Php 通过ajax返回一系列复杂的命令（或者在通过ajax返回命令后重新构造命令）

我目前有一个函数如下所示：

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();
}

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
                        }
                    }
                }

我要回报你：

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

这……有点效果。至少也就是说，它返回的数据非常大

问题是我不能再像以前那样操纵它了，如上所述：

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>

data=getEvents（$weekNumStart，$weeknummend）；
foreach（$day数据）{
对于（$i=0；$itype=''；
var monroom=“”；
$（'#mon-'）。追加(
“”+monunit+”
“+montype+”
“+monroom+”
”
);

---

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。我是否会以完全javascript的方式重写上面的所有内容？是否有一种我不知道/不理解的更符合逻辑的方法？或者是否有一种方法可以让代码正常工作（使用php和javascript完成）在不同的窗口中，让它通过ajax将所有信息发送回来，以便在信息通过ajax返回后自动处理页面而无需重写？

首先，使用json_encode（$data）就足够了，您可以从ajax获得数组

其次，用Firebug检查它，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]

for（var i=0；i

Ohhh！当然。谢谢你。这一手解决了我所有的问题。

for( var i = 0; i < html.length; i++ )
    // do something with html[i]