使用ajax成功提交表單后重新加載頁面

Question

我目前正在使用ajax制作登錄表單

<div id='error3'></div>
<input type='email' required name='email' id='email' placeholder='Email'>
<input type='password' required name='password' id='password' placeholder='password'>
<input type='submit'  name='submit99' id='submit' value='login'>

我的Java腳本是

<script>
$(document).ready(function(){
    $("#submit").click(function(){
        var emailnew = $("#email").val();
        var  password = $("#password").val();

        var dataString = '&email='+ email + '&password='+ password;
        if(emailnew==''|| password='')
        {
            document.getElementById('error3').innerHTML="Please Fill All Fields";
        }
        else
        {
            $.ajax({
                type: "POST",
                url: "process.php",
                data: dataString,
                cache: false,
                success: function(result){
                    document.getElementById('error3').innerHTML=result;
                }

            });
        }
        return false;
    });
});

而我的process.php腳本是

<?php
if(isset($_POST['email']))
{
    if (filter_var($email99, FILTER_VALIDATE_EMAIL))
    {
        $con=mysqli_connect('localhost','root','password','database');
        $query="SELECT * FROM TABLENAME WHERE USER='' AND PASSWORD=''";
        $result=mysqli_query($con,$query);
        if(mysqli_num_rows($result)!=0)
        {
            echo "You are successfully logged in";
login user
start a cookie or session
    }
        else
        {
            echo "You are bot a valid user";
        }
    }
    else
        echo "not a valid email";
}
?>

現在，如果我們收到一條消息“您已成功登錄”，則刷新當前頁面，否則僅顯示錯誤消息，而不刷新頁面。

我可以選擇添加

<meta http-equiv="refresh" content="30"> 

但這僅適用於chrome，不適用於Firefox。

Answer 1

您可以將條件放入循環success: function(result){

<script>
    $(document).ready(function () {
        $("#submit").click(function () {
            var emailnew = $("#email").val();
            var password = $("#password").val();

            var dataString = '&email=' + email + '&password=' + password;
            if (emailnew == '' || password = '') {
                document.getElementById('error3').innerHTML = "Please Fill All Fields";
            } else {
                $.ajax({
                    type: "POST",
                    url: "process.php",
                    data: dataString,
                    cache: false,
                    success: function (result) {
                        if (result == "You are successfully logged in") {
                            document.location.href = 'login.htm';
                        } else {
                            document.getElementById('error3').innerHTML = result;
                        }
                    }

                });
            }
            return false;
        });
    });
</script>

Answer 2

嘗試這個：

<script>
    $(document).ready(function () {
        $("#submit").click(function () {
            var emailnew = $("#email").val();
            var password = $("#password").val();

            var dataString = '&email=' + email + '&password=' + password;
            if (emailnew == '' || password = '') {
                document.getElementById('error3').innerHTML = "Please Fill All Fields";
            }
            else {
                $.ajax({
                    type: "POST",
                    url: "process.php",
                    data: dataString,
                    cache: false,
                    success: function (result) {
                        document.getElementById('error3').innerHTML = result;
                        if (result == "You are bot a valid user") {
                            window.location.href = "mypage.html"
                        }
                    }

                });
            }
            return false;
        });
    });
</script>

Answer 3

<script>
    $(document).ready(function(){
        $("#submit").click(function(){
            var emailnew = $("#email").val();
            var  password = $("#password").val();

            var dataString = '&email='+ email + '&password='+ password;
            if(emailnew == null || password == null)
            {
                document.getElementById('error3').innerHTML="Please Fill All Fields";
            }
            else
            {
                $.ajax({
                    type: "POST",
                    url: "process.php",
                    data: dataString,
                    cache: false,
                    success: function(result)
                    {
                        if(result == "logged in true")
                        {
                            window.setTimeout(function(){location.reload()},1000); //reload with time
                            location.reload(); //reload without time
                        }
                        else
                        {
                            document.getElementById('error3').innerHTML=result;
                        }


                    }

                });
            }
            return false;
        });
    });
</script>

Answer 4

您的process.php應該返回類似成功（true）和不成功（false）標志的結果。 或userID（例如，如果成功），或者其他一些標志（如果失敗）。 您不應在proccess.php文件中回顯通知。 那么您可以在JavaScript代碼中包含以下內容：

$.ajax({
   type: "POST",
   url: "process.php",
   data: dataString,  
   cache: false,
   success: function(result){
      if(result=='successful'){
          showMessageAndredirectFunction();
          //in this functionyou can show the message in your html page like the way you do for error and with a delay redirect the page with javascript
      else
        document.getElementById('error3').innerHTML="Bad usernaem or password";
   }});

您可以在process.php中設置會話（如果成功）。

Answer 5

window.location.reload()用於通過使用javascript重新加載頁面。 或者您可以使用window.location.href=window.location.href;