Is there any way to use a string variable instead of an actual string in quotes for labelids in service.users().messages().list?

Question

The users.messages.list method requires a string for the labelIds parameter. Is there any way to provide it a string other than entering 'the string'...I want to include this method as part of a for loop cycling through all user defined labels without knowing what they are in advance, but I get a 400 error An API error occurred: <HttpError 400 when requesting https://gmail.googleapis.com/gmail..... this is the format it likes... service.users().messages().list(userId='me',labelIds=['a string']).execute() here is what I want it to do...

 try:
    # Call the Gmail API
    service = build('gmail', 'v1', credentials=creds)
    results = service.users().labels().list(userId='me').execute()
    labels = results.get('labels', [])
    #labels is a list of all system and user defined labels
    #need to cycle through all user defined labels and store emails with 
    #each label
    for label in labels:
        if label['type']=='user':
            name = label['name']  #this is a 'string'
            results = service.users().messages().list(
                userId='me',
                labelIds=[name],
                maxResults=2  # Adjust as needed
            ).execute()
            break

Raymond · Accepted Answer · 2025-05-18 06:20:17Z

0

I think you should use label['id'] rather than label['name']. Sometimes, these two have the same value; sometimes, they don't.

Ref: https://developers.google.com/workspace/gmail/api/reference/rest/v1/users.messages/list#http-request

for label in labels:
    if label['type'] == 'user':
        label_id = label['id']  # Use the label ID, not the name
        results = service.users().messages().list(
            userId='me',
            labelIds=[label_id],  # This works!
            maxResults=2
        ).execute()

        # Do something with results here
        print(f"Messages with label {label['name']}:", results.get('messages', []))
        break  # Remove break if you want to process all labels

answered May 18 at 6:20

Raymond

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2 Comments

Dan May 18 at 23:38

Thank you, I probably should have known that labelids wanted a label id, not a label name.

Raymond May 18 at 23:49

That's OK. They are easily mixed. If you found it useful, please accept the answer.

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