我已经完成加载更多结果,但加载结果将附加在底部。如何将其添加到顶部?
因为我发送的邮件是转到底部所以我希望将旧邮件添加到顶部。
这是我的HTML:
<form name ="chatroom">
<div class="count-user-online">
<?php echo $row ?>
</div>
<div class="chatroom-upper-container" id="chatroom-upper-container">
<input type="hidden" id="result_no" value="10">
<div id="inner">
Loading Message....<img src="../images/loading.gif"/>
</div>
</div>
<input type="button" id="load" value="Load More Results">
<div class="chatroom-lower-left-container">
<textarea class="message-setting" id="area-message" placeholder="type text" name= "msg"></textarea>
</div><div class="chatroom-lower-right-container">
<button type="button" class="btn sendmessage-btn" onclick= "submitChat()">Send</button>
</div>
</form>
这是我的js:
$(document).ready(function(){
$("#load").click(function(){
loadmore();
});
});
function loadmore()
{
var val = document.getElementById("result_no").value;
$.ajax({
type: 'post',
url: 'chatloadmore.php',
data: {
getresult:val
},
success: function (response) {
var content = document.getElementById("chatroom-upper-container");
content.innerHTML = content.innerHTML+response;
// We increase the value by 2 because we limit the results by 2
document.getElementById("result_no").value = Number(val)+10;
}
});
}
这是我的loadmorefile.php:
<?php
include '../config.php';
include'login.php';
$no = $_POST['getresult'];
$username = $_SESSION['username'];
$sql1= "SELECT * FROM (
SELECT * FROM logs ORDER BY id DESC LIMIT $no,40
) sub
ORDER BY id ASC ";
$result1 = mysqli_query($connection,$sql1);
while($extract = mysqli_fetch_array($result1)){
$color = ($extract['username'] == $username) ? ' #F5F5F5' : '#DCDCDC';
$position = ($extract['username'] == $username) ? 'right' : 'left';
echo "
<div class='left-wrap-message' style='background-color:$color; float:$position;'>
<p style='text-align:$position; margin:0;'>". $extract['username']. " : </p>
<p style='text-align:$position; margin:0; text-align:left;'> " . $extract['msg']. "</p></div>
<div class='msg-dateandtime' style='text-align:$position;'> " . $extract['date']. "</div>";
}
?>
答案 0 :(得分:3)
而不是
content.innerHTML = content.innerHTML+response;
做一个
content.innerHTML = response + content.innerHTML;