Php 使引导模式使用单独的表单/sql表

我有一个表单，它修改了一个名为“invoices”的sql表。我想弹出一个引导模式，它将改变一个名为“customers”的sql表。现在，唯一正确保存的是客户名称。。。我错过什么了吗？这是我到目前为止的代码

主页：

<!-- Modal Add Customer -->
     <div class="modal fade" id="addNewCustomer" tabindex="-1" role="dialog" aria-labelledby="addNewCustomerLabel">
      <div class="modal-dialog" role="document">
        <div class="modal-content">
          <div class="modal-header">
            <button type="button" class="close" data-dismiss="modal" aria-label="Close"><span aria-hidden="true">&times;</span></button>
            <h4 class="modal-title" id="myModalLabel">Add New Customer</h4>
          </div>
           <div class="modal-body">
            <form action="<?php echo $_SERVER['PHP_SELF']; ?>" id="add_customer_form" method="post" class="form-horizontal myaccount" role="form">
            <div class="form-group col-lg-7">
                    <label for="customer_Name_modal">Customer Name (Last Name, First Name)</label>
                    <input type="text" class="form-control" id="customer_Name_modal" name="customer_Name_modal" placeholder="Last Name, First Name">
                    <span class="help-block"></span>
                </div>
            <div class="form-group col-lg-7">
                    <label for="customer_Phone_modal">Phone Number (555-555-5555)</label>
                    <input type="text" class="form-control" id="customer_Phone_modal" name="customer_Phone_modal" placeholder="555-555-5555">
                    <span class="help-block"></span>
                </div>
            <div class="form-group col-lg-7">
                    <label for="customer_Email_modal">Email (johndoe@gmail.com)</label>
                    <input type="text" class="form-control" id="customer_Email_modal" name="customer_Email_modal" placeholder="john@doe.com">
                    <span class="help-block"></span>
                </div>
            <div class="form-group col-lg-7">
                    <label for="customer_Address1_modal">Address Line 1 </label>
                    <input type="text" class="form-control" id="customer_Address1_modal" name="customer_Address1_modal" placeholder="">
                    <span class="help-block"></span>
                </div>
            <div class="form-group col-lg-7">
                    <label for="customer_Address2_modal">Address Line 2 </label>
                    <input type="text" class="form-control" id="customer_Address2_modal" name="customer_Address2_modal" placeholder="">
                    <span class="help-block"></span>
                </div>
            <input type="hidden" id="current_element_id">
        </form>
        </div>
        <div class="modal-footer">
            <button type="button" id="add_new_customer_btn" class="btn btn-primary" data-loading-text="Saving Customer...">Save changes</button>
            <button type="button" class="btn btn-default" data-dismiss="modal">Close</button>
          </div>
            </div>
            </div>
            </div>

Sql代码：

if(isset($_POST['type']) && $_POST['type'] == 'saveNewCustomer' ){
    $res = array();
    $res['success'] = false;

    if(!isset($_POST['data']) || empty($_POST['data'])){
        echo json_encode($res);exit;
    } 
    $data = $_POST['data'];

    $customerName = mysqli_real_escape_string( $db->con, trim( $data['customerName'] ) );
    $customerPhone = mysqli_real_escape_string( $db->con, trim( $data['phone'] ) );
    $customerEmail = mysqli_real_escape_string( $db->con, trim( $data['email'] ) );
    $customerAddress1 = mysqli_real_escape_string( $db->con, trim( $data['addressLine1'] ) );
    $customerAddress2 = mysqli_real_escape_string( $db->con, trim( $data['addressLine2'] ) );

    $uuid = uniqid();
    $result['operation'] = 'insert';
    $query = "INSERT INTO customers (id, customerName, phone, email, addressLine1, addressLine2, uuid) 
            VALUES (NULL, '$customerName', '$customerPhone', '$customerEmail', '$customerAddress1', '$customerAddress2', '$uuid');";


    if(mysqli_query($db->con, $query)){
        mysqli_close($db->con);
        $res['success'] = true;
    }
    echo json_encode($res);exit;


}

---

您在PHP中使用了错误的键

您正在调用这些键：

$customerName = mysqli_real_escape_string( $db->con, trim( $data['customerName'] ) );
$customerPhone = mysqli_real_escape_string( $db->con, trim( $data['phone'] ) );
$customerEmail = mysqli_real_escape_string( $db->con, trim( $data['email'] ) );
$customerAddress1 = mysqli_real_escape_string( $db->con, trim( $data['addressLine1'] ) );
$customerAddress2 = mysqli_real_escape_string( $db->con, trim( $data['addressLine2'] ) );

但你应该打电话：

$customerName = mysqli_real_escape_string( $db->con, trim( $data['customerName'] ) );
$customerPhone = mysqli_real_escape_string( $db->con, trim( $data['customerPhone'] ) );
$customerEmail = mysqli_real_escape_string( $db->con, trim( $data['customerEmail'] ) );
$customerAddress1 = mysqli_real_escape_string( $db->con, trim( $data['customerAddress1'] ) );
$customerAddress2 = mysqli_real_escape_string( $db->con, trim( $data['customerAddress2'] ) );

---

要解决这个问题，只需复制并粘贴（替换）SQL代码中的变量赋值。

我在你发帖前2秒就发现了这个问题，哈哈。这已经让我发疯一个小时了！非常感谢。

$customerName = mysqli_real_escape_string( $db->con, trim( $data['customerName'] ) );
$customerPhone = mysqli_real_escape_string( $db->con, trim( $data['customerPhone'] ) );
$customerEmail = mysqli_real_escape_string( $db->con, trim( $data['customerEmail'] ) );
$customerAddress1 = mysqli_real_escape_string( $db->con, trim( $data['customerAddress1'] ) );
$customerAddress2 = mysqli_real_escape_string( $db->con, trim( $data['customerAddress2'] ) );