php错误：试图获取非对象的属性

Question

Not able to process this error as I am getting this as my output enter code here

Notice: Trying to get property of non-object in C:\\xampp\\htdocs\\seo\\seo.php on line 161 Notice: Trying to get property of non-object in C:\\xampp\\htdocs\\seo\\seo.php on line 162

and

Notice:Trying to get property of non-object in C:\\xampp\\htdocs\\seo\\seo.php on line 163 Page Authority:0 Domain Authority:0 External Links:

and this is the code

$accessID = " xxxx ";
$secretKey = " xxxxxxxx";
$domain = "$sig";
$expire_in = time() + 500;
$SignIn = $accessID."n".$expire_in;
$binarySignature = hash_hmac('sha1', $SignIn, $secretKey, true);
$urlSafeSignature = urlencode(base64_encode($binarySignature));
$data = "103079215140";
$curlURL = "http://lsapi.seomoz.com/linkscape/url-metrics/?Cols=".$data."&AccessID=".$accessID."&Expires=".$expire_in."&Signature=".$urlSafeSignature;
$Domains = array($domain);
$Domai = json_encode($Domains);
$options = array(
    CURLOPT_RETURNTRANSFER => true,
    CURLOPT_POSTFIELDS     => $Domai
);

$ch = curl_init($curlURL);
curl_setopt_array($ch, $options);
$response = curl_exec($ch);
curl_close( $ch );

$result = json_decode($response,true);    
$pageAuthority=round($result[0]->upa,0);
$domainAuthority=round($result[0]->pda,0);
$externalLinks=$result[0]->ueid;

echo "Page Authority:".$pageAuthority."<br/>";
echo "Domain Authority:".$domainAuthority."<br/>";
echo "External Links:".$externalLinks."<br/>";

Answer 1

You are using:

$result = json_decode($response,true);
                                ^^^^ here

According to the manual :

When TRUE, returned objects will be converted into associative arrays.

So the result will be an array and there will be no objects.

So you need:

$result[0]['upa']
// etc.

Answer 2

$result[0]

is probably not an object.

try using print_r to print the contents of that variable after json_decode.