我试图将MediaUrl
(仅表示URL)的值转换为PHP上带有JSON的字符串数组。我想忽略其他一切。这是从我调用的URL返回的内容。在PHP中我需要做什么才能将这些值放入数组中?
{
"d": {
"results": [
{
"__metadata":
{
"uri": "https://api.datamarket.azure.com/Data.ashx/Bing/Search/Image?Query='12 monkeys blu-ray'&$skip=0&$top=1",
"type": "ImageResult"
},
"MediaUrl": "http://www.dvdlink.ca/images/Movies%20Covers/9779/47/1250724527_1250911533.jpg",
"Thumbnail": {
"__metadata": {
"type": "Bing.Thumbnail"
},
"MediaUrl": "http://ts4.mm.bing.net/th?id=HN.607991009242185763&pid=15.1",
"FileSize": "10424"
}
},
{
"__metadata": {
"uri": "https://api.datamarket.azure.com/Data.ashx/Bing/Search/Image?Query='12 monkeys blu-ray'&$skip=1&$top=1",
"type": "ImageResult"
},
"MediaUrl": "http://www.bluray-dvd-film-shop.at/media/images/popup/12-Monkeys-bluray.jpg",
"Thumbnail": {
"__metadata": {
"type": "Bing.Thumbnail"
},
"MediaUrl": "http://ts4.mm.bing.net/th?id=HN.608017534956014467&pid=15.1",
"FileSize": "17060"
}
}
],
"__next": "https://api.datamarket.azure.com/Data.ashx/Bing/Search/Image?Query='12%20monkeys%20blu-ray'&$skip=50"
}
}
答案 0 :(得分:1)
$result = array();
$array = json_decode($your_json_string, true);
foreach ($array['d']['results'] as $entry)
$result[] = $entry['MediaUrl'];
您需要的所有内容都不在$ result数组中,如果没有结果(缺少MediaUrl),则空($ result)将返回TRUE。
答案 1 :(得分:0)
你可以这样做
json_decode($data,true);
这将为您提供一个数组,您可以循环访问数组以获取数据。
$data = 'Your JSON string' ;