[英]Ajax Post request not working
您好,我已经使用尝试的ajax方法创建了用于登录的脚本,但这对我不起作用,请您告诉我为什么我无法使用此方法,但操作不成功
function login() {
var username = document.getElementById('username').value;
var pass = document.getElementById('pass').value;
if(username == '' && pass == '') {
alert('Fields could not be left empty');
} else {
$.ajax({
type : "POST",
url : "includes/register.php",
data : "command=login&username="+username+"&password="+pass,
datatype : "json",
success : function(data) {
if (data.status == 200) {
alert('Successfull');
var return_data = data.responseText;
document.getElementById('messagearea').innerHTML = return_data;
} else {
alert('Unsuccessful');
}
}
});
}
}
它应该成功运行警报,但是我收到警报(“不成功”),这意味着状态不等于200,任何人都可以帮我解决这个问题
这是我的PHP代码
if(isset($_POST['command']) && $_POST['command'] == 'login') {
$username = $_POST['username'];
$pass = $_POST['pass'];
$password = md5($pass);
$check = mysqli_query($connection, "SELECT * FROM users WHERE username = '$username'");
$result = mysqli_num_rows($check);
if($result != 1) {
echo "<div class='message'>Response Successful</div>";
} else {
echo "<div class='message'>Username/Password did not matched</div>";
}
}
datatype
T
应该为大写,
$.ajax({
type: "POST",
url: "includes/register.php",
data: "command=login&username=" + username + "&password=" + pass,
dataType: "json",
根据服务器代码,您的数据应如下所示
data: "command=login&username=" + username + "&pass=" + pass,
像这样更改成功处理程序中的代码,
document.getElementById('messagearea').innerHTML = data;
如果要检查xhr请求状态,则应在success function
使用3参数。
success : function(data,text,xhr) {
if (xhr.status == 200) {
alert('Successfull');
var return_data = data;
document.getElementById('messagearea').innerHTML = return_data;
} else {
alert('Unsuccessful');
}
}
对于dataType:“ json”
if(isset($_POST['command']) && $_POST['command'] == 'login') {
$username = $_POST['username'];
$pass = $_POST['pass'];
$password = md5($pass);
$check = mysqli_query($connection, "SELECT * FROM users WHERE username = '$username'");
$result = mysqli_num_rows($check);
if($result != 1) {
echo json_encode("<div class='message'>Response Successful</div>");
} else {
echo json_encode("<div class='message'>Username/Password did not matched</div>");
}
}
您可以尝试以下解决方案:
$.ajax({
type: "POST",
url: "includes/register.php",
data: {command: "login", username: username, password: pwd},
dataType: "json",
不要在脚本jQuery中使用“传递”,因为有时它不能正常工作。
再见! :D
您需要在ajax成功函数中添加两个以上的对象,这将帮助您获取请求的更多信息。
$.ajax({
type : "POST",
url : "register.php",
data : "command=login&username="+username+"&password="+pass,
datatype : "json",
success : function(data,xhr,req) {
if (req.status == 200) {
alert('Successfull');
var return_data = data.responseText;
document.getElementById('messagearea').innerHTML = return_data;
} else {
alert('Unsuccessful');
}
}
});
另外,您还需要将返回数据编码为JSON,以便您的请求返回类型与JSON匹配。 如果通过开发人员工具baar仔细查看ajax请求,您会发现还有两个对象返回,其中您的请求包含不同的内容。
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