2015-06-19 76 views
3

从表中检查电子邮件和密码后,它假设进入studentDashBoard.php文件。但每次我得到错误,并给出错误的ID或PSD错误。如果(html == true)不符合此条件

这里是我的HTML标记:

login.php

<form action="login.php" id="LoginForm" name="LoginForm" method="post" onsubmit=" return validate();"> 
       <div class="form-group"> 
        <label for="exampleInputEmail1">Email address</label> 
        <div class="input-group"> 
        <span class="input-group-addon" id="basic-addon1"><span class="glyphicon glyphicon-user"></span></span> 
        <input type="email" name="email" class="form-control" id="exampleInputEmail1" placeholder="Enter email" required> 
        </div> 
        <p id="statusEmail"></p> 

       </div> 
       <div class="form-group"> 
        <label for="exampleInputPassword1">Password</label> 
        <div class="input-group"> 
        <span class="input-group-addon" id="basic-addon1"><span class="glyphicon glyphicon-star"></span></span> 
        <input type="password" name="password"class="form-control" id="exampleInputPassword1" placeholder="Password" required> 
        </div> 
        <p id="statusPsd"></p> 

       </div> 
       <hr/> 

        <p id="status"></p> 
       <button type="button" class="btn btn-success"><span class="glyphicon glyphicon-arrow-left">Back</button> 
       <button type="submit" name="submit" id="loginButton" class="btn btn-primary"><span class="glyphicon glyphicon-lock">Login</button> 
       <p><br/></p> 
       <p id="notice"> If not registered yet</p><br/> 
      <button type="button" id="SignUp" class="btn btn-danger">SingUp</button> 

</form> 

这里是jQuery代码:

$(document).ready(function(){ 

    $("#loginButton").click(function(e){ 
     email=$("#exampleInputEmail1").val(); 
     e.preventDefault(); 
     password=$("#exampleInputPassword1").val(); 
     $.ajax({ 
      type: "POST", 
      url: "StudentLogin.php", 
      data: "email="+email+"&password="+password, 

      success:function(html) 
      { 
       if(html==true){ 
       ---> this function is not working while I am login with correct data 
       window.location="studentDashBoard.php"; 
       } 
       else{ 
       $("#status").html("<p>worng id or psd</p>"); 
       } 
      } 
     }); 
    }); 
}); 

StudentLogin.php代码

<?php 
$link = mysqli_connect('localhost','root','','users'); 
if (!$link) { 
    die('Could not connect to MySQL: ' . mysqli_error($link)); 
} 

if(isset($_POST['submit'])) 
{ 
    echo "submit"; 
    $email=$_POST['email']; 

    $password=$_POST['password']; 

    $sql = "SELECT * FROM student WHERE email='$email' AND password='$password' "; 

    $query = mysqli_query($link,$sql); 

    $result= mysqli_num_rows($query); 

    if($result > 0) 
    { 
     echo 'true'; 
    } 
    else 
    { 
     echo 'false'; 
    } 
} 

    mysqli_close($link); 
?> 

studentDashBoard.php文件只是打印“嗨”没有别的。

如果您需要任何更多的细节让我知道

+1

试试这个'成功:功能(HTML){执行console.log(HTML);}',并检查什么是你的PHP文件返回。它将显示在控制台中。 –

+0

试试'echo json_encode('false');' – Hackerman

回答

0

尝试使用下面的if条件:

if(html=="true") {}

0

替代

success:function(html) 
     { 
      if(html==true) 
      { 

      } 
     } 

喜欢的东西

success:function(html) 
     { 
      if($('#result').html(html)==true) 
      { 

      { 
     } 
1

相反的: if(isset($_POST['submit'])){ }studentLogin.php文件我改

if(isset($_POST['email']) && isset($_POST['password'])) { 和现在的作品