Covert int to hex string with a specific format in python

Asked Mar 19 '17 at 03:12

Active Mar 19 '17 at 03:31

Viewed 59 times

I need to print a range of unicode characters given in hex.

range = "<range first-cp="0030" last-cp="0039"/>"

I need to write a code in python, that takes this string range as input and prints all the code points in the range 0030 to 0039. I have come to know that I will need to convert each code point to an int first then increment it and convert it back to a hex string. For example, if

first_cp = "0030"
second_cp = int(first_cp, 16) + 1  # converts it to an integer and increments it
second_cp = hex(second_cp)  # returns '0x31'

Here, I don't know how can I convert the integer back to a hex string such that it has the format 0031 instead of 0x31. Please note that I do not want to remove 0x from the string only.

edited Mar 19 '17 at 03:31

asked Mar 19 '17 at 03:12

user7623610

[`format(48, "04x")`](https://docs.python.org/3/library/string.html#format-specification-mini-language) – jonrsharpe Mar 19 '17 at 03:17
So basically you want a zero-padded hex string? – Sumner Evans Mar 19 '17 at 03:17
seems like the first_cp has what you need but you're overwriting it. – MalcolmInTheCenter Mar 19 '17 at 03:21
I can have a code point `0673`. Using the method I described in the question, I get back `0x673`. In this case, I do not want `x` in the string. – user7623610 Mar 19 '17 at 03:21
Yes, I have edited my question to make clear that I am having difficulty in converting my integer string back to a hex string. – user7623610 Mar 19 '17 at 03:29

Covert int to hex string with a specific format in python

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