Jsonの結果を解析するには？

Question

JSON結果を"col1":"64.7020"から$resultに取得してください。

{"query":{"count":1,"created":"2016-08-28T09:50:07Z","lang":"ru-RU","diagnostics":{"publiclyCallable":"true","url":{"execution-start-time":"1","execution-stop-time":"2","execution-time":"1","content":"http://finance.yahoo.com/d/quotes.csv?e=.csv&f=c4l1&s=USDRUB=X"},"user-time":"2","service-time":"1","build-version":"0.2.48"},"results":{"row":{"col0":"RUB","col1":"64.7020"}}}} 



$tick=file_get_contents('https://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20csv%20where%20url%3D%22http%3A%2F%2Ffinance.yahoo.com%2Fd%2Fquotes.csv%3Fe%3D.csv%26f%3Dc4l1%26s%3DUSDRUB%3DX%22%3B&format=json&diagnostics=true&callback='); 


$url = $tick; 
$json = file_get_contents($url); 
$data = json_decode($json, TRUE); 

$result ?? 
echo col1 ??????? 

Answer 1

file_get_contentsを2回呼び出しました。 $ tickはyahooによって返されたjsonを持っています

$tick = file_get_contents('https://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20csv%20where%20url%3D%22http%3A%2F%2Ffinance.yahoo.com%2Fd%2Fquotes.csv%3Fe%3D.csv%26f%3Dc4l1%26s%3DUSDRUB%3DX%22%3B&format=json&diagnostics=true&callback='); 

$json_a = json_decode($tick, TRUE); 
echo $json_a["query"]["results"]["row"]["col1"];