timc3 wrote:
> I am having a problem with getting lists/dictionaries out of the
> database.
>
> If I have for instance this in a field of my database:
>
> [['Image format', 'JPEG'], ['Image mode', 'RGB'], ['Image size',
> '1440x900'], ['Compression', '21.0 times']]
>
This is a single field? That's unfortunate.
> And I call it from django, I would get the following string:
Be careful: this isn't a string, it's a list with a single dictionary in
it (but that's really just picking nits). By "call it" I assume you mean
"retrieve it". I know it's tedious of me to correct you on these terms,
but you need to understand that you are dealing with a bunch of geeks
who ascribe precise technical meaning to terms like "call" ... anyway,
enough with the linguistics. And sorry if English isn't your first language.
>
> [{'metadata': u"[['Image format', 'JPEG'], ['Image mode', 'RGB'],
> ['Image size', '1280x960'], ['Compression', '14.0 times']]", 'title':
> u'Technical Metadata'}]
>
> Because of the 'u' UTF it seems that I can't then properly use it as a
> list with dictionaries like so:
>
The u stands for "Unicode", not "UTF". But you will note that "metadata"
isn't a Unicode string, it's a regular ASCII string, so Unicode doesn't
come into that.You seem to have the basics correct, and myProblemObject[0] should indeed be a dictionary. Therefore myProblemObject[0]['metadata'] will be the (Unicode) string (ignoring any wrapping the mail system does) u"[['Image format', 'JPEG'], ['Image mode', 'RGB'], ['Image size', '1280x960'], ['Compression', '14.0 times']]" This means that > myProblemObject[0]['metadata'][0] will give you a single character, u"[", the first character from that string. > > Any ideas? I presume this is Django's handling of the data within the > database being the problem not a python problem in general? I prefer to suggest that the way you are *storing* the data is a problem, since it would really be better if your model had separate fields for image format, image mode, image size (possibly separating width and heigh) and compression. Otherwise youo will forever be fighting to extract meaningful information form the data stored on the database. If you want a really dirty solution you could try metadata = eval(myProblemObject[0]['metadata']) which will give you a list of two-element lists, but this is very fragile and extremely unsafe if anyone else can add to the database contents. regards Steve --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/django-users?hl=en -~----------~----~----~----~------~----~------~--~---
