Question

我有一个简单的ajax登录页面，如下所示。我试图让它工作4天，但不知道为什么它不起作用。即使我输入有效的电子邮件ID和名称，它也会显示无效的凭据，我也不知道我做错了什么。

这是我的代码：

的login.php

<?php 
    session_start(); 
    $mysqli  = mysqli_connect("localhost","root","","ajax1");
?>
<!DOCTYPE HTML>  
<html>

    <head>
        <title> login script with ajax</title>
        <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.0/jquery.min.js"></script>
    </head>

    <body style="background-color:#b3ffff">

        <div style="padding-left:500px ; padding-top:200px">

            Name: <input id="name" type="text" name="name" placeholder="Enter name"><br><br>
            E-mail: <input id="email" type="email" name="email" placeholder="Enter E-mail"><br><br>
            <input id="submit" name="submit" type="button" value="Log In">
            <p style="color:black">Havent Registered? <a href="index.php">Register</a>.</p><br><br>
            <div id="display" style="color:red"></div>

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

                        var datastring = 'name=' + name + '&email=' + email;

                        if(name=='' || email==''){
                            $("#display").html("Please Enter All The Fields");
                        }
                        else{
                            $.ajax({
                                type: "POST",
                                url: "success.php",
                                data: datastring,
                                cache: false,
                                success: function(result){
                                    $("#display").html(result);
                                    window.location = "welcome.php";
                                }
                            });
                        }
                        return false;
                    });
                });
            </script>
        </div>

    </body>
</html>

success.php

<?php

    session_start();
    $mysqli  = mysqli_connect("localhost","root","","ajax1");


    if (isset($_SESSION['id'])){
        header('location:welcome.php');
    }

    if (isset($_POST['submit'])){

        // removes backslashes
        $name = stripslashes($_REQUEST['name']);

        //escapes special characters in a string
        $name = mysqli_real_escape_string($mysqli,$name);

        // removes backslashes
        $email = stripslashes($_REQUEST['email']);

        //escapes special characters in a string
        $email = mysqli_real_escape_string($mysqli,$email);

        //Checking is user existing in the database or not
        $query = "SELECT * FROM `users` WHERE name='$name' and email='$email'";
        $result = mysqli_query($mysqli,$query);
        $row1 = mysqli_fetch_array($result);
        $rows = mysqli_num_rows($result);

        if($rows==1){
            $_SESSION['name'] = $name;
            $_SESSION['id']=$row1['userid'];
            echo 'Logged in Successfully '
        }
        else{
            echo ' Invalid Name or E-MAIL ';
        }
    }

?>

的welcome.php

<?php
    session_start();
    if (!isset($_SESSION['id'])) {
        header('location:login.php');
    }
?>

<!DOCTYPE html>
<html>
    <body>

        <?php
            include_once 'connect.php';
            $query=mysqli_query($mysqli,"select * from `users` where userid='".$_SESSION['id']."'");
            $row=mysqli_fetch_array($query);
            echo 'Welcome - '.$row['username'];
        ?>

        <!--br>
            <a href="logout.php">Logout</a>
        <br><br-->

    </body>
</html>

Answer 1

您确实需要尽可能简单地进行故障排除。然后当你开始工作时，慢慢地向它添加更多代码，直到它完成所有工作。一步一步。

将success.php页面更改为：

<?php

$myPostArray = array(

  'name'  => $_POST['name'],
  'email' => $_POST['email']

);


echo json_encode($myPostArray);

?>

login.php到此：

  <!DOCTYPE HTML>
<html>

  <head>
    <title> login script with ajax</title>
    <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
  </head>

  <body style="background-color:#b3ffff">

  <div style="padding-left:500px ; padding-top:200px">

    Name: <input id="name" type="text" name="name" placeholder="Enter name" required><br><br>
    E-mail: <input id="email" type="email" name="email" placeholder="Enter E-mail" required><br><br>
    <input id="submit" name="submit" type="button" value="Log In">
    <p style="color:black">Havent Registered? <a href="index.php">Register</a>.</p><br><br>

