I am trying to add more than one picture on Shopify with the Python API however, I am not able to upload 2 pictures to one product. At this time only one picture is being uploaded. How I can add more than 1 picture to Shopify API?
import shopify
API_KEY = 'dsfsdsdsdsdsad'
PASSWORD = 'sadsdasdasdas'
shop_url = "https://%s:%s#teststore.myshopify.com/admin" % (API_KEY, PASSWORD)
shopify.ShopifyResource.set_site(shop_url)
path = "audi.jpg"
path2 = "audi2.jpg"
new_product = shopify.Product()
new_product.title = "Audi pictures test "
new_product.body_html = "body of the page <br/><br/> test <br/> test"
variant = shopify.Variant({'price': 1.00, 'requires_shipping': False,'sku':'000007'})
new_product.variants = [variant]
image = shopify.Image()
image2 = shopify.Image()
with open(path, "rb") as f:
filename = path.split("/")[-1:][0]
filename2 = path2.split("/")[-1:][0]
encoded = f.read()
image.attach_image(encoded, filename=filename)
image2.attach_image(encoded, filename=filename2)
new_product.images = [image,image2]
new_product.save()
I don't have shopify account to verify the below code, but I have looked at the source code and below is what should work for you
import shopify
API_KEY = 'dsfsdsdsdsdsad'
PASSWORD = 'sadsdasdasdas'
shop_url = "https://%s:%s#teststore.myshopify.com/admin" % (API_KEY, PASSWORD)
shopify.ShopifyResource.set_site(shop_url)
path = "audi.jpg"
path2 = "audi2.jpg"
new_product = shopify.Product()
new_product.title = "Audi pictures test "
new_product.body_html = "body of the page <br/><br/> test <br/> test"
variant = shopify.Variant({'price': 1.00, 'requires_shipping': False,'sku':'000007'})
new_product.variants = [variant]
success = new_product.save()
if success:
product_id = new_product.id
image = shopify.Image({"product_id": product_id})
image2 = shopify.Image({"product_id": product_id})
filename = path.split("/")[-1:][0]
filename2 = path2.split("/")[-1:][0]
encoded = f.read()
image.attach_image(encoded, filename=filename)
image2.attach_image(encoded, filename=filename2)
image.save()
image2.save()
The idea is create the product and then attach the image
I do not known how to load a eml file in python 3.4.
I want to list all and read all of them in python.
This is how you get content of an e-mail i.e. *.eml file.
This works perfectly on Python2.5 - 2.7. Try it on 3. It should work as well.
from email import message_from_file
import os
# Path to directory where attachments will be stored:
path = "./msgfiles"
# To have attachments extracted into memory, change behaviour of 2 following functions:
def file_exists (f):
"""Checks whether extracted file was extracted before."""
return os.path.exists(os.path.join(path, f))
def save_file (fn, cont):
"""Saves cont to a file fn"""
file = open(os.path.join(path, fn), "wb")
file.write(cont)
file.close()
def construct_name (id, fn):
"""Constructs a file name out of messages ID and packed file name"""
id = id.split(".")
id = id[0]+id[1]
return id+"."+fn
def disqo (s):
"""Removes double or single quotations."""
s = s.strip()
if s.startswith("'") and s.endswith("'"): return s[1:-1]
if s.startswith('"') and s.endswith('"'): return s[1:-1]
return s
def disgra (s):
"""Removes < and > from HTML-like tag or e-mail address or e-mail ID."""
s = s.strip()
if s.startswith("<") and s.endswith(">"): return s[1:-1]
return s
def pullout (m, key):
"""Extracts content from an e-mail message.
This works for multipart and nested multipart messages too.
m -- email.Message() or mailbox.Message()
key -- Initial message ID (some string)
Returns tuple(Text, Html, Files, Parts)
Text -- All text from all parts.
Html -- All HTMLs from all parts
Files -- Dictionary mapping extracted file to message ID it belongs to.
Parts -- Number of parts in original message.
"""
Html = ""
Text = ""
Files = {}
Parts = 0
if not m.is_multipart():
if m.get_filename(): # It's an attachment
fn = m.get_filename()
cfn = construct_name(key, fn)
Files[fn] = (cfn, None)
if file_exists(cfn): return Text, Html, Files, 1
save_file(cfn, m.get_payload(decode=True))
return Text, Html, Files, 1
# Not an attachment!
# See where this belongs. Text, Html or some other data:
cp = m.get_content_type()
if cp=="text/plain": Text += m.get_payload(decode=True)
elif cp=="text/html": Html += m.get_payload(decode=True)
else:
# Something else!
# Extract a message ID and a file name if there is one:
# This is some packed file and name is contained in content-type header
# instead of content-disposition header explicitly
cp = m.get("content-type")
try: id = disgra(m.get("content-id"))
except: id = None
# Find file name:
o = cp.find("name=")
if o==-1: return Text, Html, Files, 1
ox = cp.find(";", o)
if ox==-1: ox = None
o += 5; fn = cp[o:ox]
fn = disqo(fn)
cfn = construct_name(key, fn)
Files[fn] = (cfn, id)
if file_exists(cfn): return Text, Html, Files, 1
save_file(cfn, m.get_payload(decode=True))
return Text, Html, Files, 1
# This IS a multipart message.
# So, we iterate over it and call pullout() recursively for each part.
y = 0
while 1:
# If we cannot get the payload, it means we hit the end:
try:
pl = m.get_payload(y)
except: break
# pl is a new Message object which goes back to pullout
t, h, f, p = pullout(pl, key)
Text += t; Html += h; Files.update(f); Parts += p
y += 1
return Text, Html, Files, Parts
def extract (msgfile, key):
"""Extracts all data from e-mail, including From, To, etc., and returns it as a dictionary.
msgfile -- A file-like readable object
key -- Some ID string for that particular Message. Can be a file name or anything.
