我無法從使用PHP的JSON獲取「img_url」(在第三行)值。這裏的JSON字符串:如何解析這個JSON?
編輯(這是JSON,我對張貼的var_dump道歉):
{"status_code":200,"status_txt":"OK",
"data":{"img_name":"KdxIC.png","img_url":"http:\/\/s0.uploads.im\/KdxIC.png",
"img_view":"http:\/\/uploads.im\/KdxIC.png","img_width":"504","img_height":"504",
"img_attr":"width=\"504\" height=\"504\"","img_size":"15.1 KB","img_bytes":15494,
"thumb_url":"http:\/\/s0.uploads.im\/t\/KdxIC.png","thumb_width":360,"thumb_height":360,
"source":"http:\/\/site.com\/uploads\/[email protected]","resized":"0","delete_key":"6e814d3c5201feee"}}
接受的答案的作品。謝謝
您的企圖解決方案在哪裏? – PeeHaa
我從來沒有見過像那樣的JSON ......什麼是返回這些數據? – Justin
這不是JSON(它可能是'json_parsing'在PHP中的一些JSON結果的轉儲)。 – Quentin