How can I convert a string x='0x67c31080115dDfeBa0474B3893b2caB1d567438f' into hex to be x=0x67c31080115dDfeBa0474B3893b2caB1d567438f ?
I want to use the same value of the hex but not in a string format
In other word, how can I treat x as a hex value not a string type ( without ' ' )
Thanks
There is no such thing as "hex value". 0x67c31080115dDfeBa0474B3893b2caB1d567438f belongs to type int. You can convert your string to int just with:
x = int(x, 16)
If, however, you want to print it as hex, either use:
print(hex(x))
or
print("{:x}".format(x))
(the first one adds "0x" to the beginning of the string, the other one does not).
You could do it using ast.literal_eval:
import ast
x='0x67c31080115dDfeBa0474B3893b2caB1d567438f'
hex_val = hex(ast.literal_eval(x))
print(hex_val)
Related
I want to convert a hex string to utf-8
a = '0xb3d9'
to
동 (http://www.unicodemap.org/details/0xB3D9/index.html)
First, obtain the integer value from the string of a, noting that a is expressed in hexadecimal:
a_int = int(a, 16)
Next, convert this int to a character. In python 2 you need to use the unichr method to do this, because the chr method can only deal with ASCII characters:
a_chr = unichr(a_int)
Whereas in python 3 you can just use the chr method for any character:
a_chr = chr(a_int)
So, in python 3, the full command is:
a_chr = chr(int(a, 16))
i'm trying to convert ASCII values in string to normal characters, but whatever i'll try to do, i'll get output like
this\032is\032my\032\output\000
and I need
this is my output
Thanks for your help!
I solved it with
import re
def replace(match):
return chr(int(match.group(1)))
aux = str(element.text)
regex = re.compile(r"\\(\d{1,3})")
new = regex.sub(replace, aux)
where element.text is string i need to convert
\032 is octal for the decimal number 26, so it is not in fact a space character, if you use a hex value with the correct number, you should get what you want. 0x20 == 32 below
>>> s = 'this\x20is\x20my\x20\output\00'
>>> print(s)
this is my \output
My code looks like this :
import struct
str = "AAAAAAA"
len = len(str)+32
package = struct.pack("!H",len)
print repr(package)
the result is :
"\x00'"
When I use len = len(str)
the result is \x00\x07
Why when len is larger than 32,it is not working?
You're misunderstanding the "\x00'" result. It's a mixture of a string hexadecimal character code value and a regular printable ASCII character. If it were displayed purely in hexadecimal character codes, it would be "\x00x\x27".
The \x27 in decimal is the integer 39, which is the result of len(str)+32. It's also the character code of the ' (single quote) character, which is part of what repr() is displaying.
This question already has answers here:
What's the correct way to convert bytes to a hex string in Python 3?
(9 answers)
Closed 7 years ago.
I have an ANSI string Ď–ór˙rXüď\ő‡íQl7 and I need to convert it to hexadecimal like this:
06cf96f30a7258fcef5cf587ed51156c37 (converted with XVI32).
The problem is that Python cannot encode all characters correctly (some of them are incorrectly displayed even here, on Stack Overflow) so I have to deal with them with a byte string.
So the above string is in bytes this: b'\x06\xcf\x96\xf3\nr\x83\xffrX\xfc\xef\\\xf5\x87\xedQ\x15l7'
And that's what I need to convert to hexadecimal.
So far I tried binascii with no success, I've tried this:
h = ""
for i in b'\x06\xcf\x96\xf3\nr\x83\xffrX\xfc\xef\\\xf5\x87\xedQ\x15l7':
h += hex(i)
print(h)
It prints:
0x60xcf0x960xf30xa0x720x830xff0x720x580xfc0xef0x5c0xf50x870xed0x510x150x6c0x37
Okay. It looks like I'm getting somewhere... but what's up with the 0x thing?
When I remove 0x from the string like this:
h.replace("0x", "")
I get 6cf96f3a7283ff7258fcef5cf587ed51156c37 which looks like it's correct.
But sometimes the byte string has a 0 next to a x and it gets removed from the string resulting in a incorrect hexadecimal string. (the string above is missing the 0 at the beginning).
Any ideas?
If you're running python 3.5+, bytes type has an new bytes.hex() method that returns string representation.
>>> h = b'\x06\xcf\x96\xf3\nr\x83\xffrX\xfc\xef\\\xf5\x87\xedQ\x15l7'
b'\x06\xcf\x96\xf3\nr\x83\xffrX\xfc\xef\\\xf5\x87\xedQ\x15l7'
>>> h.hex()
'06cf96f30a7283ff7258fcef5cf587ed51156c37'
Otherwise you can use binascii.hexlify() to do the same thing
>>> import binascii
>>> binascii.hexlify(h).decode('utf8')
'06cf96f30a7283ff7258fcef5cf587ed51156c37'
As per the documentation, hex() converts “an integer number to a lowercase hexadecimal string prefixed with ‘0x’.” So when using hex() you always get a 0x prefix. You will always have to remove that if you want to concatenate multiple hex representations.
But sometimes the byte string has a 0 next to a x and it gets removed from the string resulting in a incorrect hexadecimal string. (the string above is missing the 0 at the beginning).
That does not make any sense. x is not a valid hexadecimal character, so in your solution it can only be generated by the hex() call. And that, as said above, will always create a 0x. So the sequence 0x can never appear in a different way in your resulting string, so replacing 0x by nothing should work just fine.
The actual problem in your solution is that hex() does not enforce a two-digit result, as simply shown by this example:
>>> hex(10)
'0xa'
>>> hex(2)
'0x2'
So in your case, since the string starts with b\x06 which represents the number 6, hex(6) only returns 0x6, so you only get a single digit here which is the real cause of your problem.
What you can do is use format strings to perform the conversion to hexadecimal. That way you can both leave out the prefix and enforce a length of two digits. You can then use str.join to combine it all into a single hexadecimal string:
>>> value = b'\x06\xcf\x96\xf3\nr\x83\xffrX\xfc\xef\\\xf5\x87\xedQ\x15l7'
>>> ''.join(['{:02x}'.format(x) for x in value])
'06cf96f30a7283ff7258fcef5cf587ed51156c37'
This solution does not only work with a bytes string but with really anything that can be formatted as a hexadecimal string (e.g. an integer list):
>>> value = [1, 2, 3, 4]
>>> ''.join(['{:02x}'.format(x) for x in value])
'01020304'
By hex-string, it is a regular string except every two characters represents some byte, which is mapped to some ASCII char.
So for example the string
abc
Would be represented as
979899
I am looking at the binascii module but don't really know how to take the hex-string and turn it back into the ascii string.
Which method can I use?
Note: I am starting with 979899 and want to convert it back to abc
You can use ord() to get the integer value of each character:
>>> map(ord, 'abc')
[97, 98, 99]
>>> ''.join(map(lambda c: str(ord(c)), 'asd'))
'979899'
>>> ''.join((str(ord(c)) for c in 'abc'))
'979899'
You don't need binascii to get the integer representation of a character in a string, all you need is the built in function ord().
s = 'abc'
print(''.join(map(lambda x:str(ord(x)),s))) # outputs "979899"
To get the string back from the hexadecimal number you can use
s=str(616263)
print "".join([chr(int(s[x:x+2], 16)) for x in range(0,len(s),2)])
See http://ideone.com/dupgs