Javascript 表单提交，将数据插入数据库，但不在div中显示内容

我有一个有表格的div。提交表单时，我希望将数据插入数据库，并将数据库的结果显示在表单下方，以便再次提交，并添加另一条记录。正在填充表单和数据库，但结果不会显示在div中

包含初始表单、将包含刷新结果列表和新表单的div如下所示：

<div id=activitylist>
    <?
    $q2a="SELECT activitynumber, title, description, leaders, time FROM activities where activities.meetingid='$id' AND activities.unitid='$input2'";
    $r2a=mysqli_query($dbc,$q2a) or die(mysqli_error($dbc));
    echo "<table class='layouttable'><tr><td>ActivityNumber</td><td>Title</td><td>Description</td><td>Leaders</td><td>Time</td><td>Edit</td>";
    while($row2a   =mysqli_fetch_assoc($r2a))   
    {
        echo "
        <tr><td>"  . $row2a['activitynumber'] .  "</td>
        <td>" . $row2a['title'] . " </td>
        <td>" . $row2a['description'] . " </td>
        <td>" . $row2a['leaders'] . " </td>
        <td>" . $row2a['time'] . " </td>
        <td><input type='button' value='editactivity' onclick='editactivity(" . $row2a['activityid'] . ")'></td></tr>.
        ";
    }
    echo"</table>"; ?><br><Br>

    <form id='newactivity' method="post" onSubmit="return submitForm();">
        <input type =hidden name='meetingid' value=<?echo$id?>>
        <b>Activity Number:</b><input type=text name='activitynumber' class='textborder'>   
        <b>Title:</b><input type=text name='activitytitle' class='textborder'>
        <b>Time (mins):</b> <input type=text name='activitytime' class='textborder'>
        <b>Leaders:</b> <input type=text name='leaders' class='textborder'><br>
        <b>Description:</b><textarea name='activitydescription' class='textareaborder'></textarea>
        <input type='submit' value='Submit'></form></div>

该表格通过以下方式提交：

<script name='addactivity'>
function submitForm() {
    $.ajax({type:'POST', url: 'activity_new.php', data:$('#newactivity').serialize(), success: function(response) {
        $('#newactivity').find(".activitylist").html(response);
    }});

    return false;
}
</script>

和activity_new.php，它是插入到数据库中的内容，我希望在完成插入后在div中显示它是：

<?php
session_start();
$input2=$_SESSION[ 'unitid' ];

$meetingid=$_POST['meetingid'];
$activitynumber=$_POST['activitynumber'];
$activitytitle=$_POST['activitytitle'];
$activitytime=$_POST['activitytime'];
$leaders=$_POST['leaders'];
$activitydescription=$_POST['activitydescription'];

include 'connect_db.php'; 

$q1c="INSERT into activities (meetingid, unitid, activitynumber, title, description, time, leaders) VALUES ('$meetingid', '$input2', '$activitynumber','$activitytitle', '$activitydescription', '$activitytime', '$leaders')";
$r1c = mysqli_query($dbc,$q1c)or die(mysqli_error($dbc)); 


        $q2a="SELECT activitynumber, title, description, leaders, time FROM activities where activities.meetingid='$meetingid' AND activities.unitid='$input2'";
        $r2a=mysqli_query($dbc,$q2a) or die(mysqli_error($dbc));
        echo "<table class='layouttable'><tr><td>ActivityNumber</td><td>Title</td><td>Description</td><td>Leaders</td><td>Time</td><td>Edit</td>";
        while($row2a   =mysqli_fetch_assoc($r2a))   
        {
            echo "
            <tr><td>"  . $row2a['activitynumber'] .  "</td>
            <td>" . $row2a['title'] . " </td>
            <td>" . $row2a['description'] . " </td>
            <td>" . $row2a['leaders'] . " </td>
            <td>" . $row2a['time'] . " </td>
            <td><input type='button' value='editactivity' onclick='editactivity(" . $row2a['activityid'] . ")'></td></tr>.
            ";
        };
    echo"</table>"; ?><br><Br>
    <form id='newactivity' method="post" onSubmit="return submitForm();">
    <input type =hidden name='meetingid' value=<?echo$id?>>
    <b>Activity Number:</b><input type=text name='activitynumber' class='textborder'>   
    <b>Title:</b><input type=text name='activitytitle' class='textborder'>
    <b>Time (mins):</b> <input type=text name='activitytime' class='textborder'>
    <b>Leaders:</b> <input type=text name='leaders' class='textborder'><br>
    <b>Description:</b><textarea name='activitydescription' class='textareaborder'></textarea>
    <input type='submit' value='Submit'>

---

编辑Jquery选择器到ID选择器，因为您的div没有ID类。也可以省略newactivity！它不是activitylist的父级，而是它的子级。这就足够了：

$("#activitylist").html(response);

---

$'newactivity'.find.activitylist在表单中搜索class=activitylist的元素。你在任何地方都没有这样的类。有没有办法让它进入一个div，它不在表单中，但在表单所在的页面中？你可以把它放在任何你想要的地方。只需编写一个与所需位置匹配的选择器。我已将div更改为，但它仍然没有进入其中。表单在我试图发送到的div中是否重要？只有在使用选择器将事件处理程序绑定到要替换的元素时才重要。那就看