问题
I'm using Factory in AngularJS, The Script is
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:
My PHP Script:
<?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
[18-Mar-2016 21:40:40 America/Denver] PHP Notice: Trying to get property of non-object in /home/sample.php
I tried both $objData->token
and $objData->token[0]
. But I got the same error notice. Kindly assist me...
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. 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. Kindly assist me...
回答1:
The AngularJS is not sending any Object instead it passes the GET element.
Just access the Value by simply using $_GET['uid']
回答2:
$objData doesn't seem to be an object.
try $objData['token'].
Or echo the object $objData and try to figure out what is its structure.
来源:https://stackoverflow.com/questions/36097824/php-notice-trying-to-get-property-of-non-object-error