[英]Ajax Post request not working
您好,我已經使用嘗試的ajax方法創建了用於登錄的腳本,但這對我不起作用,請您告訴我為什么我無法使用此方法,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 {
$.ajax({
type : "POST",
url : "includes/register.php",
data : "command=login&username="+username+"&password="+pass,
datatype : "json",
success : function(data) {
if (data.status == 200) {
alert('Successfull');
var return_data = data.responseText;
document.getElementById('messagearea').innerHTML = return_data;
} else {
alert('Unsuccessful');
}
}
});
}
}
它應該成功運行警報,但是我收到警報(“不成功”),這意味着狀態不等於200,任何人都可以幫我解決這個問題
這是我的PHP代碼
if(isset($_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
應該為大寫,
$.ajax({
type: "POST",
url: "includes/register.php",
data: "command=login&username=" + username + "&password=" + pass,
dataType: "json",
根據服務器代碼,您的數據應如下所示
data: "command=login&username=" + username + "&pass=" + pass,
像這樣更改成功處理程序中的代碼,
document.getElementById('messagearea').innerHTML = data;
如果要檢查xhr請求狀態,則應在success function
使用3參數。
success : function(data,text,xhr) {
if (xhr.status == 200) {
alert('Successfull');
var return_data = data;
document.getElementById('messagearea').innerHTML = return_data;
} else {
alert('Unsuccessful');
}
}
對於dataType:“ json”
if(isset($_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>");
}
}
您可以嘗試以下解決方案:
$.ajax({
type: "POST",
url: "includes/register.php",
data: {command: "login", username: username, password: pwd},
dataType: "json",
不要在腳本jQuery中使用“傳遞”,因為有時它不能正常工作。
再見! :D
您需要在ajax成功函數中添加兩個以上的對象,這將幫助您獲取請求的更多信息。
$.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,您還需要將返回數據編碼為JSON,以便您的請求返回類型與JSON匹配。 如果通過開發人員工具baar仔細查看ajax請求,您會發現還有兩個對象返回,其中您的請求包含不同的內容。
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.