使用 XPATH 读取 XML PHP DOM

Question

我有一个问题。

我有 xml

http://maps.googleapis .com/maps/api/geocode/xml?address=new+york&sensor=true

我想读取示例

GeocodeResponse/result/geometry/location/lat

和

GeocodeResponse/result/geometry/location/lng

可能使用 XPATH，这就是我到目前为止所拥有的...

<?php

$Address = "new+york";
$Query = "http://maps.googleapis.com/maps/api/geocode/xml?address=".$Address."&sensor=true";
$XmlResponse = file_get_contents($Query);

$doc = new DOMDocument();
$doc->loadXML($XmlResponse);

$root = $doc->getElementsByTagName( "GeocodeResponse" );
foreach( $root as $val )
{
    $hrefs = $val->getElementsByTagName( "status" );
    $status = $hrefs->item(0)->nodeValue;

    foreach( $hrefs as $val2 )
    {
        $hrefs2 = $val2->getElementsByTagName( "type" );
        $type = $hrefs2->item(0)->nodeValue;

        echo "Type is: $type  <br>";
    }

    echo "Status is: $status  <br>";
}


?>

我可以有一些建议吗？

也许我可以使用

$xpath = new DOMXPath($xml);
$hrefs = $xpath->evaluate("/page");

更新！

我已经设法通过这个得到结果......

$xpath = new DOMXPath($doc);
$res = $xpath->evaluate('//GeocodeResponse/result/geometry');

$root = $doc->getElementsByTagName( "location" );
foreach( $root as $val )
{
    $hrefs = $val->getElementsByTagName( "lat" );
    $status = $hrefs->item(0)->nodeValue;

    echo "Status is: $status  <br>";
}

但我想要一些没有 foreach 的东西，比如

$xpath = new DOMXPath($doc);
$res = $xpath->evaluate('//GeocodeResponse/result/geometry');
$hrefs = $val->getElementsByTagName( "lat" );
$status = $hrefs->item(0)->nodeValue;

echo "Status is: $status  <br>";

这可能吗？

Answer 1

您可能希望切换到更容易解析的 JSON：

http://maps.googleapis。 com/maps/api/geocode/json?address=new+york&sensor=true

$json = file_get_contents('http://maps.googleapis.com/maps/api/geocode/json?address=new+york&sensor=true');

$geocodeResponse = json_decode($json, true);

foreach($geocodeResponse['results'] as $result){
   echo $result['geometry']['location']['lat'].', '.$result['geometry']['location']['lng'] .'<br>';
}