Question

所以我班上的一个项目是创建一个包含数据库的网站。我无法使登录功能正常工作。我在php中编写了登录和登录脚本。我登录时发生的所有事情都是页面刷新到“localhost / login.php”？而不是“index.php”任何帮助表示赞赏，提前谢谢。

我的login.php

<?php
 include 'nav.php';
?>

<!DOCTYPE html>
<html lang="en">
<head>
  <title>UCF Events</title>
  <meta charset="utf-8">
  <meta name="viewport" content="width=device-width, initial-scale=1">

  <link rel="stylesheet" href="css/login.css">
  <link rel="stylesheet" href="http://maxcdn.bootstrapcdn.com/bootstrap/3.2.0/css/bootstrap.min.css">
  <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
  <script src="http://maxcdn.bootstrapcdn.com/bootstrap/3.2.0/js/bootstrap.min.js"></script>
  <link rel="stylesheet" href="css/navstuff.css">


  <?php
    include 'setUniversityColors.php';
    setColors();
  ?>

    <script>
    function call() {

    //we will POST to this php file on the server
    //it will process what we send and can return back JSON information
    var request = $.post("login-script.php",

      {
        //these are defined in the inputs within our form
        //each input is defined by their id attribute in the HTML
        email: $("#inputEmail").val(),
        password: $("#inputPassword").val()
      }

      //this function is called when we get a response back from the server
     function(json){

      //write back what we get in the "message" field to the response div defined in this HTML
      //the response div is located right below the "Register" button
      $("div.response").html(json.message);

      //on success redirect to the index page
      if(json.success === "success")
        self.location="index.php";

    }

    //defines that we are expecting JSON back from the server
    "json");

    }
  </script>
</head>

<body>

<?php 
    displayNav('login.php' ,'Login');
?>

<div class="container">
   <form class="form-signin">
        <label for="inputEmail" class="sr-only">Email address</label>
        <input type="email" id="inputEmail" class="form-control" placeholder="Email address" required autofocus>
        <label for="inputPassword" class="sr-only">Password</label>
        <input type="password" id="inputPassword" class="form-control" placeholder="Password" required>

        <button class="btn btn-lg btn-primary btn-block" onClick="call();">Sign in</button>
      </form>
</div>


</body>
</html>

我的login-script.php

    <?php

    //sets the following variables:
    $mysqluser = 'root';
    $mysqlpassword = '';
    $mysqldbname = 'db';

    //now connect to the database
    $mysqli = new mysqli("localhost", $mysqluser, $mysqlpassword, $mysqldbname);

    //here we can extract information from the client's POST request
    //  this was submitted by the jQuery function
    //always use mysql_real_escape_string when taking in user input
    //  this prevents SQL injection
    $email = $mysqli->real_escape_string($_POST['UserEmail']);
    $password = $mysqli->real_escape_string($_POST['Password']);

    $success = " ";
    $message = " ";

    $sql = "SELECT Password, isAdmin, isSuperAdmin, UserID FROM user WHERE UserEmail='$email'";

    //run the query and check if it was successful
    $result = $mysqli->query( $sql );
        if($result){

        //get an associative array from the result
        //retrieve specific attribute values by $row['tableAttribute']
        $row = $result->fetch_assoc();

        //check entered password against the hash
        if (password_verify($password, $row['Password'])) {

          $message = "Login Successful, redirecting...";
          $success = "success";

          //start the session and set some identifying variables
          //these are save across pages
          //we will end the session when the user logs out
          session_start();
          $_SESSION['UserEmail'] = $email;
          $_SESSION['isAdmin'] = $row['isAdmin'];
          $_SESSION['isSuperAdmin'] = $row['isSuperAdmin'];
          $_SESSION['id'] = $row['UserID'];



        } else {
          $message = "There was a problem with your user name or password.";
          $success = "fail";
        }
    } else {
        $message = "Error accessing database: " . $mysqli->error;
        $success = "fail";
    }


    $return = array('message' => $message, 'success' => $success);

    echo json_encode($return);

?>

Answer 1

你的代码非常奇怪。我看到你使用jquery，这是一个例子。

$.ajax({
  url: "your.php",
  method: "POST",
  data: { email : $('#email').val(), password: $('#pass').val() }
});

现在，在你的php中

$_POST['email'];
$_POST['password'];

现在使用JSON。

//Prepare your JSON
var myJson = {};
myJson['email'] = $('#email').val();
myJson['password'] = $('#pass').val();
$.ajax({
 url: "your.php",
 method: "POST",
 data: { email : $('#email').val(), password: $('#pass').val() },
 dataType: "json"
});

在你的PHP中

 echo "The array ".print_r($_POST['paramet'],1);
 $data = json_decode($_POST['paramet']);
 echo $data['email'];
 echo $data['password'];

使用PHP登录网站

1 个答案: