Json_decode throws "Trying to get property of non-object" - php

Question

I'm trying to use the simple weather API, but somehow it doesn't recognize it as an object whatever I do. This is my code:
$url = "https://www.amdoren.com/api/weather.php?api_key=za8LEJ8F9mcHK8SvLxdM98rM9mNFjW&lat=40.7127837&lon=-74.0059413";
$curl = curl_init($url);
$curl_response = curl_exec($curl);
$jsonobj = json_decode($curl_response);
$msg = "Temperature in ".$city."will be: ". $jsonobj->forecast->max_c;
and this is the data I'm trying to reach with $jsonojb->forecast->max_c:
{
"error" : 0,
"error_message" : "-",
"forecast":[
{"date":"2016-12-02",
"avg_c":8,
"min_c":5,
"max_c":11,
"avg_f":46,
"min_f":41,
"max_f":52,
(...)
but it doesn't work. What am I doing wrong guys?

Answer 1

forecast is an array, so you have to use like this:
$forecast = $jsonobj->forecast;
$forecast[0]->max_c;

Answer 2

You can try out this code:
$url = "https://www.amdoren.com/api/weather.php?api_key=za8LEJ8F9mcHK8SvLxdM98rM9mNFjW&lat=40.7127837&lon=-74.0059413";
$curl = curl_init($url);
curl_setopt($curl, CURLOPT_HTTPHEADER, array('Content-Type: application/json'));
curl_setopt($curl, CURLOPT_RETURNTRANSFER, 1);
$curl_response = curl_exec($curl);
$jsonobj = json_decode($curl_response);
$msg = "Temperature in ". $city . " will be: ". $jsonobj->forecast[0]->max_c;
echo $msg;
Hope it helps!

Answer 3

The "forecast" is an array, you have to add [0] to get the first one and then you will be able to get your "max_c".
Also, your API does not give you the city. You would have to use the Google Geocoding API to convert the lon,lat into a City Name.
$api_key = "za8LEJ8F9mcHK8SvLxdM98rM9mNFjW";
$lon = -74.0059413;
$lat = 40.7127837;
$url = "https://www.amdoren.com/api/weather.php?api_key=".$api_key."&lat=".$lat."&lon=".$lon."";
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, $url);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
$response = json_decode(curl_exec($ch), true);
$max_c = $response['forecast'][0]['max_c'];
// Your API doesn't return the city name.
$city = "City Name";
$msg = "Temperature in ".$city."will be: ". $max_c;
echo $msg;
To get your City Name, here's the function you can use. You will need to replace the API KEY.
// Geocode
function geocode($lat,$lon){
$details_url = "https://maps.googleapis.com/maps/api/geocode/json?latlng=".$lat.",".$lon."&key=YOUR_API_KEY";
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, $details_url);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
$response = json_decode(curl_exec($ch), true);
// If Status Code is ZERO_RESULTS, OVER_QUERY_LIMIT, REQUEST_DENIED or INVALID_REQUEST
if ($response['status'] != 'OK') {
return null;
}
$formatted_address = $response['results'][0]['formatted_address'];
$geometry = $response['results'][0]['geometry'];
$longitude = $geometry['location']['lat'];
$latitude = $geometry['location']['lng'];
$array = array(
'lat' => $geometry['location']['lng'],
'lon' => $geometry['location']['lat'],
'location_type' => $geometry['location_type'],
'formatted_address' => $formatted_address
);
return $array;
}