All the connect call does is this:
int rc = Sqlite3.sqlite3_open(database, out this.db);
if(rc != Sqlite3.SQLITE_OK)
throw GetSqliteError(this.db, null);
Which makes me thing that sqlite3_open is returning an error code.
What happens if you do the same in CPython?
On Fri, Feb 7, 2014 at 6:01 AM, Doug Blank <doug.bl...@gmail.com> wrote:
>
> On Fri, Feb 7, 2014 at 7:47 AM, Slide <slide.o....@gmail.com> wrote:
>
>> Do you get a specific error message?
>>
>
> Sorry, yes:
>
> Traceback (most recent call last):
> File "<string>", line 1, in <module>
> _sqlite3.OperationalError: unable to open database file
>
> Does this work for you? I get this error running latest ipy.exe (Mono
> 2.10, Ubuntu 13.10) and IronPython in Calico.
>
> -Doug
>
>
>
>>
>>
>> On Fri, Feb 7, 2014 at 5:37 AM, Doug Blank <doug.bl...@gmail.com> wrote:
>>
>>> Just recently discovered the sqlite3 port for IronPython, but having
>>> some basic trouble:
>>>
>>> This works fine:
>>>
>>> import sqlite3
>>> sqlite3.connect("test.db")
>>>
>>> but this doesn't:
>>>
>>> import sqlite3
>>> sqlite3.connect("/home/dblank/test.db")
>>>
>>> Any ideas?
>>>
>>> -Doug
>>>
>>>
>>> _______________________________________________
>>> Ironpython-users mailing list
>>> Ironpython-users@python.org
>>> https://mail.python.org/mailman/listinfo/ironpython-users
>>>
>>>
>>
>>
>> --
>> Website: http://earl-of-code.com
>>
>
>
--
Website: http://earl-of-code.com
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