I'm new to python and was playing around with how to change my Instagram profile picture. The part I just can't get past is how I can put my image into the program. This is my code:
from instagram_private_api import Client, ClientCompatPatch
user_name = 'my_username'
password = 'my_password'
api = Client(user_name, password)
api.change_profile_picture('image.png')
Now, from what I read on the API Documentation, I can't just put in an image. It needs to be byte data. On the API documentation, the parameter is described like this:
photo_data – byte string of image
I converted the image on an encoding website and now I have the file image.txt with the byte data of the image. So I changed the last line to this:
api.change_profile_picture('image.txt')
But this still doesn't work. The program doesn't read it as byte data. I get the following error:
Exception has occurred: TypeError
a bytes-like object is required, not 'str'
What is the right way to put in the picture?
The error is telling you that "input.txt" (or "image.png") is a string, and it's always going to say that as long as you pass in a filename because filenames are always strings. Doesn't matter what's in the file, because the API doesn't read the file.
It doesn't want the filename of the image, it wants the actual image data that's in that file. That's why the parameter is named photo_data and not photo_filename. So read it (in binary mode, so you get bytes rather than text) and pass that instead.
with open("image.png", "rb") as imgfile:
api.change_profile_picture(imgfile.read())
The with statement ensures that the file is closed after you're done with it.
if you have .png or .jpeg or ... then use this.
with open("image.png", "rb") as f:
api.change_profile_picture(f.read())
and if you have a .txt file then use this.
with open("image.txt", "rb") as f:
api.change_profile_picture(f.read())
Related
i have a small problem and it caused me a lot of trubble. basicly i want to convert an immage to bytes than store string wersion of those bytes in an txt file and than read file contents and transform it into bytes and than into image. i've goten first part of this kinda ready (it works but it's made quickly and badly) but the conversion from string to byte gives me problem.
when i read image bytes it's something like this: b'GIF89aP\x00P\x00\xe3'
but when i read it from txt by 'rb' or just transform str to byte it gives me this: b'GIF89aP\\x00P\\x00\\xe3'
and with this i can't write it to an immage.
so i've tried to read and learn anything about this but i couldn't find anything that would help.
the code is here and i know it's really messy but i just need it to work
file = open('p.gif', 'rb')
image = file.read()
str_b = str(image)
leng = len(str_b)
print(leng)
str_b = str_b[:0] + str_b[0+2:]
leng =- 1
str_b = str_b[:leng]
print(image)
#a = open('bytearray', 'w+')
#a.write(str_b)
#a.close
a = open('bytearray', 'r')
a = a.read()
temp = a.encode('utf-8')
print(temp)
#b = open('check', 'w+')
#b.write(str(string))
#print(string)
image_result = open('decoded.jpg', 'wb') # create a writable image and write the decoding result
image_result.write(temp)
basicly my goal right now is to get bytes that look like this: b'GIF89aP\x00P\x00\xe3'
Please do not use eval like suggested above, eval has serious security vulnerabilities and will execute any python code you pass within it. You could accidentally read a text file that has code to reformat the disk and it will just execute, this is just an example but you get my point its bad practice and just results in more problems see https://nedbatchelder.com/blog/201206/eval_really_is_dangerous.html if you want some examples on why eval is bad
anyways lets try to fix your code
instead of converting your byte array to string by wrapping it in the str() method I would suggest you use .decode() and .encode()
Fixed Code:
with open('p.gif', 'rb') as file:
image = file.read() # read file bytes
str_image = image.decode("utf-8") #using decode we changed the bytes to a string
with open('image.txt', 'w') as file:
file.write(str_image) # write image as string to a text file
with open('image.txt', 'r') as file
str_from_file = file.read() # read the text file and store the string
file_bytes = str_from_file.encode("utf-8") # encode the image str back to bytes
print(type(str_from_file)) #type is str
print(type(file_bytes)) # types is bytes
I hope this fixes your issue and also doesn't include vulnerabilties in what your building
for example, this is my code:
#extract the object from "lastringa.pickle" and save it
extracted = ""
with open("lastringa.pickle","rb") as f:
extracted = pickle.load(f)
Where "lasting.pickle" contains a string object with some text.
So if I type extracted. before the opening of the file, I'm able to get the code suggestion as shown in the picture:
But then, after this operation extracted = pickle.load(f), if I type extracted. I don't get code suggestion anymore.
Can somebody explain me why is that and how to solve this?
Pickle reads and writes objects as binary files. You can confirm this by the open('lastringa.pickle', 'rb'), command where you are using the rb option, i.e. read binary.
Your IDE doesn't know the type of the object that the pickle is expected to read, so that it can suggest the string methods (e.g. .split(), .read())
On the other hand, in the first photo, your IDE knows that expected is a string and it knows what to suggest.
