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Get file’s content by link

Заметка

[Since 2021.03]

A request to get file content by the public link.

POST /api/v1/public/{linkUid}/content HTTP/1.1
  1. Path parameters
ParameterDescription
linkUidThe UID of the link for the file
  1. Example request
curl 'http://localhost:8080/api/v1/public/b2e5947d-b3fb-4a07-ba37-fc0236a31df3/content?accessToken=hjdivzoeylmnenutdziizjxsduhhwugh' –I \
-X GET \
-H 'X-co-auth-token: 8e8a2cd607788b22765b3bc3409dbb90' \
-H 'If-None-Match: b2e5947d-b3fb-4a07-ba37-fc0236a31df3'
  1. Example response
HTTP/1.1 200 OK
X-co-request-id: 506734466131427175447930366d6548
Content-Disposition: attachment; filename*=UTF-8''testfile.txt
Content-Length: 20
Content-Type: text/plain
[ content of the response is skipped ]