在页面contacts.php上我有这段代码:
<form method="post" action="request.php">
<label for="bike">Enter Model: </label>
<input type="text" name="bike" value=''/>
<input type="submit" name="submit" value="submit" />
</form>
在页面request.php上我有这段代码:
if (!empty($_POST['bike']))
{
//
$name=$_POST['bike'];
$price=get_price($name);
if (empty($price))
deliver_response(200,"Bike Not Found",NULL);
else
deliver_response(200,"Bike Found",$price);
}
else
{
deliver_response(400,"Invalid Request",NULL);
}
function deliver_response($status,$status_message,$data)
{
header("HTTP/1.1 $status $status_message");
$response['status']=$status;
$response['status_message']=$status_message;
$response['data']=$data;
$json_response=json_encode($response);
echo $json_response;
}
这回应了页面上的json格式化响应&#34; request.php&#34;。 我的问题是,如何将该响应($ json_response)传递回&#34; contacts.php&#34;网页?
以下是我的回答:
$json_response=json_encode($response);
$_SESSION['bike']=$json_response;
header('Location:contacts.php');
然后是contacts.php;
echo $_SESSION['bike'];
答案 0 :(得分:0)
获取request.php页面上的代码并将其放在运行表单的页面顶部。然后改变动作=&#34; request.php&#34; to action =&#34; contacts.php&#34;