多值發送search.php 發送兩個值search_word和search_word1。但兩者值不顯示在ajax發送多個值
var search_word = $("#search_box").val();
var search_word1 = $("#check_id").val();
var dataString = 'search_word='+ search_word;
if(search_word=='')
{
}
else
{
$.ajax({
type: "GET",
url: "include/search.php",
data: dataString,
的search.php
if(isset($_POST['search_word'])){
$serach_word = $_POST['search_word'];
$serach_word1 = $_POST['search_word1'];
echo "$serach_word<br/>$serach_word1";
}
,請幫助。
就放哪裏'dataString'搜索詞? – Patrick
我已添加。所以請什麼樣的代碼是在 – user3329459