通过 API [PHP] 从返回的参数中获取下载 URL mega.nz


get download url from returned parameter by mega.nz API [PHP]

我使用这个简单的PHP代码将文件上传到云 mega.nz 并且代码工作正常。

代码返回如下数组:

h => GxsyWLTK
t => 0
a => KxaM1xBhlW8oAc1zECKfnk7uSulGtHy0FKagNNF2iFo
k => 6ffMqPn8b0Y:fsN980rUUZhmDNBnDZM7PjM2S5rQe-7oTjLVbssQ2F4
p => jwsyAbxY
ts => 1422434349
u => 6ffMqPn8b0Y
s => 30751

但我想知道:如何获取与此链接类似的上传文件的下载链接?https://mega.nz/#!SttF0BiC!BXQIrgxB9Q5qWmccdjMnsISUCyRV4Hr6f-v5RjQxB_w

对于每个节点,结果包含以下信息:

h: The ID of the node ; --> <ID-NODE>
p: The ID of the parent node (directory) ;
u: The owner of the node ;
t: The type of the node:
  0: File
  1: Directory
  2: Special node: Root (“Cloud Drive”)
  3: Special node: Inbox
  4: Special node: Trash Bin
a: The attributes of the node. Currently only contains its name.
k: The key of the node; --> <KEY-NODE>
s: The size of the node ;
ts: The time of the last modification of the node

通用超级下载链接:

https://mega.nz/#! <ID-NODE> ! <KEY-NODE>

示例(使用您的数据):

https://mega.nz/#!GxsyWLTK!6ffMqPn8b0Y:fsN980rUUZhmDNBnDZM7PjM2S5rQe-7oTjLVbssQ2F4

希望这有帮助。