如何从这段代码加载xml请

时间:2012-04-03 17:37:53

标签: php xml load

我想从这个xml代码中加载内部值:

    <?xml version="1.0" encoding="UTF-8"?>
<geoPlugin>
    <geoplugin_city>Salt Lake City</geoplugin_city>
    <geoplugin_region>UT</geoplugin_region>
    <geoplugin_areaCode>801</geoplugin_areaCode>
    <geoplugin_dmaCode>770</geoplugin_dmaCode>
    <geoplugin_countryCode>US</geoplugin_countryCode>
    <geoplugin_countryName>United States</geoplugin_countryName>
    <geoplugin_continentCode>NA</geoplugin_continentCode>
    <geoplugin_latitude>40.700199127197</geoplugin_latitude>
    <geoplugin_longitude>-111.94339752197</geoplugin_longitude>
    <geoplugin_regionCode>UT</geoplugin_regionCode>
    <geoplugin_regionName>Utah</geoplugin_regionName>
    <geoplugin_currencyCode>USD</geoplugin_currencyCode>
    <geoplugin_currencySymbol>&#36;</geoplugin_currencySymbol>
    <geoplugin_currencyConverter>1</geoplugin_currencyConverter>
</geoPlugin>

如果你能看到geoplugin_city等 我想将这些值加载到php

$location = 'http://www.geoplugin.net/xml.gp?ip='.$_SERVER['REMOTE_ADDR'];
$xml = simplexml_load_file($location);

没用。 请帮帮我。

opps php i ment xml

5 个答案:

答案 0 :(得分:2)

它不输出XML,而是输出序列化的PHP数据:

$location = 'http://www.geoplugin.net/php.gp?ip=192.168.1.1';
$file = unserialize(file_get_contents($location));
print_r($file);

答案 1 :(得分:1)

如果要调用xml API,则URL为

$location = 'http://www.geoplugin.net/xml.gp?ip='.$_SERVER['REMOTE_ADDR'];
$xml = simplexml_load_file($location);

输出样本

<?xml version="1.0" encoding="UTF-8"?>
<geoPlugin>
    <geoplugin_city>Lagos</geoplugin_city>
    <geoplugin_region>Lagos</geoplugin_region>
    <geoplugin_areaCode>0</geoplugin_areaCode>
    <geoplugin_dmaCode>0</geoplugin_dmaCode>
    <geoplugin_countryCode>NG</geoplugin_countryCode>
    <geoplugin_countryName>Nigeria</geoplugin_countryName>
    <geoplugin_continentCode>AF</geoplugin_continentCode>
    <geoplugin_latitude>6.4531002044678</geoplugin_latitude>
    <geoplugin_longitude>3.395800113678</geoplugin_longitude>
    <geoplugin_regionCode>05</geoplugin_regionCode>
    <geoplugin_regionName>Lagos</geoplugin_regionName>
    <geoplugin_currencyCode>NGN</geoplugin_currencyCode>
    <geoplugin_currencySymbol>&#8358;</geoplugin_currencySymbol>
    <geoplugin_currencyConverter>157.6899963379</geoplugin_currencyConverter>
</geoPlugin>

编辑1

echo "<pre>";
foreach($xml as $key => $value)
{
    echo $key , " = " , $value , "\n" ;
}

输出

geoplugin_city = Lagos
geoplugin_region = Lagos
geoplugin_areaCode = 0
geoplugin_dmaCode = 0
geoplugin_countryCode = NG
geoplugin_countryName = Nigeria
geoplugin_continentCode = AF
geoplugin_latitude = 6.4531002044678
geoplugin_longitude = 3.395800113678
geoplugin_regionCode = 05
geoplugin_regionName = Lagos
geoplugin_currencyCode = NGN
geoplugin_currencySymbol = ₦
geoplugin_currencyConverter = 157.6899963379

由于

:)

答案 2 :(得分:1)

您将需要使用XML api而不是PHP api。

$location = 'http://www.geoplugin.net/xml.gp?ip='.$_SERVER['REMOTE_ADDR'];
$xml = simplexml_load_file($location);

答案 3 :(得分:1)

如果你想要PHP中的值,我建议你根本不使用XML库。

$data = unserialize(file_get_contents('http://www.geoplugin.net/php.gpip='.$_SERVER['REMOTE_ADDR'])));
//Then access the data directly.
echo $data['geoplugin_city'];
echo $data['geoplugin_region'];

答案 4 :(得分:1)

将您的代码更改为第一个

$location = 'http://www.geoplugin.net/xml.gp?ip='.$_SERVER['REMOTE_ADDR'];
$xml = simplexml_load_file($location);

之后,进行Foreach循环以获取该数据的所有数据。

foreach ($xml as $keys => $values) {
     $data[] = $values; // $data array have all your data independently.
}

现在,打印数组并测试值。

echo "<pre>"; print_r($data); echo "</pre>";

但是,尝试转储变量,以便获得数据类型..

var_dump($xml);