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Fix bug when XDG directories are not detected

master
Matteo Cypriani 3 years ago
parent
commit
fc66f256c8
1 changed files with 6 additions and 3 deletions
  1. +6
    -3
      kaabot.py

+ 6
- 3
kaabot.py View File

@@ -85,7 +85,8 @@ class KaaBot(sleekxmpp.ClientXMPP):

If `database` is empty, the file name is generated from the MUC's name
`muc` in the first "kaabot" XDG data dir (usually
`$HOME/.local/share/kaabot/`).
`$HOME/.local/share/kaabot/`). If the XDG data dir can't be located, the
database is created/open in the work directory.

If it contains a value, it is assumed to be the path to the database. If
the name contains the string "{muc}", it will be substituted with the
@@ -98,7 +99,7 @@ class KaaBot(sleekxmpp.ClientXMPP):

data_dir = xdg.BaseDirectory.save_data_path("kaabot")
database = "{muc}.db".format(muc=muc)
return "{}/{}".format(data_dir, database)
return os.path.join(data_dir, database)

@staticmethod
def init_vocabulary(vocabulary_file):
@@ -115,6 +116,8 @@ class KaaBot(sleekxmpp.ClientXMPP):
Error handling:
- If the user-specified file can't be opened, the program will crash.
- Ditto if the XDG-found file exists but can't be opened.
- If the XDG directory cannot be detected, "vocabulary.json" is searched
in the work directory.
- In case of parsing error (the file exists but is invalid JSON),
minimalistic vocabulary is set.
- Ditto if the user didn't use --vocabulary and no vocabulary file is
@@ -122,7 +125,7 @@ class KaaBot(sleekxmpp.ClientXMPP):
"""
if not vocabulary_file:
config_dir = xdg.BaseDirectory.load_first_config("kaabot")
full_path = config_dir + "/vocabulary.json"
full_path = os.path.join(config_dir, "vocabulary.json")
if os.path.exists(full_path):
vocabulary_file = full_path


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