提交表單時保持表單窗口打開

Question

考慮這個圖像，當用戶點擊一個鏈接時這是一個iframe窗口。

我的問題

當用戶點擊存款的形式被提交和窗口關閉，因此用戶不知道，如果存款是成功還是失敗。

我想做

我正在尋找一種方式來什麼保持iframe的窗口打開形式提交後，顯示相應的消息

表HTML

<form name="depForm" action="" id="register_form" method="post"> 
      User Name<br /><input type="text" name="uname" value="" /><br /> 
      Credit Card Nr<br /><input type="text" name="cc" value="" /><br /> 
      CSV Nr<br /><input type="text" name="csv" value="" /><br /> 
      Amount<br /> <input type="text" name="amount" value="" /><br /> 
      <input type="submit" value="deposit" name="deposit" class="buttono" /> 
      </form> 

PHP代碼

if(isset($_POST['deposit'])){ 
      if(isset($_SESSION['FBID'])){ 
       $uid=$_SESSION['FBID']; 
       $amount = $_POST['amount']; 
       $cc = $_POST['cc']; 
       $csv = $_POST['csv']; 
       //new bal 
       $bal = getBal($uid); 
       $newBal = $bal+$amount; 
       $sql="UPDATE members SET balance='$newBal' WHERE member_nr='$uid'"; 
       $result = mysql_query($sql) or die("error please try again"); 
        if($result){ 

        } 
      } 

如果有人可以告訴我如何保持iframe在表單提交後打開它將不勝感激。

Answer 1

您需要更改表單提交以使用AJAX。在響應中，您可以包含一個狀態標誌以向UI指示請求是否成功並且適當地執行。事情是這樣的：

$('#register_form').submit(function(e) { 
    e.preventDefault(); // stop the standard form submission 
    $.ajax({ 
     url: this.action, 
     type: this.method, 
     data: $(this).serialize(), 
     success: function(data) { 
      if (data.success) { 
       // show success message in UI and/or hide modal 
      } else { 
       // it didn't work 
      } 
     }, 
     error: function(xhr, status, error) { 
      // something went wrong with the server. diagnose with the above properties 
     } 
    }); 
}); 

$success = false; 
if (isset($_POST['deposit'])) { 
    if (isset($_SESSION['FBID'])) { 
     $uid = $_SESSION['FBID']; 
     $amount = $_POST['amount']; 
     $cc = $_POST['cc']; 
     $csv = $_POST['csv']; 

     $bal = getBal($uid); 
     $newBal = $bal + $amount; 
     $sql = "UPDATE members SET balance='$newBal' WHERE member_nr='$uid'"; 
     $result = mysql_query($sql) or die("error please try again"); 
     if ($result) { 
      $success = true; 
     } 
    } 
} 

echo json_encode(array('success' => $success));