'python image (.jpeg) to hex code
I operate with a thermal printer, this printer is able to print images, but it needs to get the data in hex format. For this I would need to have a python function to read an image and return a value containing the image data in hex format. I currently use this format to sent hex format to the printer:
content = b"\x1B\x4E"
Which is the simplest way to do so using Python2.7? All the best;
Solution 1:[1]
I don't really know what you mean by "hex format", but if it needs to get the whole file as a sequence of bytes you can do:
with open("image.jpeg", "rb") as fp:
img = fp.read()
If your printer expects the image in some other format (like 8bit values for every pixel) then try using the pillow library, it has many image manipulation functions and handles a wide range of input and ouput formats.
Solution 2:[2]
How about this:
with open('something.jpeg', 'rb') as f:
binValue = f.read(1)
while len(binValue) != 0:
hexVal = hex(ord(binValue))
# Do something with the hex value
binValue = f.read(1)
Or for a function, something like this:
import re
def imgToHex(file):
string = ''
with open(file, 'rb') as f:
binValue = f.read(1)
while len(binValue) != 0:
hexVal = hex(ord(binValue))
string += '\\' + hexVal
binValue = f.read(1)
string = re.sub('0x', 'x', string) # Replace '0x' with 'x' for your needs
return string
Note: You do not necessarily need to do the re.sub portion if you use struct.pack to write the bits, but this will get it into the format that you need
Solution 3:[3]
Read in a jpg and make a string of hex values. Then reverse the procedure. Take a string of hex and write it out as a jpg file...
import binascii
with open('my_img.jpg', 'rb') as f:
data = f.read()
print(data[:10])
im_hex = binascii.hexlify(data)
# check out the hex...
print(im_hex[:10])
# reversing the procedure
im_hex = binascii.a2b_hex(im_hex)
print(im_hex[:10])
# write it back out to a jpg file
with open('my_hex.jpg', 'wb') as image_file:
image_file.write(im_hex)
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | anonymous_user_13 |
| Solution 2 | |
| Solution 3 | Harley |
