How to create file object using filename in Box

Question

I need to rename a file in Box using the Python SDK API. I know the filename, but I guess to rename it I need a file object. How do I create a file object when all I have is the name of the file? The file is located in the root folder.

Answer 1

consider this python script:

from boxsdk import JWTAuth, Client

class CFG:
    """config class"""
    JWT_CONFIG_FILE = ".jwt.config.json"
    AS_USER = "18622116055"
    PARENT_FOLDER_ID = "165803865043"  # folder id 0 is root folder


def get_box_client(as_user: bool = False):
    """get a box client"""
    auth = JWTAuth.from_settings_file(CFG.JWT_CONFIG_FILE)
    service_client = Client(auth)
    if not as_user:
        return service_client
    user = service_client.user(CFG.AS_USER)
    return service_client.as_user(user)


def print_items(items):
    """print items"""
    print("\n")
    print("Type\tID\tName")
    print("----\t--\t----")
    for item in items:
        print(f"{item.type}\t{item.id}\t{item.name}\t")


def main():
    """main function"""

    client = get_box_client(as_user=True)

    # print current user info
    user = client.user().get()
    print(f"Current User: {user.name}\tid:{user.id}")

    # list files in root folder
    items = client.folder(CFG.PARENT_FOLDER_ID).get_items()
    print_items(items)

In the above example we have a box app, configured with JWT authentication. We are grabbing a client for which the service user of the app is impersonating the as-user who actually has access to the content we want to locate.

I'm also using another parent folder, but you can replace this with 0 which always represent the user root folder.

This initial part, outputs the items inside the parent folder, this test checks if we actually have access to the file.

In my case the output is:

Type    ID      Name
----    --      ----
file    974207525964    Audio.mp3
...
file    974229112148    Single Page.docx
file    974225001875    ZIP.zip

So let's locate file audio.mp3 in this parent folder using the search:

    # get parent folder
    parent_folder = client.folder(CFG.PARENT_FOLDER_ID).get()

    # search for audio.mp3 text string
    file_name_exact_match = '"audio.mp3"' # note the exact match double quotes
    ancestor_folders = (parent_folder,) 
    result_type = ("file",) # what type of result we are looking for
    content_types = ("name",) # look only in file name for query string
    
    items = client.search().query(
        file_name_exact_match,
        ancestor_folders=ancestor_folders,
        result_type=result_type,
        content_types=content_types,
    )
    print_items(items)

And the output is:

Type    ID      Name
----    --      ----
file    974207525964    Audio.mp3

And now we just rename the file:

# get the first item of type file (there shoul dbe only 1 or none)
item = next((item for item in items if item.type == "file"), None)

# rename the file
if item is not None:
    item.update_info(data={"name": "Audio_renamed.mp3"})

And we can see the file renamed:

# list files in parent folder
items = client.folder(CFG.PARENT_FOLDER_ID).get_items()
print_items(items)

Type    ID      Name
----    --      ----
file    974207525964    Audio_renamed.mp3
...
file    974229112148    Single Page.docx
file    974225001875    ZIP.zip

Please note that the search engine does not index the file immediately, a new file takes a few minutes to show up on the search.

If you know the parent folder, you can implement a manual name match using the items() method on the folder.

# list files in parent folder
items = client.folder(CFG.PARENT_FOLDER_ID).get_items()

item = next((item for item in items if item.name == "ZIP.zip"), None)
if item is not None:
    item.update_info(data={"name": "ZIP_renamed.zip"})

# list items in parent folder
items = client.folder(CFG.PARENT_FOLDER_ID).get_items()
print_items(items)

Resulting in:

Type    ID      Name
----    --      ----
file    974207525964    Audio_renamed.mp3
...
file    974225084827    Preview SDK Sample Excel.xlsx
file    974229112148    Single Page.docx
file    974225001875    ZIP_renamed.zip