Question

如何alert json_encode()消息并重新加载页面？以下功能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

您正在尝试获取未定义的索引response

前提是您的PHP脚本返回：

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

在您的javascript中，您可以这样做：

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

Answer 2

以这种方式使用代码

transferProduct.php

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

代码页

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

或者您可以使用$ .getJSON代替$ .ajax

$("#transfer").click(function() {
$.getJSON("transferProduct.php",function (data){
alert(data.msg);    
   location.reload();
});
});

Answer 3

我认为您应该将返回格式设置为在后台正确识别：

{ “成功”：真， “消息”： “---------”}

然后在JavaScript中：data.message

JSON编码警报消息

3 个答案: