使用重音符号将utf-8编码为base64

Question

I have some data like this:

data1 = ['Agos', '30490349304']
data2 = ['Desir\xc3\xa9','9839483948']

I'm using an API that expects the data encoded in base64, so what I do is:

data = data1
string = base64.b64encode("Hi, %s! Your code is %s" % (data[0], data[0]))
myXMLRPCCall(string)

Which works fine with data1. With data2 the encoding goes ok, but then the XMLRPC returns an error, since it expects (from the API docs) only ISO-8859-1 (Latin1) characters.
My question is: how can I transform my string into Latin1 so that the API accepts it?

Answer 1

First make sure you're not confused about encodings, etc. Read, for example, this .

Then notice that the main problem isn't with the base64 encoding, but with the fact that you're trying to put byte string (normal string in Python 2.x) inside a Unicode string. I believe you can fix this by removing the "u" from the last string in your example code.

Answer 2

base64.b64encode("Hi, %s! Your code is %s" % (data[0].decode('utf8').encode('latin1'), data[0]))

Answer 3

This seem to work:

...

data = data2
base64.b64encode("Hi, %s! Your code is %s" % (data[0], data[0]))
# => 'SGksIERlc2lyw6khIFlvdXIgY29kZSBpcyBEZXNpcsOp'

# I can't test the XMLRPC parts, so this is just a hint ..
for_the_wire = base64.b64encode("Hi, %s! Your code is %s" % (data[0], data[0]))
latin_1_encoded = for_the_wire.encode('latin-1')

# send latin_1_encoded over the wire ..

Some python (2.X) unicode readings:

• http://evanjones.ca/python-utf8.html
• http://www.amk.ca/python/howto/unicode
• http://boodebr.org/main/python/all-about-python-and-unicode