使用 PHP SimpleXml 从 xmlfeed 选择特定 id

Question

我有一个使用 ID 或类型的 XML Feed。我需要选择特定节点并将其转换为变量。

XML Feed 示例：

<tour>
<tourName>Amazon Riverboat Adventure</tourName>
<dossierCode>PVIIA</dossierCode>
<tripDetails>
<tripDetail type="StartFinish">ex Lima</tripDetail>
<tripDetail type="What's Included">Lots of stuff </tripDetail>
</tripDetails>

我一直在使用以下方法提取此数据：

<?php
if(!$xml=simplexml_load_file('xmlfeed.xml')){
    trigger_error('Error reading XML file',E_USER_ERROR);
}
foreach ($xml->tourName as $tourname)
foreach ($xml->dossierCode as $agentcode)
?>

但是我不确定如何将 提取为 $startfinish。

有人可以帮忙吗？

非常感谢！

Answer 1

http://www.php.net/manual/en/simplexmlelement.attributes.php

<?php
echo $xmlAsString = <<<XML
<tour>
<tourName>Amazon Riverboat Adventure</tourName>
<dossierCode>PVIIA</dossierCode>
<tripDetails>
<tripDetail type="StartFinish">ex Lima</tripDetail>
<tripDetail type="What's Included">Lots of stuff</tripDetail>
</tripDetails>
</tour>


XML;
$xml = simplexml_load_string($xmlAsString);
foreach ($xml->tourName as $tourname) {
    var_dump($tourname);
}
foreach ($xml->dossierCode as $agentcode) {
    var_dump($agentcode);
}

foreach ($xml->tripDetails as $tripDetails) {
    foreach ($tripDetails as $tripDetail) {
        $attributes = $tripDetail->attributes();
        echo 'Content of the type: ' . $attributes['type'] . PHP_EOL .
             'Content of the whole tag: ' . $tripDetail  . PHP_EOL;
    }
}