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[APP SUBMITTED]: sqlite3.OperationalError: near "WHERE": syntax error

Open XxUnkn0wnxX opened this issue 2 years ago • 0 comments

INFO

Python Version: 3.8.12 (default, Oct 13 2021, 06:42:42) [Clang 13.0.0 (clang-1300.0.29.3)] Operating System: macOS-11.6.1-x86_64-i386-64bit Locale: UTF-8 Branch: feature/pp-check-corruption Database: 44.18 Commit: pymedusa/Medusa@dee6639d07521015926db6886ceb3eaeba652e55 Link to Log: https://gist.github.com/982b364f91429525e25f49eac8000161

ERROR

2021-12-14 12:26:18 ERROR    TORNADO :: [dee6639] Uncaught exception in APIv2: OperationalError('near "WHERE": syntax error')
Request: GET /api/v2/history?page=1&limit=25&sort[]=%7B%22field%22:%22date%22,%22type%22:%22desc%22%7D&filter=%7B%22columnFilters%22:%7B%22provider.id%22:%22er%22,%22resource%22:%22rank%22%7D%7D (127.0.0.1)
Traceback (most recent call last):
  File "/Users/ovidijus/Apps/Medusa/ext/tornado/web.py", line 1704, in _execute
    result = await result
  File "/Users/ovidijus/Apps/Medusa/ext/tornado/gen.py", line 769, in run
    yielded = self.gen.throw(*exc_info)  # type: ignore
  File "/Users/ovidijus/Apps/Medusa/medusa/server/api/v2/base.py", line 64, in async_call
    content = yield IOLoop.current().run_in_executor(executor, prepared)
  File "/Users/ovidijus/Apps/Medusa/ext/tornado/gen.py", line 762, in run
    value = future.result()
  File "/usr/local/Cellar/[email protected]/3.8.12_1/Frameworks/Python.framework/Versions/3.8/lib/python3.8/concurrent/futures/thread.py", line 57, in run
    result = self.fn(*self.args, **self.kwargs)
  File "/Users/ovidijus/Apps/Medusa/medusa/server/api/v2/history.py", line 133, in get
    results = db.DBConnection().select(sql_base, params)
  File "/Users/ovidijus/Apps/Medusa/medusa/db.py", line 315, in select
    sql_results = self.action(query, args, fetchall=True)
  File "/Users/ovidijus/Apps/Medusa/medusa/db.py", line 285, in action
    sql_results = self._execute(query, args, fetchall=fetchall, fetchone=fetchone)
  File "/Users/ovidijus/Apps/Medusa/medusa/db.py", line 115, in _execute
    sql_results = cursor.execute(query, args)
sqlite3.OperationalError: near "WHERE": syntax error

STAFF NOTIFIED: @pymedusa/support @pymedusa/moderators

XxUnkn0wnxX avatar Dec 14 '21 01:12 XxUnkn0wnxX