javascript - JSON 编码警报消息

Question

如何提醒 json_encode() 消息并重新加载页面？下面的函数仅显示未定义的警报消息

if($result1 == false)
 $response['msg'] = "Transfer failed due to a technical problem. Sorry.";
else
 $response['msg'] = "Successfully transferred";
echo json_encode($response);

$("#transfer").click(function() {
 $.ajax({
 type : "POST",
 url : "transferProduct.php",
 data : {},
 success : function(data) {                     
   data = $.parseJSON(data);
   alert(data.response);    
   location.reload();         
  }
 });
});

Answer 1

您正在尝试获取 undefined index 响应

假设您的 PHP 脚本返回:

{
    "msg": "<your-message-here>"
}

在你的 JavaScript 中你可以做到这一点:

$.ajax({
    type : "POST",
    url: "transferProduct.php",
    dataType: 'json',
    success : function(response) {                     
        alert(response.msg);
        location.reload();         
    }
});