通过AJAX/PHP传递两个变量？

Question

用更新的代码编辑:我在这里遗漏了一些东西; jQuery正在工作,但变量没有传递给查询.以下是所有更新的代码:

<span class="accepted"><a href="#" class="accept" id="<?php echo $id1; ?>" data-order="<?php echo $name; ?>"><input type="button" title="accept" value="Accept" /></a></span>

AJAX脚本

<script type="text/javascript">
$(function() {
$(".accept").click(function(){
var element = $(this);
var del_id = element.attr("id");
var order_id = element.attr("data-order");

//if(confirm("Are you sure you want to delete this?"))
//{

     $.ajax({
       type: "POST",
       url: "accept.php",
       //data: info,
       data: {id:del_id,order_id:order_id},
       success: function(){}
    //
    });
      $(this).parents(".show").animate({ backgroundColor: "#003" }, "slow")
      .animate({ opacity: "hide" }, "slow");
     //}
    //return false;
    });
    });
    </script>

新的php文件

include('db.php');
$id = $_POST['id'] ;
$name = $_POST['order_id'] ;
$sql = "UPDATE mgap_orders SET mgap_status = 1 WHERE mgap_ska_id = :id AND mgap_ska_report_category = :name"; 
$stmt = $pdo->prepare($sql); 

$stmt->execute(array(
':id' => '$id',
':name' => '$name'
));

Answer 1

而不是使用var info = 'id=' + del_id;和data: info,你应该发送如下变量:

data: {id:del_id}

未来更多,如果你想传递更多变量,你可以像下面这样:

data: {id:del_id,order:ORDER_ID}

希望这可以帮助.