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PHP注意:尝试获取非对象错误的属性

[英]PHP Notice: Trying to get property of non-object error

I'm using Factory in AngularJS, The Script is 我在AngularJS中使用Factory,脚本是

app.factory('GetCountryService', function ($http, $q) {
        return {
            getCountry: function(str) {
                // the $http API is based on the deferred/promise APIs exposed by the $q service
                // so it returns a promise for us by default
                var url = "https://www.bbminfo.com/sample.php?token="+str;
                return $http.get(url)
                    .then(function(response) {
                        if (typeof response.data.records === 'object') {
                            return response.data.records;
                        } else {
                            // invalid response
                            return $q.reject(response.data.records);
                        }

                    }, function(response) {
                        // something went wrong
                        return $q.reject(response.data.records);
                    });
            }
        };

    });

My Output Response Screen Shot: 我的输出响应屏幕截图:

在此处输入图片说明

My PHP Script: 我的PHP脚本:

<?php



header("Access-Control-Allow-Origin: *");

header("Content-Type: application/json; charset=UTF-8");



session_start();


$ip = $_SERVER['REMOTE_ADDR'];

$dbname = "xyzData";
$link = mysql_connect("localhost", "Super", "Super") or die("Couldn't make connection.");
$db = mysql_select_db($dbname, $link) or die("Couldn't select database");

$uid = "";
$txt = 0;
$outp = "";

$data    = file_get_contents("php://input");
$objData = json_decode($data);

if (isset($objData->token))
    $uid = mysql_real_escape_string($objData[0]->token, $link);
else if(isset($_GET['uid']))
    $uid = mysql_real_escape_string($_GET['uid']);
else
    $txt += 1;

$outp ='{"records":[{"ID":"' . $objData->token . '"}]}';
echo($outp);

?>

I got error_log message is 我收到error_log消息是

[18-Mar-2016 21:40:40 America/Denver] PHP Notice: Trying to get property of non-object in /home/sample.php [2016年3月18日21:40:40美国/丹佛] PHP注意:尝试获取/home/sample.php中非对象的属性

I tried both $objData->token and $objData->token[0] . 我尝试了$objData->token$objData->token[0] But I got the same error notice. 但是我得到了同样的错误通知。 Kindly assist me... 请帮助我...

I tried the Solution provided in the Post Notice: Trying to get property of non-object error , But it fails so, I raised 50 Bounty Points for this post. 我尝试了“发布通知”中提供的解决方案:尝试获取非对象错误的属性 ,但失败了,因此,我为此帖子提高了50个赏金积分。 I tried to update my requirement in that post Question https://stackoverflow.com/review/suggested-edits/11690848 , But the Edit was Rejected so I posted my requirement as a new Question. 我试图在那条问题https://stackoverflow.com/review/suggested-edits/11690848中更新我的要求,但是编辑被拒绝了,所以我将我的要求发布为新问题。 Kindly assist me... 请帮助我...

$objData doesn't seem to be an object. $ objData似乎不是一个对象。

try $objData['token']. 尝试$ objData ['token']。

Or echo the object $objData and try to figure out what is its structure. 或回显对象$ objData并尝试找出其结构。

The AngularJS is not sending any Object instead it passes the GET element. AngularJS没有发送任何对象,而是传递了GET元素。

Just access the Value by simply using $_GET['uid'] 只需使用$_GET['uid']访问值

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