首先要說明的是我正在做的事情:我正在製作一個簡單的表單,其中包含一個用於更新公司的ajax請求(在表單提交按鈕上)信息,我是相當新的PHP,但想要在ajax中使用數據傳輸對象(就像我以前在其他語言中做過的那樣),但我沒有越來越多如何在SQL更新中使用DTO對象上的值
好吧,這可能是一個非常簡單的題。就像我上面說的,我之前使用過AJAX數據傳輸對象,只是沒有使用PHP ...我得到了一點勉強,因爲我似乎無法選擇對象數據在目標頁面的sql更新中使用。我可以在console.log中看到對象看起來完全形成幷包含數據,但我不能爲我的生活獲取選擇數據的語法!幫助
Ajax請求:
var $CompanyName = $("#CompanyName").val(),
$TelNumberAreaCode = $("#TelNumberAreaCode").val(),
$TelNumber = $("#TelNumber").val(),
$EmailAddress = $("#EmailAddress").val(),
$AddressLine1 = $("#AddressLine1").val(),
$AddressLine2 = $("#AddressLine2").val(),
$AddressCity = $("#AddressCity").val(),
$AddressCounty = $("#AddressCounty").val(),
$AddressPostCode = $("#AddressPostCode").val();
var CompanyDetailsDTO = {};
CompanyDetailsDTO.CompanyName = $CompanyName;
CompanyDetailsDTO.TelNumberAreaCode = $TelNumberAreaCode;
CompanyDetailsDTO.TelNumber = $TelNumber;
CompanyDetailsDTO.EmailAddress = $EmailAddress;
CompanyDetailsDTO.AddressLine1 = $AddressLine1;
CompanyDetailsDTO.AddressLine2 = $AddressLine2;
CompanyDetailsDTO.AddressCity = $AddressCity;
CompanyDetailsDTO.AddressCounty = $AddressCounty;
CompanyDetailsDTO.AddressPostCode = $AddressPostCo;
var DTO = { 'request': CompanyDetailsDTO }
$.ajax({
type: "POST",
url: "php/updates/update-businessdetails.php",
data: JSON.stringify(DTO),
contentType: "application/json; charset=utf-8",
dataType: "json",
success: CompanyDetailsSuccess,
error: CompanyDetailsError
});
PHP更新/目標文件代碼:
<?php
// DB Connection
// How can i pick the below vars out the object???
// I've tried loads of different syntaxes that i think should/could be right
$CompanyName = ??
$TelNumberAreaCode = ??
$TelNumber = ??
$EmailAddress = ??
$AddressLine1 = ??
$AddressLine2 = ??
$AddressCity = ??
$AddressCounty = ??
$AddressPostCo = ??
$sql="UPDATE `businessdetails` SET `TelNumberAreaCode` = '$TelNumberAreaCode',`EmailAddress` = '$EmailAddress',`AddressLine1` = '$AddressLine1',`AddressLine2` = '$AddressLine2',`AddressCity` = '$AddressCity',`AddressCounty` = '$AddressCounty',`TelNumber` = '$TelNumber',`AddressPostCode` = '$AddressPostCode',`CompanyName` = '$CompanyName'";
$result = $conn->query($sql);
if ($result)
{
header('Content-Type: application/json');
print json_encode($result);
}
?>
我必須做一些錯誤的,或愚蠢的我似乎無法要麼找到答案?!
首先你的SQL是不是安全的。 http://bobby-tables.com/ – Xidh
As或AJAX是一個'POST'請求,你可以使用'$ _POST ['var_name']' – Xidh
'來取得這個值,而不是使用'JSON.stringify',你可以放一個' var postdata = {'plataforma':「valhere」};' – Xidh