Question

我正在使用模态框按钮的id向login.php提交值。我的jquery代码是：

$('#login1').on('click', function () {
  $.post('login.php', {
      'u': $('#email').val(),
      'p': $('#password').val(),
      'r': $('#role').val()
  }, function (data) {
      if (data == 'success') {
          $('#myModal1').modal('show');
      }
      else {
          $('#myModal2').modal('show');
      }
  });
  return false;
});

的login.php

   <?php
   $u=$_POST['u'];
   $p=$_POST['p'];
   $r=$_POST['r'];
   $p=md5($p);
   $mysqli=new mysqli('localhost','root','password','user_details');
   if (mysqli_connect_errno()) {
     printf("Connect failed: %s\n", mysqli_connect_error());
     exit();
   }
   $q=$mysqli->query("select * from register_info where Email='{$u}' ,     Password='{$p}' and Role='{$r}'");
   $e=$q->num_rows;
   if($e>0)
   {
     $loggedIn=true;
     $result='success';
     $_SESSION['logged_in']=true;
     header('location: profile.php?email='.$u);
   }
   else
   {
     $loggedIn=false;
     $result='fail';
   }
   ?>

模态表单代码：

    <div class="modal-body">
        <form class="form-horizontal" action='login.php' method="POST">
            <br>
            <fieldset>
                <div id="legend">
                    <legend class="">Login Details</legend>
                </div>
                <div class="control-group">
                    <!-- Username -->
                    <label class="control-label"  for="email">Email</label>
                    <div class="controls">
                        <input type="text" id="email" name="email" placeholder="Enter Email" class="input-xlarge">
                    </div>
                </div>


                <div class="control-group">
                    <!-- Password-->
                    <label class="control-label" for="password">Password</label>
                    <div class="controls">
                        <input type="password" id="password" name="password" placeholder="Enter Password" class="input-xlarge">
                    </div>
                </div>
                <div class="control-group">
                    <!-- Password-->
                    <label class="control-label" for="role">Role</label>
                    <div class="controls">
                        <select name="role"  class="form-control input-sm" id="role"   placeholder="Enter role">
                        </select>
                    </div>
                </div>
                <br>
                <br>
                <button type="button" class="btn btn-success btn-lg" id="login1" data-dismiss="modal">Login</button>
            </fieldset>
        </form>
    </div>

    </div>
    <div class="modal-footer">


    </div>
</div>

我希望当我点击登录模式框的按钮成功检查电子邮件和密码时，它应该将我重定向到profile.php页面。请帮我纠正代码

Answer 1

您尝试将数据从表单发送到数据库。您的PHP代码没有返回值。用户登录时重定向到另一个页面。因此我修改了可以帮助你的PHP代码...

<?php
       $u=$_POST['u'];
       $p=$_POST['p'];
       $r=$_POST['r'];
       $p=md5($p);
       $mysqli=new mysqli('localhost','root','password','user_details');
       if (mysqli_connect_errno()) {
         printf("Connect failed: %s\n", mysqli_connect_error());
         exit();
       }
       $q=$mysqli->query("select * from register_info where Email='{$u}' ,     Password='{$p}' and Role='{$r}'");
       $e=$q->num_rows;
       if($e>0)
       {
         $loggedIn=true;
         $result='success';
         $_SESSION['logged_in']=true;

       }
       else
       {
         $loggedIn=false;
         $result='fail';
       }
    echo $result; // based on this open your modal window 
       ?>

Answer 2

您在SQL查询中遇到一个问题。

另一个问题是你的jQuery需要遵循ajax post响应的重定向，以下是未经测试的代码：

$('#login1').on('click', function () {
  $.post('login.php', {
      'u': $('#email').val(),
      'p': $('#password').val(),
      'r': $('#role').val()
  }, function (e, xhr) {
     location.href = '/profile.php';
  });
  return false;
});

jquery post值登录.php

2 个答案: