumEnd){
    $mysqli = new mysqli(DB_SERVER, DB_USER, DB_PASSWORD, DB_NAME);
    if (!$mysqli) {
        die('There was a problem connecting to the database.');
    }
    else {
        if ($weekNumEnd == '') {
            $weekNumEnd = $weekNumStart;
        }
        $Group = $_SESSION['Group'];
        $query = $mysqli->prepare("SELECT EventID, DAYOFWEEK(Start) AS wday, Events.Start, HOUR(Start) AS sHour, HOUR(End) AS eHour, Events.End, Events.Group, Events.Unit, Type, Room, Lecturer, Cancelled, StartName FROM Events, Week WHERE StartName >= '$weekNumStart' AND StartName <= '$weekNumEnd' AND Events.Start >= StartWeek AND Events.Start <=EndWeek AND (Events.Group = '$Group' OR Events.Group = '');");
        $query->execute();
        $query->bind_result($eventID, $dayofweek, $startDateTime, $startHour, $endHour, $endDateTime, $group, $unit, $type, $room, $lecturer, $cancelled, $weekName);
        $data_arr = array();
        while ($query->fetch()){
            $data_arr[] = array(
               $eventID, $dayofweek, $startDateTime, $startHour, $endHour, $endDateTime, $group, $unit, $type, $room, $lecturer, $cancelled, $weekName
            );
        }
        return $data_arr; //
    }
    $mysqli->close();
}

此功能按我的意愿工作。我目前这样操作它：

if (!isset($weekNumEnd)){
        $weekNumEnd = '';
    }
    $data = getEvents($weekNumStart, $weekNumEnd);
    foreach($data as $day) {
        for($i=0;$i<7;$i++){
            for($j=9;$j<=18;$j++){
                if ($day[1] == $i && $day[3] == $j){
                    $unit = $day[7];
                    $type = $day[8];
                    $room = $day[9];
                    if ($i == 2){
                    ?>
                    <script>
                    var mon<?=$j?>unit = '<?=$unit?>';
                    var mon<?=$j?>type = '<?=$type?>';
                    var mon<?=$j?>room = '<?=$room?>';
                    $('#mon-<?=$j?>').append(
                        "<p>"+mon<?=$j?>unit+"<br>"+mon<?=$j?>type+"<br>"+mon<?=$j?>room+"</p>"
                    );
                    </script>
                    <?php
                    if (($day[4] - $day[3]) > 1) {
                        for($q=($day[3]+1);$q<$day[4];$q++){
                        ?>
                        <script>
                            var mon<?=$q?>unit = '<?=$unit?>';
                            var mon<?=$q?>type = '<?=$type?>';
                            var mon<?=$q?>room = '<?=$room?>';
                            $('#mon-<?=$q?>').append(
                                "<p>"+mon<?=$q?>unit+"<br>"+mon<?=$q?>type+"<br>"+mon<?=$q?>room+"</p>"
                            );
                        </script>
                        <?php
                        }
                    }
                }

问题是。我显然无法实时运行它以在每周或任何更改时不断更新事件列表。

所以为了解决这个问题，我决定尝试使用 Ajax。到目前为止，我有这个：

function getEvents() {
        var data = 'Events=Yes';
        $.ajax({
            type: "POST",
            url: "functions/updateWeek.php", 
            data: data,
            cache: false,
            dataType: 'json',
            success: function(html) {
                    alert(html.returned_val);
            }
    });
}

我有这个要返回：

if($_POST['Events'] == "Yes"){
    $weekNumStart = $_SESSION['startWeek'];
    $weekNumEnd = $_SESSION['endWeek'];
    $data = getEvents($weekNumStart, $weekNumEnd);
    echo json_encode(array('returned_val' => $data));
}

这......有点作品。至少也就是说它以一个丑陋的大块返回数据。

问题是我不能再像上面那样操作它：

data = getEvents($weekNumStart, $weekNumEnd);
    foreach($data as $day) {
        for($i=0;$i<7;$i++){
            for($j=9;$j<=18;$j++){
                if ($day[1] == $i && $day[3] == $j){
                    $unit = $day[7];
                    $type = $day[8];
                    $room = $day[9];
                    if ($i == 2){
                    ?>
                    <script>
                    var mon<?=$j?>unit = '<?=$unit?>';
                    var mon<?=$j?>type = '<?=$type?>';
                    var mon<?=$j?>room = '<?=$room?>';
                    $('#mon-<?=$j?>').append(
                        "<p>"+mon<?=$j?>unit+"<br>"+mon<?=$j?>type+"<br>"+mon<?=$j?>room+"</p>"
                    );
                    </script>

也就是说-当我通过ajax调用信息时，我不知道如何重新操作它。我会以完全 javascript 的方式重写以上所有内容吗？有没有我不知道/理解的更合乎逻辑的方法？或者有没有一种方法可以让代码在那个不同的窗口中工作（用php和javascript完成），并让它全部发回，通过ajax更正，自动在页面上工作，而不必在信息后重写它是通过ajax拉回来的？

4

1 回答 1

1

首先，使用 json_encode($data) 就足够了，您将从 AJAX 获取数组。

其次，使用 Firebug 进行检查，因为您应该获得结构化数据，而不是大块数据。

alert() 只是尝试将所有内容打印为单个字符串，仅此而已

for( var i = 0; i < html.length; i++ )
    // do something with html[i]
于 2013-03-24T23:50:44.743 回答