數據未插入到數據庫中

Question

所以我在考慮jQuery ajax的序列化部分存在問題。它不會將任何信息放入數據庫，我不知道爲什麼！很顯然，網頁輸入控件的變量沒有被傳遞到php處理頁面。我在這裏做錯了什麼？在這裏與這些人需要一些幫助！這對我來說是非常新的。

網頁：

<form name="main_form" id="main_form" method="post"> 
    <div id="ChangeAddressDialog" title="Change of Address"> 
     <p>Mailing Address: <input type="text" id="Address1" name="Address1" /></p> 
     <p>Mailing Address 2: <input type="text" id="Address2" name="Address2" /></p> 
     <p>City: <input type="text" id="City" name="City" /></p> 
     <p>State: <input type="text" id="State" name="State" maxlength="2" /></p> 
     <p>Zip Code: <input type="text" id="Zip" id="Zip" maxlength="10" /></p> 
     <p>Country: <input type="text" id="County" name="Country" /></p> 
     <input type="hidden" id="change_of_address_form" name="change_of_address_form" /> 
    </div> 
    </form> 

    $('#ChangeOfAddress').click(function() { 
     //change of address dialog 
     $("#ChangeAddressDialog").dialog({ 
      width:500, 
      modal:true, 
      closeOnEscape:true, 
      buttons: [ 
       { text: "Ok", type: "submit", click: function() { 
         $.ajax({ 
          url: "classes/add-address.php", 
          type: "POST", 
          data: $("#main_form").serialize(), 
          dataType: 'json', 
          error: function(SMLHttpRequest, textStatus, errorThrown){ 
           alert("An error has occurred making the request: " + errorThrown) 

          }, 
          success: function(result){ 
           //do stuff here on success such as modal info 
           //$("#main_form").submit(); 
           $(this).dialog("close"); 
          } 
         }) 
        } 
       }, 
       { text: "Close", click: function() { $(this).dialog("close"); } } ] 
     });//end dialog 
    }); 

PHP處理頁面：

<?php 
require_once('../config.php'); 

//$sqlCheck = ''; 
$parcel_id = isset($_POST['ParcelId']) ? $_POST['ParcelId'] : null; 
$address1 = isset($_POST['Address1']) ? $_POST['Address1'] : null; 
$address2 = isset($_POST['Address2']) ? $_POST['Address2'] : null; 
$city = isset($_POST['City']) ? $_POST['City'] : null; 
$state = isset($_POST['State']) ? $_POST['State'] : null; 
$zip = isset($_POST['Zip']) ? $_POST['Zip'] : null; 
$country = isset($_POST['Country']) ? $_POST['Country'] : null; 

$db = new ezSQL_mysql(DB_USER, DB_PASSWORD, DB_NAME, DB_HOST); 
$result = $db->query("INSERT INTO change_of_address (parcel_id, address_1, address_2, City, State, Zip, Country) VALUES ('" . $parcel_id . "','" . $address1 . "','" . $address2 . "','" . $city . "','" . $state . "','" . $zip . "','" . $country . "')"); 
if ($result == 1) { 
    echo '{"success":true}'; 
} else { 
    echo '{"success":false}'; 
} 

//$sqlCheck = "INSERT INTO change_of_address (parcel_id, address_1, address_2, City, State, Zip, Country) VALUES ('" . $parcel_id . "','" . $address1 . "','" . $address2 . "','" . $city . "','" . $state . "','" . $zip . "','" . $country . "')"; 

//echo json_encode($sqlCheck); 


?> 

Answer 1

data: $("#main_form").serialize(), 

的表單元素是命名main_form。您的jquery選擇器正在尋找ID。你可以在你的表單標籤上更改name="main_form"到id="main_form"，它應該修復它。