我需要實時更新消息的建議,這裏是我的代碼:實時更新的收件箱中的郵件
<script>
function fetch_messages(){
var user_id = "1" // example id
$.ajax({
url: "do_fetch.php",
type: "POST",
data: { user_id : user_id },
dataType: "json",
success: function(data) {
for (var i = 0; i < data.length; i++) {
$("#messages_list").append('<div id="'+data.id+'">'+data.message+'</div>');
}
setTimeout("fetch_messages()", 3000);
}
});
}
$(function(){
fetch_messages();
});
</script>
並沒有DIV消息:
<div id="messages">
</div>
PHP代碼:
<?php
include('connection.php')
$user_id = $_POST['user_id'];
$my_id = $_SESSION['user_id'];
$sql = 'SELECT * FROM messages WHERE user_id = '.mysql_real_escape_string($user_id).' AND my_id = '.mysql_real_escape_string($my_id);
$result = mysql_query($sql, $connection);
echo json_encode($result);
?>
此代碼有效,但我不想閱讀所有消息並顯示它們,我只需要加載新消息,那麼我該怎麼做?我是否需要在時間戳中放置一個隱藏的字段,並且檢查Ajax的時間?
'mysql_ *'已棄用 – 2012-07-27 14:03:26