Question

我正在使用AJAX HTTP请求将登录数据传递给我的php登录脚本。传递数据时，php脚本会在错误处理中检查后输入用户名和密码变量为空。我正在使用数据库中的有效登录详细信息。当用户使用password_hash（）注册时，密码已经过哈希处理;方法也许这就是问题？

这是登录PHP脚本：

   <?php
     session_start();

     if ( isset( $_POST[ 'submit' ] ) ) {
        //Establish DB Connection
        include_once 'DBConnection.php';
        //Store Username and Password from Login Form
        $Username = mysqli_real_escape_string( $conn, $_POST[ 'Username' ] );
        $Password = mysqli_real_escape_string( $conn, $_POST[ 'Password' ] );

       //Error Handlers
      //Check if inputs are empty
      if ( empty( $Username ) || empty( $Password ) ) {
        header( "Location: ../index.php?login=empty" );
        exit();
       } else {
            $sql = "SELECT * FROM User WHERE Username='" . $Username . "'";
            $result = mysqli_query( $conn, $sql );
            $resultCheck = mysqli_num_rows( $result );
            if ( $resultCheck < 1 ) {
                header( "Location: ../index.php?login=error" );
                exit();
             } else {
                if ( $row = mysqli_fetch_assoc( $result ) ) {
                    //Dehash Password
                    $hashedPasswordCheck = password_verify( $Password, $row[ 'Password' ] );
                  if ( $hashedPasswordCheck == false ) {
                     header( "Location: ../index.php?login=error" );
                     exit();
                  } elseif ( $hashedPasswordCheck == true ) {
                    //Log in user
                    $_SESSION[ 'u_id' ] = $row[ 'User_ID' ];
                    $_SESSION[ 'u_first' ] = $row[ 'Forename' ];
                    $_SESSION[ 'u_last' ] = $row[ 'Surname' ];
                    $_SESSION[ 'u_email' ] = $row[ 'Email' ];
                    $_SESSION[ 'u_user' ] = $row[ 'Username' ];
                    header( "Location: ../index.php?login=success" );
                    exit();
                }
             }
          }
        }
    } else {
      header( "Location: ../index.php?login=error" );
      exit();
 }

这是表格代码：

   <form class="form" role="form" method="post" action="../Assets/SignIn.php" accept-charset="UTF-8" id="login-nav">
    <div class="form-group">
     <label class="sr-only" for="exampleInputEmail2">Email Address</label>
     <input type="email" class="form-control" id="Username" placeholder="Email Address/Username" required>
     </div>
    <div class="form-group">
     <label class="sr-only" for="exampleInputPassword2">Password</label>
     <input type="password" class="form-control" id="Password" placeholder="Password" required>
     <div class="help-block text-right"><a href="">Forgot Password?</a>
     </div>
    </div>
    <div class="form-group">
     <button type="submit" name="submit" class="btn btn-primary btn-block" onclick="ajax_Login()">Sign in</button>
   </div>
  <div class="checkbox">
    <label>
    <input type="checkbox"> keep me logged-in
    </label>
  </div>
</form>

这是AJAX请求：

    function ajax_Login(){
      "use strict";
      //Create XMLHttpRequest object
      var request = new XMLHttpRequest();
     //Create variables to store data to be sent to PHP file
     var url = "signIn.php";
     var uname = document.getElementById("Username").value;
     var pword = document.getElementById("Password").value;
     var vars = "Username="+uname+"&Password="+pword;
     request.open("POST", url, true);
     //Set content type header info for sending url encoded variables in the request
    request.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
    //Access the onreadystatechange event for XMLHttpRequest object
    request.onreadystatechange = function(){
    if(request.readyState === 4 && request.status === 200){
        var return_data = request.responseText;
        document.getElementById("status").innerHTML = return_data;
      }
   };
  //Send data to PHP file
  request.send(vars);

}

Answer 1

请检查您的提交按钮名称，因为没有设置提交错误

的原因

Answer 2

当您提交POST方法表单时，它会向服务器发送POST请求，然后设置全局Tree。

让我们来看看你的代码：您使用$_POST检查是否已提交登录表单。这意味着您必须拥有一个带有isset( $_POST[ 'submit' ] )名称的HTML表单元素（输入，复选框等）。如果不这样做，只需简单地检查submit以查询数据库或继续工作。

isset($_POST)

希望这个帮助

登录PHP无法正常工作

2 个答案: