我不知道谁导出API
所以URL是：
https://en.wikipedia.org/w/api.php?action=query&prop=pageimages&format=json&髓核大小＝200&title=让-克洛德%20Van%20Damme
我得到
{"batchcomplete":"","query":{"pages":{"89265": {"pageid":89265,"ns":0,"title":"Jean-Claude Van Damme","thumbnail":{"source":"https://upload.wikimedia.org/wikipedia/commons/thumb/2/27/Jean-Claude_Van_Damme_2012.jpg/141px-Jean-Claude_Van_Damme_2012.jpg","width":141,"height":200},"pageimage":"Jean-Claude_Van_Damme_2012.jpg"}}}}
谁只能导出源？
现在我有了这个代码：
$json=file_get_contents("https://en.wikipedia.org/w/api.php?action=query&prop=pageimages&format=json&pithumbsize=200&titles=$title");
$details=json_decode($json);
echo $details['thumbnail']['source'];
那么我是谁呢？
您可以使用json_decode并将true作为第二个参数传递。
来自文件：
当为TRUE时，返回的对象将转换为关联数组。
那么我想你要找的是：
$json=file_get_contents("https://en.wikipedia.org/w/api.php?action=query&prop=pageimages&format=json&pithumbsize=200&titles=Jean-Claude%20Van%20Damme");
$details=json_decode($json, true);
$source = $details['query']['pages']['89265']['thumbnail']['source'];
echo $source;
将导致：
https://upload.wikimedia.org/wikipedia/commons/thumb/2/27/Jean-Claude_Van_Damme_2012.jpg/141px-Jean-Claude_Van_Damme_2012.jpg
试试这个。
$json = file_get_contents("https://en.wikipedia.org/w/api.php?action=query&prop=pageimages&format=json&pithumbsize=200&titles=Jean-Claude%20Van%20Damme");
$details = json_decode($json);
$pages = $details->query->pages;
$output = '';
foreach ($pages as $pageid => $content){
$image_url = $content->thumbnail->source;
$title = $content->title;
$output .= '<img src="'.$image_url.'"/>'.$title;
}
echo $output;