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我想從這個字符串獲取圖像的URL http://www.example.com/xx.jpg
和http://www.example.com/yy.jpg
如何使用php獲取此字符串中的數據?
{
"attachments": {
"data": [
{
"subattachments": {
"data": [
{
"description": "this is description",
"media": {
"image": {
"height": 720,
"src": "http://www.example.com/xx.jpg",
"width": 405
}
}
},
{
"media": {
"image": {
"height": 720,
"src": "http://www.example.com/yy.jpg",
"width": 701
}
},
"target": {
"id": "123456789",
"url": "http://www.example.com"
},
"type": "text",
"url": "http://www.example.com"
}
]
},
"target": {
"id": "123456",
"url": "http://www.example.com"
},
"title": "This is title",
"type": "number",
"url": "http://www.example.com"
}
]
},
"id": "123456789"
}
,所以我用這個代碼
$jason_data = file_get_contents('http://www.test.com');
$result_jason_data = json_decode($jason_data);
echo $result_jason_data->attachments->data->media[0]->image;
echo $result_jason_data->attachments->data->media[1]->image;
但我沒有任何數據,你能告訴我,我該怎麼辦?
的可能的複製[如何解析JSON(https://stackoverflow.com/questions/2591098/how-to-parse-json) –
是的,這是可以做到一步一步,並檢查導致之間。我想你錯過了一個陣列。 –
'echo $ result_jason_data-> attachments-> data [0] - > subattachments-> data [0] - > media-> image-> src;' –