显示XML feed内容

Question

我正在尝试从xml提要中获取内容并将其显示在列表中。我下面的测试只是为了获得职位空缺。

$feed = file_get_contents('https://www.jobs.nhs.uk/search_xml?client_id=120650');
$xml = simplexml_load_string($feed);

$items = $xml->nhs_search->vacancy_details;

foreach($items as $item) {
    $job_title = $item->job_title;
    echo $job_title;
}

这是xml提要的摘录

<nhs_search>
    <vacancy_details>
    <id>915854585</id>
    <job_title>Band 5 Speech and Language Therapist</job_title>
    </vacancy_details>
</nhs_search>

什么都没有显示，没有错误。

Answer 1

工作正常：

$feed = file_get_contents('https://www.jobs.nhs.uk/search_xml?client_id=120650');

$xml = simplexml_load_string($feed);

$items = $xml->vacancy_details;

foreach ($items as $item) {
    $id = $item->id;
    $job_title = $item->job_title;
    echo $id; echo '<br />';
    echo $job_title;
}

将$items = $xml->nhs_search->vacancy_details;更改为$items = $xml->vacancy_details;