I am trying to stream a csv to azure blob storage, the csv is generated directly from python scripts without local copy, i have the following code, df is the csv file:
with open(df,'w') as f:
stream = io.BytesIO(f)
stream.seek(0)
block_blob_service.create_blob_from_stream('flowshop', 'testdata123', stream)
then i got the error massage:
stream = io.BytesIO(f) TypeError: a bytes-like object is required, not '_io.TextIOWrapper'
i think the problem has been the format incorrect, can you please identify the problem. thanks.
You opened df for write, then tried to pass the resulting file object as the initializer of io.BytesIO (which is supposed to to take actual binary data, e.g. b'1234'). That's the cause of the TypeError; open files (read or write, text or binary) are not bytes or anything similar (bytearray, array.array('B'), mmap.mmap, etc.), so passing them to io.BytesIO makes no sense.
It looks like your goal is to read from df, and you shouldn't need io.BytesIO at all for that. Just change the mode from (text) write, 'w', to binary read, 'rb'. Then pass the resulting file object to your API directly:
with open(df, 'rb') as f:
block_blob_service.create_blob_from_stream('flowshop', 'testdata123', f)
Update: Apparently df was your actual data, not a file name to open at all. Given that, you should really skip the stream API (which is pointless if the data is already in memory) and just use the bytes based API directly:
block_blob_service.create_blob_from_bytes('flowshop', 'testdata123', df)
or if df is str, not bytes:
block_blob_service.create_blob_from_bytes('flowshop', 'testdata123', df.encode('utf-8'))
I feel like this is a simple question but nonetheless I cannot find a straightforward answer.
I have an email (an .eml file) that I need to parse. This email has a data table in the body that I need to export to my database. I have been successful parsing data out of txt file emails and attached PDF files, so I understand concepts like mapping to where the data is stored as well as RegularExpressions, but these eml files I can't seem to figure out.
In my code below I have three blocks of code essentially trying to do the same thing (two of them are comments now). I am simply attempting to capture any, or all, of the data in the email. Each block of code produces the same error though:
TypeError: initial_value must be str or None, not _io.TextIOWrapper
I have read that this error is most likely due to Python expecting to receive a string but receives bytes instead, or vice versa. So I followed up those attempts by trying to implement io.StringIO or io.BytesIO but neither worked. I would like to be able to recognize and parse specific data out of the email.
Thank you for any help, as well as question asking criticism.
My code:
import email
#import io
import os
import re
path = 'Z:\\folderwithemlfile'
for filename in os.listdir(path):
file_path = os.path.join(path, filename)
if os.path.isfile(file_path):
with open(file_path, 'r', encoding="utf-8") as f:
b = email.message_from_string(f)
if b.is_multipart():
for paylod in b.get_payload():
print(payload.get_payload())
else:
print(b.get_payload())
#b = email.message_from_string(f)
#bbb = b['from']
#ccc = b['to']
#print(f)
#msg = email.message_from_string(f)
#msg['from']
#msg['to']
Picture of email:
I have a database with some of the data is binary (blob datatype in MySQL), which was actually webpages scrapped and gzipped. Now I want to extract them and write each record into a gzip file, which I'd assume to be doable - after all they are gzipped-data right?
The question is, however, how would I do that? By searching I could find a million of examples on how to write gzip file from original data, not gzipped one. Writing the gzipped string directly into a file doesn't result in a gzip file, not to mention I got a load of "ordinal not in range" exceptions.
Could you guys help? Thanks in advance. I'm a newbie to Python...
Edit: Here is the method I used:
def store_cache(self, content, news_id):
if not content:
return
# some of the records may contain normal data (not gzipp-ed), hence this try block
try:
content = self.gunzip(content)
except:
return
import gzip
with gzip.open('static/cache/%s' % (self.base36encode(news_id), ), 'wb') as f:
f.write(content)
f.close()
This causes an exception:
<type 'exceptions.UnicodeEncodeError'> at /migrate
'ascii' codec can't encode character u'\u1edb' in position 186: ordinal not in range(128)
And this is the innermost traceback:
E:\Python27\lib\gzip.py in write
self.crc = zlib.crc32(data, self.crc) & 0xffffffffL
You said it yourself: extract them and then write them into a gzip file. There is nothing special about writing "from gzipped data": you un-gzip the data to get the original data, and then write the original data as if it were original data (because it is). The documentation shows you how to do these things.
However, gzip is just a compression format, not an archive format. It is not built to handle multiple files, so you must use something else to create a single file from the multiple inputs. Typically this is done by making a tar archive which is then gzipped. You can do this in Python using the tarfile module. Since your data will come from gzip-decompression streams, you will want to use the TarFile.addfile(tarinfo, fileobj) method to add them to the archive. You should be able to use the gzip.GzipFile instance as the fileobj to add this way.