如何从php中的知识图api中仅提取文章正文？

Question

我试图只从 php 中的 json-ld 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    }  ]}我只需要文章正文“泰勒·艾莉森·斯威夫特是美国创作歌手兼演员……”。我如何实现这一目标？

Answer 1

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']