单击提交按钮时,应返回所选奖牌的 ID,并通过 Ajax 将其插入数据库中。但这似乎不起作用。有人知道发生了什么或我做错了什么吗?这是表格:
<form action="" method="post">
<?php
while($medal = $allMedals->fetch_assoc())
{
echo "<img src='" . $medal['imagepath'] . "' />";
echo "<input type='radio' name='radio' class='selectmedal' value='" . $medal['medalid'] . "' />";
}
?>
<input type='submit' value='Nominate' id='submitknop' onclick='geklikt()'/>
</form>
这是geklikt的功能:
function geklikt(){
alert($("#selectmedal").val());
var medalid = $(".selectmedal").val();
var data = { selectmedal:medalid };
$.ajax({
type: "POST",
url: "ajax/medal_save.php",
data: data
}).done(function( msg ) {
$(".feedback").text(msg.message);
});
return false;
正在显示警报,但显示与选择的每个奖牌相同的 id (id 3)。这是来自 Medal_save.php 的代码:
<?php
include('../classes/Nomination.class.php');
if(!empty($_POST['selectmedal']))
{
$nomination = new Nomination();
$nomination->Medal = $_POST['selectmedal'];
try
{
$nominate->Save();
$feedback ['message'] = "Nomination succes!";
$feedback ['status'] = "success";
}
catch(Exception $e)
{
$feedback ['message'] = $e->getMessage();
$feedback ['status'] = "error";
}
}
else
{
$feedback ['message'] = "Please select a medal";
$feedback ['status'] = "error";
}
header('Content-type: application/json');
echo json_encode($feedback);
?>
提前致谢!