我试图从 php 中的 json-ld 中仅获取文章正文,但我不明白如何实现。
我不太熟悉从 php 编码和解码 json,所以似乎没有任何作用。
"@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
}
]
}
我只需要文章正文“泰勒·艾莉森· swift 是一位美国创作歌手和 Actor ......”。我该如何实现这一目标?
最佳答案
您必须使用 json_decode 解码该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'];
对于 articleBody
应该是:
$arr['itemListElement'][0]['result']['detailedDescription']['articleBody']
关于javascript - 如何从php中的知识图谱api中仅提取文章正文?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/57835150/