问题描述
我有 json-ld 脚本,用于显示 Google 招聘信息部分的招聘信息。
<script type="application/ld+json">
{
"@context" : "https://schema.org/","@type" : "JobPosting","title" : "<?PHP echo($title); ?>","description" :"<?PHP echo($description); ?>","hiringOrganization" : {
"@type" : "Organization","name" : "<?PHP echo($name); ?>","logo" : "example.com/images/<?PHP echo($id);?>.jpg"
},"jobLocation": {
"@type": "Place","address": {
"@type": "PostalAddress","streetAddress": "MW","addressLocality": "Moscow","addressRegion": "Russia","addressCountry": "RU","postalCode": ""
}
},"baseSalary": {
"@type": "MonetaryAmount","currency": "RUB","value": {
"@type": "QuantitativeValue","value": "1500","unitText": "HOUR"
}
},"datePosted" : "2021-06-21","validThrough" : "2021-08-18T00:00","employmentType": "FULL_TIME"
}
</script>
对于第 1 篇文章,我可以在 Google 招聘信息部分找到空缺职位。但是,当我添加更多招聘信息时,它没有显示,我对其进行了测试here 说没问题。
robots.txt 内容:站点地图:https://example.com/sitemap.xml sitemap.xml 内容:
<?xml version="1.0" encoding="UTF-8"?>
<urlset xmlns="http://www.sitemaps.org/schemas/sitemap/0.9">
<url>
<loc>https://example.com</loc>
<lastmod>2021-06-21</lastmod>
</url>
</urlset>
职位发布网址为 https://example.com/posts/71.php 72,依此类推。 谁能帮忙解决这个问题?
解决方法
为职位列表创建结构化数据与Google's guidelines相反:
尽可能将结构化数据放在最详细的叶子页面上。不要添加 结构化数据到旨在呈现作业列表的页面(对于 例如,搜索结果页)。
关于结构化数据的测试有效性。自动化测试工具可以跳过某些 Google 要求的内容合规性。阅读更多General structured data guidelines:
使用自动化工具不容易测试这些指南。
,发现的 URL 170