Javascript 如何在php中仅从知识图api中提取文章正文？

我试图在php中仅从json ld获取文章正文，但我不明白如何获取

我不太熟悉从php对json进行编码和解码，所以似乎什么都不管用

  "@context": {
    "@vocab": "http://schema.org/",
    "goog": "http://schema.googleapis.com/",
    "resultScore": "goog:resultScore",
    "detailedDescription": "goog:detailedDescription",
    "EntitySearchResult": "goog:EntitySearchResult",
    "kg": "http://g.co/kg"
  },
  "@type": "ItemList",
  "itemListElement": [
    {
      "@type": "EntitySearchResult",
      "result": {
        "@id": "kg:/m/0dl567",
        "name": "Taylor Swift",
        "@type": [
          "Thing",
          "Person"
        ],
        "description": "Singer-songwriter",
        "image": {
          "contentUrl": "https://t1.gstatic.com/images?q=tbn:ANd9GcQmVDAhjhWnN2OWys2ZMO3PGAhupp5tN2LwF_BJmiHgi19hf8Ku",
          "url": "https://en.wikipedia.org/wiki/Taylor_Swift",
          "license": "http://creativecommons.org/licenses/by-sa/2.0"
        },
        "detailedDescription": {
          "articleBody": "Taylor Alison Swift is an American singer-songwriter and actress. Raised in Wyomissing, Pennsylvania, she moved to Nashville, Tennessee, at the age of 14 to pursue a career in country music. ",
          "url": "http://en.wikipedia.org/wiki/Taylor_Swift",
          "license": "https://en.wikipedia.org/wiki/Wikipedia:Text_of_Creative_Commons_Attribution-ShareAlike_3.0_Unported_License"
        },
        "url": "http://taylorswift.com/"
      },
      "resultScore": 896.576599
    }
  ]
}

---

我只需要这篇文章的主体是“泰勒·艾莉森·斯威夫特是一位美国歌手、作曲家和女演员……”。如何实现这一点？

您必须先解码此字符串，然后它才能从数组中提取您的需要。例如

$j = '{"@context": {"@vocab": "http://schema.org/", "goog": "http://schema.googleapis.com/", "resultScore": "goog:resultScore" }}';

$arr = json_decode($j, true);

echo $arr['@context']['goog'];

对于

articleBody

，它应该是：

$arr['itemListElement'][0]['result']['detailedDescription']['articleBody']

---

你必须解码这个字符串，然后它只是从数组中提取你的需要。例如

$j = '{"@context": {"@vocab": "http://schema.org/", "goog": "http://schema.googleapis.com/", "resultScore": "goog:resultScore" }}';

$arr = json_decode($j, true);

echo $arr['@context']['goog'];

对于

articleBody

，它应该是：

$arr['itemListElement'][0]['result']['detailedDescription']['articleBody']

---

它说“未定义的索引：@itemListElement”，这是我修改过的代码：但现在它显示“未定义的索引：结果”。问题可能是由什么引起的？请尝试

['itemListElement'][0]['result']['detailedDescription']['articleBody']

它说的是“未定义的索引：@itemListElement”这是我修改过的代码：但现在它显示的是“未定义的索引：结果”。问题可能是由什么引起的？请尝试

['itemListElement'][0]['result']['detailedDescription']['articleBody']