尝试在MySQL表格中向所有用户显示消息＆＃39;屏幕使用xmlhttp，toastr alert和php。
我使用xmlhttp.open("GET","includes/ajax/user/general/messages.php?username=<?php echo $_SESSION['username']; ?>",true);这是SQL所在的地方，我把它放在5秒循环上，所以它每5秒检查一次messages.php。
然后我这样做：
xmlhttp.send();
$.toast({
heading: <?php echo json_encode($messageInfo['title']); ?>,
text: <?php echo json_encode($messageInfo['message']); ?>,
position: 'top-right',
loaderBg: '#ff6849',
icon: <?php echo json_encode($messageInfo['type']); ?>,
hideAfter: 6000,
stack: 6
});
然而，这仅输出＆＃39; null＆＃39;作为标题和＆＃39; null＆＃39;作为文本。
messages.php：
<?php
if (!isset($_SERVER['HTTP_REFERER'])){
die;
}
ob_start();
require_once '../../../app/config.php';
require_once '../../../app/init.php';
if (!empty($maintaince)){
die();
}
if (!($user -> LoggedIn()) || !($user -> notBanned($odb)) || !($user->isAdmin($odb))){
die();
}
$username = $_GET['username'];
if(empty($username)){
die();
}
$SQLGetInfo = $odb -> prepare("SELECT * FROM `live_messages` WHERE id = 1");
$SQLGetInfo -> execute(array($_SESSION['ID']));
$messageInfo = $SQLGetInfo -> fetch(PDO::FETCH_ASSOC);
$type = $messageInfo['type'];
$title = $messageInfo['title'];
$message = $messageInfo['message'];
?>
编辑：如果我将messages.php中的PHP代码直接放入运行此脚本的文件中，那么它会成功从数据库中取出，但是这没有任何帮助，因为我希望在实时。
编辑： 完整的脚本
<script>
window.setInterval(function(){
var xmlhttp;
if (window.XMLHttpRequest){
xmlhttp = new XMLHttpRequest();
}
else{
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.open("GET","includes/ajax/user/general/messages.php?username=<?php echo $_SESSION['username']; ?>",true);
xmlhttp.send();
$.toast({
heading: <?php echo json_encode($title); ?>,
text: <?php echo json_encode($message); ?>,
position: 'top-right',
loaderBg: '#ff6849',
icon: <?php echo json_encode($type); ?>,
hideAfter: 6000,
stack: 6
});
}, 5000);
</script>
答案 0 :(得分：0)
我认为您对如何使用XHR请求有轻微的误解。 XHR请求的工作方式与浏览器请求类似，使用您向我们展示的代码，您没有为响应注册处理程序。
请参阅MDN上的POST示例：https://developer.mozilla.org/en-US/docs/Web/API/XMLHttpRequest/send
现在，为了访问由main处理的数据，您将不得不发回适当的JSON响应。
即。
messages.php
然后将其传递回您的XHR请求加载处理程序。
W3学校有一个“好的”例子，说明它们如何组合在一起：https://www.w3schools.com/js/js_json_php.asp
编辑：我提供了一个工作代码示例，它实质上是w3schools链接中内容的直接复制和粘贴。
header('Content-Type: application/json');
echo json_encode($messageInfo);
test.html <!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<meta http-equiv="X-UA-Compatible" content="ie=edge">
<title>Document</title>
<script>
var xmlhttp = new XMLHttpRequest();
// When the XHR Request receives a response from the server
// this function will be called.
xmlhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
// This function is necessary to parse the JSON sent back
// from the server.
// Note: It can fail, so you'll want to make sure it actually parsed
// the response.
var message = JSON.parse(this.responseText);
// Here is an example of how you would use it with your $.toast call
// $.toast({
// heading: message.heading,
// text: message.body,
// position: 'top-right',
// loaderBg: '#ff6849',
// icon: message.type,
// hideAfter: 6000,
// stack: 6
// });
document.getElementById("toast_title").innerHTML = message.title;
document.getElementById("toast_body").innerHTML = message.body;
}
};
xmlhttp.open("GET", "demo_file.php", true);
xmlhttp.send();
</script>
</head>
<body>
<div id="toast_title"></div>
<div id="toast_body"></div>
</body>
</html>
demo_file.php答案 1 :(得分：0)
可能是
<script>
var username = "<?php echo $_SESSION['username']; ?>";
var title = JSON.parse("<?php echo json_encode($title); ?>");
var message = JSON.parse("<?php echo json_encode($message); ?>");
var type = JSON.parse("<?php echo json_encode($type); ?>");
window.setInterval(function(){
var xmlhttp;
if (window.XMLHttpRequest){
xmlhttp = new XMLHttpRequest();
}
else{
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.open("GET","includes/ajax/user/general/messages.php?username="+username ,true);
xmlhttp.send();
$.toast({
heading: title,
text: message,
position: 'top-right',
loaderBg: '#ff6849',
icon: type,
hideAfter: 6000,
stack: 6
});
}, 5000);
</script>
答案 2 :(得分：0)
我不明白为什么那些甚至不知道问题是什么的人都在随意尝试答案，但是：
<?php
if (!isset($_SERVER['HTTP_REFERER'])){
header("HTTP/1.0 404 Not Found")
die('HTTP_REFERER is not set');
}
require_once '../../../app/config.php';
require_once '../../../app/init.php';
if (!empty($maintaince)){
header("HTTP/1.0 404 Not Found")
die('maintaince is not set');
}
if (!($user -> LoggedIn()) || !($user -> notBanned($odb)) || !($user->isAdmin($odb))){
header("HTTP/1.0 404 Not Found")
die('User not logged in, banned, or isn\'t an admin');
}
$username = $_GET['username'];
if(empty($username)){
header("HTTP/1.0 404 Not Found")
die('username is not set');
}
$SQLGetInfo = $odb -> prepare("SELECT * FROM `live_messages` WHERE id = 1");
$SQLGetInfo -> execute(array($_SESSION['ID']));
$messageInfo = $SQLGetInfo -> fetch(PDO::FETCH_ASSOC);
$type = $messageInfo['type'];
$title = $messageInfo['title'];
$message = $messageInfo['message'];
echo json_encode(array(
'type' => $type,
'title' => $title,
'message' => $message,
));
exit();
?>
和
<script>
window.setInterval(function(){
var xmlhttp;
if (window.XMLHttpRequest){
xmlhttp = new XMLHttpRequest();
} else {
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.open("GET","includes/ajax/user/general/messages.php?username=<?php echo $_SESSION['username']; ?>",true);
xmlhttp.addEventListener("load", function () {
if (xmlhttp.status === 200) {
var res = JSON.parse(xmlhttp.responseText);
$.toast({
heading: res.title,
text: res.message,
position: 'top-right',
loaderBg: '#ff6849',
icon: res.type,
hideAfter: 6000,
stack: 6
});
} else {
$.toast({
heading: 'ERROR',
text: 'AN ERROR OCCURED' + xmlhttp.responseText,
position: 'top-right',
loaderBg: '#ff6849',
icon: '',
hideAfter: 6000,
stack: 6
});
}
};
xmlhttp.send();
}, 5000);
</script>