Damien Churchill schrieb:
after several attempts I have finally succeeded in developing a urlencode()
function to encode text correctly like defined in RFC 1738.
Now i have a big problem: how to decode the text?
Example:
# SELECT urlencode('Hellö World!');
urlencode
-----------------------
Hell%C3%B6%20World%21
Does anybody know a way to convert '%21' back to '!' and '%C3%B6' to 'ö'?
I've extracted the unquote method [0] from urllib in the python stdlib
that decodes urlencoded strings. Hopefully be some use!
Not directly, but it gives me some helpful hints. For example i'm now
able to decode some basic characters, for example:
# SELECT chr(x'21'::int);
chr
-----
!
(1 row)
But i clearly have a missunderstanding of other chars, like umlauts or
utf-8 chars. This, for example, should return a 'ö':
# SELECT chr(x'C3B6'::int);
chr
-----
쎶
(1 row)
Also i'm not sure how to figure out, when to decode '%C3' and when to
decode '%C3%B6'.
Thanks for your help,
Torsten
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