用PHP和json回溯YQL

你好家伙我想获取这个YQL查询的天气预报的代码元素，问题是这会让我返回null .. 任何人都可以帮助，请e。我在PHP用PHP和json回溯YQL

     php 
      var_dump(getResultFromYQL()); 

     function getResultFromYQL() { 

      $yql_query_url = "http://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20weather.forecast%20where%20woeid%3D946738&format=json&diagnostics=true&callback=?"; 

      $session = curl_init($yql_query_url); 
      curl_setopt($session, CURLOPT_RETURNTRANSFER, true); 

      $json = curl_exec($session); 
      curl_close($session); 

      return json_decode($json); 
     } 
     ?>

OK是小白我找到了解决办法 PHP

  $Jakarta = 946738; /* Jakarta */ 


      /* Use cURL to query the API for some XML */ 
      $ch = curl_init(); 
      curl_setopt($ch, CURLOPT_URL, 'http://weather.yahooapis.com/forecastrss?w='.$Jakarta.'&u=f'); 
      curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1); 
      $weather_rss = curl_exec($ch); 
      curl_close($ch); 

      /* Create an object of the XML returned */ 
      $weather = new SimpleXMLElement($weather_rss); 

      /* 
      * Since I don't want to figure out an image system, I'll use Weather.com's (what Yahoo does) 
      * by pulling the image directly out of the returned API request. This could be done better. 
      */ 
      $weather_contents = $weather->channel->item->description; 
      // preg_match_all('/<img[^>]+>/i',$weather_contents, $img); 
      // $weather_img = $img[0][0]; 

      /* Get clean parts */ 

      $weather_cond = $weather->channel->item->xpath('yweather:condition'); 


      /* Function to convert Wind Direction from given degree units into Cardinal units */ 

      ?> 


          hp 

             print $weather_cond[0]->attributes()->code; 

           ?> 
        </div> 
       </div> 
      </div> 
      </div>

来源

2013-07-15 user2304067

降callback=?查询部分。它旨在用于JavaScript回调函数。

工作URL应该是：$yql_query_url = "http://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20weather.forecast%20where%20woeid%3D946738&format=json&diagnostics=true";

UPDATE

工作代码：（PHP的27年3月5日和5.4.17测试）

<?php 
function getResultFromYQL() { 
    $yql_query_url = "http://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20weather.forecast%20where%20woeid%3D946738&format=json&diagnostics=true"; 

    $session = curl_init($yql_query_url); 
    curl_setopt($session, CURLOPT_RETURNTRANSFER, true); 

    $json = curl_exec($session); 
    curl_close($session); 
    return json_decode($json); 
} 

var_dump(getResultFromYQL()); 
?>

来源

2013-07-15 12:42:41 Fracsi

我甚至有越来越再次返回null 'callback =？'删除...有一些错误... – user2304067

不幸的是不工作我得到NULL，你会得到什么？我甚至复制粘贴你的代码...我在php 5+服务器上测试 – user2304067

我的代码中的错字。它现在应该工作。 – Fracsi

用PHP和json回溯YQL

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