    <div id="display" style="color:red"></div>

  </div>

  <script>

  $(document).ready(function(){

    $("#submit").click(function(){

      var name = $("#name").val();
      var email = $("#email").val();

      if(name  != '' && email != ''){

        var datastring = 'name=' + name + '&email=' + email;

        $.ajax({

          url: "success.php",
          type: "POST",
          cache: false,
          data: datastring,
          dataType: 'json',
          success: function(data){
            console.log(data); //This will show what you get back from success.php.
            if (data) {
              //It works, do something.
              alert('Here is your response from the php page:' + response);
              console.log('Here is your response from the php page:' + 'Name: ' + data['name'] + ' ' + 'Email: ' + data['email']);

            } else{
              //It does not work, do something.
              alert('Your ajax failed to get the info from your php page. Keep troubleshooting');
              console.log('Your ajax failed to get the info from your php page. Keep troubleshooting');

              }

          }

        });

      }else{

        alert('You must enter a valid name and email.');

        }

    });

  });

  </script>

  </body>

</html>

你真的需要使用你的控制台来帮助你解决这个问题。您可以获得更详细的信息。停止使用alert（）进行故障排除。

Answer 2

您的AJAX和PHP代码都需要一些修复。还尝试在登录时使用密码而不是仅使用电子邮件和名称。尝试修复您的AJAX代码和PHP代码。

在AJAX中使用以下代码。

echo '1'

PHP中的

：

而不是$ _REQUEST ['name']，$ _REQUEST ['email']尝试使用$ _POST ['name']和$ _POST ['email']

然后echo '0'表示成功，或import 'package:flutter/material.dart'; void main() => runApp(new MyApp()); class MyApp extends StatelessWidget { // This widget is the root of your application. @override Widget build(BuildContext context) { return new MaterialApp( title: 'Football App', // app name here home: new TesterClass(), // home page loader here ); } } class TesterClass extends StatefulWidget{ TesterClassState createState() => new TesterClassState(); } class TesterClassState extends State<TesterClass>{ int index = 0; @override Widget build(BuildContext context) { // TODO: implement build return new Scaffold( appBar: new AppBar( title: new Text("Tester"), centerTitle: true, ), bottomNavigationBar: new BottomNavigationBar( type: BottomNavigationBarType.shifting, currentIndex: index, items: [ index == 0? new NavigationButtonView(Icons.home, "Home").activeButtonBuilder():new NavigationButtonView(Icons.home, "Home").inactiveButtonBuilder(), index == 1? new NavigationButtonView(Icons.search, "PL").activeButtonBuilder():new NavigationButtonView(Icons.search, "PL").inactiveButtonBuilder(), index == 2? new NavigationButtonView(Icons.search, "Fantasy").activeButtonBuilder():new NavigationButtonView(Icons.search, "Fantasy").inactiveButtonBuilder(), ], onTap: (int i){setState(() { index = i; });}, ), ); } } class NavigationButtonView{ IconData _icon; String _textLabel; TextStyle activeTextStyle; TextStyle inactiveTextStyle; Color inactiveIconColor; Color activeIconColor; Color activeBackgroundColor; Color inactiveBackgroundColor; NavigationButtonView( IconData icon, String textLabel, ){ _icon = icon; _textLabel = textLabel; } BottomNavigationBarItem activeButtonBuilder(){ return new BottomNavigationBarItem( icon:new Icon(_icon, color: activeIconColor,), title:new Text( _textLabel, style: activeTextStyle, ), backgroundColor: activeBackgroundColor ); } BottomNavigationBarItem inactiveButtonBuilder(){ return new BottomNavigationBarItem( icon: new Icon( _icon, color: inactiveIconColor, ), title: new Text(_textLabel, style: inactiveTextStyle, ), backgroundColor: inactiveBackgroundColor ); } }`表示失败。

尝试在块bby块中构建它并在开发时检查错误

使用有效凭据登录时出错

2 个答案: