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我试图从 php 中的 json-ld 只获取文章正文，但我不明白如何。

我对从 php 编码和解码 json 不太熟悉，所以似乎没有任何效果。

  "@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“Taylor Alison Swift 是一位美国创作歌手和女演员......”的文章正文。我如何实现这一目标？

4

1 回答 1

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您必须使用json_decode解码此字符串，然后它只是从数组中获取您的需求。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'];

因为articleBody它应该是：

$arr['itemListElement'][0]['result']['detailedDescription']['articleBody']
于 2019-09-07T15:55:41.013 回答