Question

Держите получаю ошибку пространства имен при использовании кода ниже

<?php 

    // error_reporting(0); 

    $BASE_URL = "http://query.yahooapis.com/v1/public/yql"; 

    $yql_query = 'select * from weather.forecast where woeid in (select woeid from geo.places(1) where text="austin, tx")'; 
    $result = file_get_contents($BASE_URL . "?q=" . urlencode($yql_query) . "&format=xml"); 

    if ($result == true) { 
     $xml = simplexml_load_string($result); 
     $xml->registerXPathNamespace('yweather', 'http://xml.weather.yahoo.com/ns/rss/1.0'); 
     $location = $xml->results->channel; 

     if(!empty($location)){ 
      foreach($xml->results->channel->item as $item){ 
       $current = $item->xpath('yweather:condition'); 
       $temp = $current['temp']; 

       echo $temp; 
      } 
     } 
     else{ 
      echo '<h1>No weather for today</h1>'; 
     } 
    }else{ 
     echo '<p>Weather service is down</p>'; 
    } 
?> 

Answer 1

В соответствии с просьбой, вот решение с использованием JSON вместо XML:

$BASE_URL = "http://query.yahooapis.com/v1/public/yql"; 

$yql_query = 'select * from weather.forecast where woeid in (select woeid from geo.places(1) where text="austin, tx")'; 
$result = json_decode(file_get_contents($BASE_URL . "?q=" . urlencode($yql_query) . "&format=json"), true); 

if(is_array($result)) { 
    $location = $result['query']['results']['channel']; 

    if(!empty($location)) { 
     $temp = $result['query']['results']['channel']['item']['condition']['temp']; 
     echo $temp; 
    } else { 
     echo '<h1>No weather for today</h1>'; 
    } 
} else { 
    echo '<p>Weather service is down</p>'; 
} 

Испытано локально моя машина.