Problems with library ID when running zotcli configure a second time

[joelostblom]

When I run zotcli configure the first time, everything works as expected. However, if I try to run it a second time, I am prompted to enter my library ID (this prompt did not occur the first time). I have tried entering my Zotero login, but the configuration always errors with the following traceback:

Traceback (most recent call last):
  File "/home/joel/anaconda3/lib/python3.5/site-packages/pyzotero/zotero.py", line 284, in _retrieve_data
    self.request.raise_for_status()
  File "/home/joel/anaconda3/lib/python3.5/site-packages/requests/models.py", line 893, in raise_for_status
    raise HTTPError(http_error_msg, response=self)
requests.exceptions.HTTPError: 500 Server Error: Internal Server Error for url: https://api.zotero.org/users/My%20Library/items/top?limit=100&format=json&since=1602

During handling of the above exception, another exception occurred:

Traceback (most recent call last):
  File "/home/joel/anaconda3/bin/zotcli", line 11, in <module>
    sys.exit(cli())
  File "/home/joel/anaconda3/lib/python3.5/site-packages/click/core.py", line 722, in __call__
    return self.main(*args, **kwargs)
  File "/home/joel/anaconda3/lib/python3.5/site-packages/click/core.py", line 697, in main
    rv = self.invoke(ctx)
  File "/home/joel/anaconda3/lib/python3.5/site-packages/click/core.py", line 1066, in invoke
    return _process_result(sub_ctx.command.invoke(sub_ctx))
  File "/home/joel/anaconda3/lib/python3.5/site-packages/click/core.py", line 895, in invoke
    return ctx.invoke(self.callback, **ctx.params)
  File "/home/joel/anaconda3/lib/python3.5/site-packages/click/core.py", line 535, in invoke
    return callback(*args, **kwargs)
  File "/home/joel/anaconda3/lib/python3.5/site-packages/zotero_cli/cli.py", line 155, in configure
    zot = ZoteroBackend(config['api_key'], config['library_id'], 'user')
  File "/home/joel/anaconda3/lib/python3.5/site-packages/zotero_cli/backend.py", line 156, in __init__
    self.synchronize()
  File "/home/joel/anaconda3/lib/python3.5/site-packages/zotero_cli/backend.py", line 160, in synchronize
    new_items = tuple(self.items(since=self._index.library_version))
  File "/home/joel/anaconda3/lib/python3.5/site-packages/zotero_cli/backend.py", line 193, in items
    items = self._zot.makeiter(query_fn(**query_args))
  File "/home/joel/anaconda3/lib/python3.5/site-packages/pyzotero/zotero.py", line 127, in wrapped_f
    retrieved = self._retrieve_data(func(self, *args))
  File "/home/joel/anaconda3/lib/python3.5/site-packages/pyzotero/zotero.py", line 286, in _retrieve_data
    error_handler(self.request)
  File "/home/joel/anaconda3/lib/python3.5/site-packages/pyzotero/zotero.py", line 1342, in error_handler
    raise ze.HTTPError(err_msg(req))
pyzotero.zotero_errors.HTTPError:
Code: 500
URL: https://api.zotero.org/users/MyZoteroID/items/top?limit=100&format=json&since=1602
Method: GET
Response: An error occurred

[astholkohtz]

Hi, go to zotero.com and login. Then click on Settings > Feeds/API.
You will see "Your userID for use in API calls is ID".
That's the ID you have to use, not your login.
I hope this helps.

[joelostblom]

Yes, that works, thank you!

[aetherfaerber]

I ran into the same problem. Thanks for the solution!

As I had to dig it up here and others may not be able to invest the time maybe it would be a good idea to put this in some kind of FAQ or clarify the prompt?

[jbaiter]

A more clear prompt is probably the way to go. Do I understand correctly that this is about the Please enter your library ID? message? How about Please enter your user ID for API calls instead?

[joelostblom]

Yes, that is definitely clearer.

I think part of of the problem is that if I enter "zotero user id for api calls" into a search engine, I don't get any immediately useful hints of where to go to find it. So, it might be worthwhile adding a line to the readme, spelling out how to get it.

If possible, getting the ID automatically somehow given the username would be a good way to go about it in my opinion. Maybe something here can help https://forums.zotero.org/discussion/42131/get-userid-from-username

[jbaiter]

I will look into it, thanks for the link!