我試圖只從 php 中的 json-ld 獲取文章正文，但我不明白如何。我對從 php 編碼和解碼 json 不太熟悉，所以似乎沒有任何效果。 "@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·斯威夫特是美國創作歌手兼演員……”。我如何實現這一目標？
1 回答
慕俠2389804
TA貢獻1719條經驗 獲得超6個贊
您必須使用json_decode解碼此字符串，然后它只是從數組中獲取您的需求。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'];
因為articleBody它應該是：
$arr['itemListElement'][0]['result']['detailedDescription']['articleBody']
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