Convert JSON string to PHP Variable

Question

I have a JSON array with Yahoo Weather API data:

"query":{
  "count":1,
  "created":"2015-09-08T15:33:25Z",
  "lang":"en-US",
  "results":{
    "channel":{
      "item":{
        "condition":{
          "code":"30",
          "date":"Tue, 08 Sep 2015 11:13 am EDT",
          "temp":"81",
          "text":"Partly Cloudy"
        }
      }
    }
  }
}

I just need to get temp and text and save them as variables... how do I do this?

I've tried encode, decode, subtr, and a few other methods, but can't seem to get the syntax right. I've tried instructions from Convert JSON string to PHP Array

Here's the code on my site:

  $BASE_URL = "http://query.yahooapis.com/v1/public/yql";
  $yql_query = 'select item.condition from weather.forecast  where woeid in (select woeid from geo.places(1) where text="'.$city.', '.$state.'")';
  $yql_query_url = $BASE_URL . "?q=" . urlencode($yql_query) . "&format=json";
  // Make call with cURL
  $session = curl_init($yql_query_url);
  curl_setopt($session, CURLOPT_RETURNTRANSFER,true);
  $json = curl_exec($session);
  // Convert JSON to PHP object
  $phpObj =  json_decode($json);

    echo '<br><br><br><br>';

    echo $json;

Answer 1

First the result of a json_decode() should be an object or an array so to view it dont use echo try using print_r() or var_dump()

$phpObj =  json_decode($json);
print_r($phpObj);

To get the 2 values you are interested in as all the data structures in your data are objects use :-

echo $phpObj->query->result->channel->item->temp;
echo $phpObj->query->result->channel->item->text;

If you are not sure that the json_decode() is working, possibly the json string is badly formed then test the result of the json_decode() for any errors like so :-

$phpObj =  json_decode($json);
if ( json_last_error() !== 0 ) {
    echo json_last_error_msg();
} else {
    print_r($phpObj);
}