標題不重定向到另一個頁面ajax php

-1

我正在使用ajax概念登錄表單請求。下面我發佈了與登錄相關的代碼。它不會重定向到myaccount.php頁面。如果我使用這個echo'<script>window.location="myaccount.php";</script>'; JavaScript函數，它將正常工作。但是，在這裏我想用php做。 它將myaccount.php結果作爲ajax消息返回。標題不重定向到另一個頁面ajax php

Error image

我得到使用此<div id="loginreturn"></div>的錯誤消息。以上我發佈了一個錯誤圖片。 myaccount.php頁面的結果也顯示在<div id="loginreturn"></div>裏面。這是我的問題。它不會重定向到另一個PHP頁面。如何解決這個錯誤？

的login.php

<?php 
    ob_start(); 
    include('config.php'); 
    error_reporting(E_ALL | E_WARNING | E_NOTICE); 
    ini_set('display_errors', TRUE); 
    if($_SERVER['REQUEST_METHOD'] == "POST") 
    { 
     $user_email = $_POST['email']; 
     $user_password = $_POST['password'];  
     $_SESSION['ses_uemail'] = $user_email; 
     if(empty($user_email) || empty($user_password)) 
     { 
      echo "You must provide your email id and password"; 
      die(); 
     }  
     try 
     { 
      $stmt = $conn->prepare("SELECT * FROM table_name WHERE EmailID = ? AND Password = ?"); 
      $conn->errorInfo(); 
      $stmt->bindParam('1', $user_email, PDO::PARAM_STR); 
      $stmt->bindParam('2', $user_password, PDO::PARAM_STR); 
      $stmt->execute(); 
      while($row = $stmt->fetch()) 
      { 
       $rename = $row['Name']; 
       $reemail = $row['EmailID']; 
       $repassword = $row['Password']; 
      }  
      if($reemail == $_SESSION['ses_uemail'] && $repassword == $user_password) 
      { 
       $_SESSION['ses_name'] = $rename; 
       $_SESSION['ses_email'] = $reemail; 
       $_SESSION['ses_password'] = $repassword; 
       header('Location:myaccount.php'); 
      } 
      else 
      { 
       echo "Incorrect email id and password"; 
      } 
     } 
     catch(PDOException $e) 
     { 
      'Error : ' .$e->getMessage(); 
     } 
    } 
    ob_flush();  
?>

ajax.js

// ajax signin 
    jQuery(document).ready(function() { 
     jQuery("#registertologin").click(function() { 

       jQuery("#loginreturn").html("<img src='img/loading.gif'/>"); 
       var postData = $(this).serializeArray(); 
       var formURL = $(this).attr("action"); 
       $.ajax({ 
        url : 'login.php', 
        type: "POST", 
        data : $('#loginform').serialize(), 
        success:function(data, textStatus, jqXHR) { 
         jQuery("#loginreturn").html('<pre><code class="returndata">'+data+'</code></pre>'); 
         $("#loginform").submit_login(); //SUBMIT FORM 
        }, 
        error: function(jqXHR, textStatus, errorThrown) 
        { 
         $("#loginreturn").html('<pre><code class="returndata">AJAX Request Failed<br/> textStatus='+textStatus+', errorThrown='+errorThrown+'</code></pre>'); 
        } 
       }); 
       e.preventDefault(); //STOP default action 
      }); 
    });

的index.php

<form name="loginform" id="loginform" > 
     <table class="outlineborder" width="500" border="0" align="center" cellpadding="0" cellspacing="0"> 
      <tr> 
      <td height="35" colspan="2"> 
       <div class="signinform">SignIn Form</div> 
      </td> 
      </tr> 
      <tr> 
      <td height="35" colspan="2">&nbsp;</td> 
      </tr> 
      <tr> 
      <td width="150" height="35" class="rightalign">Email Id : </td> 
      <td width="348"><input class="inputfield" type="text" name="email" /></td> 
      </tr> 
      <tr> 
      <td height="35" class="rightalign">Password : </td> 
      <td><input class="inputfield" type="password" name="password" /></td> 
      </tr> 
      <tr> 
      <td height="35" colspan="2" align="right"> 
       <span class="forgetpassword">Forget Password ?</span> 
      </td> 
      </tr> 
      <tr> 
      <td height="35" colspan="2" align="center"> 
       <div id="loginreturn"></div> 
      </td> 
      </tr> 
      <tr> 
      <td height="35" colspan="2" align="center"> 
       <input class="btns" type="button" name="submit_login" id="registertologin" value="SignIn" /> 
       <input class="btns" type="button" value="Close" onclick="document.getElementById('login_signup').style.display='none'; document.getElementById('fade').style.display='none'" /> 
      </td> 
      </tr> 
     </table> 
    </form>

