Get Directory File
beta.directories.files.get(strdirectory_file_id, FileGetParams**kwargs) -> FileGetResponse
GET/api/v1/beta/directories/{directory_id}/files/{directory_file_id}
Get a file by its directory_file_id within the specified directory. If you’re trying to get a file by its unique_id, use the list endpoint with a filter instead.
Get Directory File
import os
from llama_cloud import LlamaCloud
client = LlamaCloud(
api_key=os.environ.get("LLAMA_CLOUD_API_KEY"), # This is the default and can be omitted
)
file = client.beta.directories.files.get(
directory_file_id="directory_file_id",
directory_id="directory_id",
)
print(file.id){
"id": "id",
"directory_id": "directory_id",
"display_name": "x",
"project_id": "project_id",
"unique_id": "x",
"created_at": "2019-12-27T18:11:19.117Z",
"data_source_id": "data_source_id",
"deleted_at": "2019-12-27T18:11:19.117Z",
"file_id": "file_id",
"metadata": {
"foo": "string"
},
"updated_at": "2019-12-27T18:11:19.117Z"
}Returns Examples
{
"id": "id",
"directory_id": "directory_id",
"display_name": "x",
"project_id": "project_id",
"unique_id": "x",
"created_at": "2019-12-27T18:11:19.117Z",
"data_source_id": "data_source_id",
"deleted_at": "2019-12-27T18:11:19.117Z",
"file_id": "file_id",
"metadata": {
"foo": "string"
},
"updated_at": "2019-12-27T18:11:19.117Z"
}