I have an HTML page that takes user input from HTML page and display it. I am calling the PHP using AJAX. so the output is display in the same HTML. There are input in the database that are in Chinese. When the data is fetched, it changes to ?????. Below is the code:
HTML:
<script>
function PostData() {
var online = navigator.onLine;
if(online){
// 1. Create XHR instance - Start
var xhr;
if (window.XMLHttpRequest) {
xhr = new XMLHttpRequest();
}
else if (window.ActiveXObject) {
xhr = new ActiveXObject("Msxml2.XMLHTTP");
}
else {
throw new Error("Ajax is not supported by this browser");
}
// 1. Create XHR instance - End
// 2. Define what to do when XHR feed you the response from the server - Start
xhr.onreadystatechange = function () {
if (xhr.readyState === 4) {
if (xhr.status == 200 && xhr.status < 300) {
document.getElementById('div1').innerHTML = xhr.responseText;
}
}
}
// 2. Define what to do when XHR feed you the response from the server - Start
var userid = document.getElementById("userid").value;
var pid = document.getElementById("pid").value;
// var image = document.getElementById("image").value;
// 3. Specify your action, location and Send to the server - Start
xhr.open('POST', 'login3.php');
//xhr.open('POST', 'config.php');
xhr.setRequestHeader("Content-Type", "application/x-www-form-urlencoded");
xhr.send("userid=" + userid + "&pid=" + pid);
//xhr.send("&pid=" + pid);
// 3. Specify your action, location and Send to the server - End
}
else{
alert("You are offline");
}
}
</script>
</head>
<body>
<form>
<label for="userid">User ID :</label><br/>
<input type="text" name ="userid" id="userid" /><br/>
<label for="pid">Password :</label><br/>
<input type="password" name="password" id="pid" /><br><br/>
<div id="div1">
<input type="button" value ="Login" onClick="PostData()" />
</div>
</form>
</body>
PHP:
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "dtable";
mysql_query("SET NAMES 'utf8'");//this is what i tried after searching on google
//session_start();
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
if(isset($_POST['userid'],$_POST['pid']))
{
$userid = trim($_POST["userid"]);
$pid = trim($_POST["pid"]);
$sql = "SELECT * FROM demo WHERE username = '$userid' and password = '$pid'";
$result = mysqli_query($conn,$sql);
$row = mysqli_fetch_array($result);
echo $row['week'].'<br/>'.'<br/>';
echo '<a href="2ndHTML.html"/>'.$row['day1'].'</a>'.'<br/>';
?>