在Python中将已解码（来自北印度语）参数添加到给定的url

Question

I have this url = ' http://www.bhaskar.com/uttar_pradesh/lucknow/= '. after the "=" sign one Hindi word which denotes the word searched for is given. I want to be able to add that as a parameter to this url, so that I will only need to change the word each time and not the whole url. I tried to use this:

>>> url = 'http://www.bhaskar.com/uttar_pradesh/lucknow/='
>>> word = 'word1'
>>> conj = url + word

but this gives me the Hindi word in unicode. like this:

>>> conj
'http://www.bhaskar.com/uttar_pradesh/lucknow/=\xe0\xa6\xb8\xe0\xa6\xb0'

Can anyone help?

Answer 1

but this gives me the Bengali word in unicode

No, it does not :)

When you type temp in the terminal, it displays an unique interpretation of the string. When you type print(temp) , however, you are getting a more user-friendly representation of the same string. In the end, however, the string pointed by temp is the same all the time, it is only presented in different ways. See, for example, if you get the second one and put it in a variable and print it:

>>> temp2 = 'http://www.cfilt.iitb.ac.in/indowordnet/first?langno=3&queryword=\xe0\xa6\xb8\xe0\xa6\xb0'
>>> print(temp2)
http://www.cfilt.iitb.ac.in/indowordnet/first?langno=3&queryword=সর

Actually, you can create the string by using escaped values in all characters, not only the Bengali one:

>>> temp3 = '\x68\x74\x74\x70\x3a\x2f\x2f\x77\x77\x77\x2e\x63\x66\x69\x6c\x74\x2e\x69\x69\x74\x62\x2e\x61\x63\x2e\x69\x6e\x2f\x69\x6e\x64\x6f\x77\x6f\x72\x64\x6e\x65\x74\x2f\x66\x69\x72\x73\x74\x3f\x6c\x61\x6e\x67\x6e\x6f\x3d\x33\x26\x71\x75\x65\x72\x79\x77\x6f\x72\x64\x3d\xe0\xa6\xb8\xe0\xa6\xb0'
>>> print(temp3)
http://www.cfilt.iitb.ac.in/indowordnet/first?langno=3&queryword=সর

In the end, all these strings are the same:

>>> temp == temp2
True
>>> temp == temp3
True

So, don't worry, you have the correct string in the variable. You are only getting a problem if the escaped string is displayed elsewhere. Finish your program, run it until the end and you'll see there will be no errors.