Php 使用yahoo weather查询api时未定义名称空间前缀
使用下面的代码时,不断获取命名空间错误Php 使用yahoo weather查询api时未定义名称空间前缀,php,api,yahoo,weather,Php,Api,Yahoo,Weather,使用下面的代码时,不断获取命名空间错误 <?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")'; $re
<?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>';
}
?>
根据要求,这里有一个使用JSON而不是XML的解决方案:
$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>';
}
在我的机器上进行了本地测试。如果知道xml是什么样子就好了。旁注:将&format=xml更改为&format=json,然后只使用json解码结果并使用简单的PHP数组!这是因为您可能使用了错误的名称空间。在文件顶部定义您正在使用的名称空间,或包含此脚本运行名称空间所需的文件。无法粘贴xml。。使用url查看xml@jszobody,您是否介意显示和示例?