PHP 처리 페이지에서 AJAX 요청을 수신하지 못합니다.

Question

PHP 처리 페이지를 가져와야하는 AJAX 함수가 호출 된 제출 단추를 클릭하여 데이터를 게시하는 양식이 있습니다.

//the form 


<form name="quicktransfer" action="" method="post"> 
<td style="width: 214px;padding-left: 5px;padding-right: 5px; cell-spacing:10px "> 
    <dl id='accService'> 
      <dt id='accHeader1'>Quick transfer</dt> 
     <dt class='accLinks1'>From.. </dt> 
    <dt class='accLinks1'> 
<!-- Creates a select drop down box with useres bank account numbers as the options --> 

<?php 
    $acc_tbl='accounts'; 
    $cust = $_SESSION['custno']; 
    $result = mysql_query("SELECT accountNo FROM $acc_tbl WHERE custNo = '$cust'"); 
    echo '<select name="acc_from">'; 
    echo "<option value=\" \"></option>"; 
    while($array = mysql_fetch_assoc($result)) 
    { 
    echo "<option value=\"{$array['accountNo']}\">{$array['accountNo']}</option>\n"; 
    } 
    echo '</select>'; 
?> 

<!-- end of the select drop down --> 

    </dt> 
     <dt class='accLinks1'>To... </dt> <dt class='accLinks1'> 

<!-- Creates a select drop down box with useres bank account numbers as the options --> 

<?php 
    $acc_tbl='accounts'; 
    $cust = $_SESSION['custno']; 
    $result = mysql_query("SELECT accountNo FROM $acc_tbl WHERE custNo = '$cust'"); 
    echo '<select name="acc_to">'; 

    echo "<option value=\" \"></option>"; 
    while($array = mysql_fetch_assoc($result)) 
    { 
    echo "<option value=\"{$array['accountNo']}\">{$array['accountNo']}</option>\n"; 
    } 
    echo '</select>'; 
?> 

<!-- end of the select drop down --> 

    </dt> 
    <dt class='accLinks1'>Amount.. </dt> 
    <dt class='accLinks1'><input type="input" name="amount" id="amount" size="10"/></dt> 
     <dt class='accLinks1'><input type="submit" name="transfer" value="transfer" onclick="loadQuickTransfer()"><br/><br/></dt> 
     </dl>  
    </form> 


//the AJAX script 


function loadQuickTransfer() 
{ 
var xmlhttp; 
if (window.XMLHttpRequest) 
    {// code for IE7+, Firefox, Chrome, Opera, Safari 
    xmlhttp=new XMLHttpRequest(); 
    } 
else 
    {// code for IE6, IE5 
    xmlhttp=new ActiveXObject("Microsoft.XMLHTTP"); 
    } 
xmlhttp.onreadystatechange=function() 
    { 
    if (xmlhttp.readyState==4 && xmlhttp.status==200) 
    { 
    document.getElementById("QuickTransfer").innerHTML=xmlhttp.responseText; 
    } 
    } 
xmlhttp.open("GET","quicktransfer.php",true); 
xmlhttp.send(); 
} 




// the form processing page getting called 


if(isset($_POST['transfer']) && $_POST['amount'] == "") 
{  
    echo '<b><i>Select account (From..To..)<br/><br/>Type amount to transfer.</i></b>'; 
} 
else 
{ 
    $transFrom = $_POST['acc_from']; 
    $transTo = $_POST['acc_to']; 
    $amount = $_POST['amount']; 
    $acc_tbl = 'accounts'; 
    $trans_tbl = 'transactions'; 
    $cust = $_SESSION['custno']; 

    $transfer = mysql_query("UPDATE $acc_tbl SET accountBalance = accountBalance -'$amount' WHERE custNo = '$cust' AND accountNo = '$transFrom'"); 
    if($transfer) 
    { 
     $transfer2 = mysql_query("UPDATE $acc_tbl SET accountBalance = accountBalance + '$amount' WHERE custNo = '$cust' AND accountNo = '$transTo'"); 
    } 
    else 
    { 

     echo '<b>Error.. Could not transfer<br/> Please try again!</b>'; 
     exit(); 

    } 
     if($transfer2) 
     { 

      echo '<b>Transfer complete..</b> '; 
      exit(); 
     } 

} 

Answer 1

난 그냥 간단히 살펴했다, 그러나 당신이합니다 (XMLHttpRequest.send에 사후 변수를 보내지 않을 것 같다 처리 페이지를 호출하기 또는 게시 수신하지 않는 것 그러나 data..the 코드 조각은 다음과 같습니다) 전화.