來源

2014-02-12 Karuppiah

爲什麼你把標題：在登錄頁面（「位置myaccount.php」），你應該把在成功的函數調用Ajax –

爲什麼downvotes？ – Karuppiah

試試這個，

個

的login.php

<?php 
    ob_start(); 
    include('config.php'); 
    error_reporting(E_ALL | E_WARNING | E_NOTICE); 
    ini_set('display_errors', TRUE); 
    $msg  = array(); 

    if($_SERVER['REQUEST_METHOD'] == "POST") 
    { 
     $user_email = $_POST['email']; 
     $user_password = $_POST['password'];  
     $_SESSION['ses_uemail'] = $user_email; 
     if(empty($user_email) || empty($user_password)) 
     { 
      echo "You must provide your email id and password"; 
      die(); 
     }  
     try 
     { 
      $stmt = $conn->prepare("SELECT * FROM table_name WHERE EmailID = ? AND Password = ?"); 
      $conn->errorInfo(); 
      $stmt->bindParam('1', $user_email, PDO::PARAM_STR); 
      $stmt->bindParam('2', $user_password, PDO::PARAM_STR); 
      $stmt->execute(); 
      while($row = $stmt->fetch()) 
      { 
       $rename = $row['Name']; 
       $reemail = $row['EmailID']; 
       $repassword = $row['Password']; 
      }  
      if($reemail == $_SESSION['ses_uemail'] && $repassword == $user_password) 
      { 
       $_SESSION['ses_name'] = $rename; 
       $_SESSION['ses_email'] = $reemail; 
       $_SESSION['ses_password'] = $repassword; 
       //header('Location:myaccount.php'); //no need 
       return json_encode(array('error'=> false)); 

      } 
      else 
      { 
       //echo "Incorrect email id and password"; 
       return json_encode(array('error'=> true, 'msg'=> "Incorrect email id and password")); 
      } 
     } 
     catch(PDOException $e) 
     { 
      //'Error : ' .$e->getMessage(); 
      return json_encode(array('error'=> true, 'msg'=> $e->getMessage())); 
     } 
    } 
    ob_flush();  
?>

ajax.js

// ajax signin 
    jQuery(document).ready(function() { 
     jQuery("#registertologin").click(function() { 

       jQuery("#loginreturn").html("<img src='img/loading.gif'/>"); 
       var postData = $(this).serializeArray(); 
       var formURL = $(this).attr("action"); 
       $.ajax({ 
        url : 'login.php', 
        type: "POST", 
        data : $('#loginform').serialize(), 
        dataType:"json", 
        success:function(data, textStatus, jqXHR) { 
         jQuery("#loginreturn").html('<pre><code class="returndata">'+data+'</code></pre>'); 
         $("#loginform").submit_login(); //SUBMIT FORM 

         if(data.error) 
         { 
          console.log('data.msg'); 
         } 
         else 
         { 
          window.location.href = 'myaccount.php'; 
         } 
        }, 
        error: function(jqXHR, textStatus, errorThrown) 
        { 
         $("#loginreturn").html('<pre><code class="returndata">AJAX Request Failed<br/> textStatus='+textStatus+', errorThrown='+errorThrown+'</code></pre>'); 
        } 
       }); 
       e.preventDefault(); //STOP default action 
      }); 
    });

來源

2014-02-12 06:39:19

它總是顯示loading.gif圖像.. – Karuppiah

@Beginner試圖檢查控制檯，它說了什麼？ –

我收到了這個錯誤'未捕獲的ReferenceError：未定義的json' – Karuppiah

在 '的login.php'

while($row = $stmt->fetch()) 
{ 
     $rename = $row['Name']; 
     $reemail = $row['EmailID']; 
     $repassword = $row['Password']; 
}  
if($reemail == $_SESSION['ses_uemail'] && $repassword == $user_password) 
{ 
     $_SESSION['ses_name'] = $rename; 
     $_SESSION['ses_email'] = $reemail; 
     $_SESSION['ses_password'] = $repassword; 
     return true; 
}else 
{ 
     return false; 
}

變化 'ajax.js'

變化

success:function(data, textStatus, jqXHR) { 
if(data == true) 
{ 
    window.location.href="myaccount.php" 
}else 
{ 
    jQuery("#loginreturn").html('<pre><code class="returndata">'+data+'</code></pre>'); 
    $("#loginform").submit_login(); //SUBMIT FORM 
} 

},

也許它幫助你......

感謝

來源

2014-02-12 06:41:41

我試過這段代碼。仍然不會重定向到myaccount.php頁面。 – Karuppiah

你得到任何錯誤？檢查日誌並向我發送錯誤？ –

我試過了。但它不會重定向到下一頁。它會返回一個空的消息。我得到這樣的錯誤'未捕獲的ReferenceError：e未定義' – Karuppiah

標題不重定向到另一個頁面ajax php

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