Returns dict()
Keys: from, to, subject, date, text, html, parts[, files]
Key files will be present only when message contained binary files.
For more see __doc__ for pullout() and caption() functions.
"""
m = message_from_file(msgfile)
From, To, Subject, Date = caption(m)
Text, Html, Files, Parts = pullout(m, key)
Text = Text.strip(); Html = Html.strip()
msg = {"subject": Subject, "from": From, "to": To, "date": Date,
"text": Text, "html": Html, "parts": Parts}
if Files: msg["files"] = Files
return msg
def caption (origin):
"""Extracts: To, From, Subject and Date from email.Message() or mailbox.Message()
origin -- Message() object
Returns tuple(From, To, Subject, Date)
If message doesn't contain one/more of them, the empty strings will be returned.
"""
Date = ""
if origin.has_key("date"): Date = origin["date"].strip()
From = ""
if origin.has_key("from"): From = origin["from"].strip()
To = ""
if origin.has_key("to"): To = origin["to"].strip()
Subject = ""
if origin.has_key("subject"): Subject = origin["subject"].strip()
return From, To, Subject, Date
# Usage:
f = open("message.eml", "rb")
print extract(f, f.name)
f.close()
I programmed this for my mailgroup using mailbox, that is why it is so convoluted.
It never failed me. Never any junk. If message is multipart, output dictionary will contain a
key "files" (a sub dict) with all filenames of extracted other files that were not text or html.
That was a way of extracting attachments and other binary data.
You may change it in pullout(), or just change the behaviour of file_exists() and save_file().
construct_name() constructs a filename out of message id and multipart message
filename, if there is one.
In pullout() the Text and Html variables are strings. For online mailgroup it was OK to get any text or HTML packed into multipart that wasn't an attachment at once.
If you need something more sophisticated change Text and Html to lists and append to them and add them as needed.
Nothing problematic.
Maybe there are some errors here, because it is intended to work with mailbox.Message(),
not with email.Message(). I tried it on email.Message() and it worked fine.
You said, you "wish to list them all". From where? If you refer to the POP3 mailbox or a mailbox of some nice open-source mailer, then you do it using mailbox module.
If you want to list them from others, then you have a problem.
For example, to get mails from MS Outlook, you have to know how to read OLE2 compound files.
Other mailers rarely refer to them as *.eml files, so I think this is exactly what you would like to do.
Then search on PyPI for olefile or compoundfiles module and Google around for how to extract an e-mail from MS Outlook inbox file.
Or save yourself a mess and just export them from there to some directory. When you have them as eml files, then apply this code.
I found this code much simpler
import email
import os
path = './'
listing = os.listdir(path)
for fle in listing:
if str.lower(fle[-3:])=="eml":
msg = email.message_from_file(open(fle))
attachments=msg.get_payload()
for attachment in attachments:
try:
fnam=attachment.get_filename()
f=open(fnam, 'wb').write(attachment.get_payload(decode=True,))
f.close()
except Exception as detail:
#print detail
pass
Posting this here for anyone looking to just extract text from an email and get a list of .eml files - took me forever to find a good answer to this online. NOTE: This will not get attachments to emails, just the text from email.
import email
from email import policy
from email.parser import BytesParser
import glob
import os
path = '/path/to/data/' # set this to "./" if in current directory
eml_files = glob.glob(path + '*.eml') # get all .eml files in a list
for eml_file in eml_files:
with open(eml_file, 'rb') as fp: # select a specific email file from the list
name = fp.name # Get file name
msg = BytesParser(policy=policy.default).parse(fp)
text = msg.get_body(preferencelist=('plain')).get_content()
fp.close()
text = text.split("\n")
print (name) # Get name of eml file
print (text) # Get list of all text in email
Credit to some of the code from this post: Reading .eml files with Python 3.6 using emaildata 0.3.4
Python 3 version of Dalen's answer. Basically syntax issue fixes. (Can't comment due to lack of reputation, also clearer as an answer).
# To have attachments extracted into memory, change behaviour of 2 following functions:
def file_exists (f):
"""Checks whether extracted file was extracted before."""
return os.path.exists(os.path.join(path, f))
def save_file (fn, cont):
"""Saves cont to a file fn"""
file = open(os.path.join(path, fn), "wb")
file.write(cont)
file.close()
def construct_name (id, fn):
"""Constructs a file name out of messages ID and packed file name"""
id = id.split(".")
id = id[0]+id[1]
return id+"."+fn
def disqo (s):
"""Removes double or single quotations."""
s = s.strip()
if s.startswith("'") and s.endswith("'"): return s[1:-1]
if s.startswith('"') and s.endswith('"'): return s[1:-1]
return s
def disgra (s):
"""Removes < and > from HTML-like tag or e-mail address or e-mail ID."""
s = s.strip()
if s.startswith("<") and s.endswith(">"): return s[1:-1]
return s
def pullout (m, key):
"""Extracts content from an e-mail message.
This works for multipart and nested multipart messages too.
m -- email.Message() or mailbox.Message()
key -- Initial message ID (some string)
Returns tuple(Text, Html, Files, Parts)
Text -- All text from all parts.
Html -- All HTMLs from all parts
Files -- Dictionary mapping extracted file to message ID it belongs to.
Parts -- Number of parts in original message.
"""
Html = ""
Text = ""
Files = {}
Parts = 0
if not m.is_multipart():
if m.get_filename(): # It's an attachment
fn = m.get_filename()
cfn = construct_name(key, fn)
Files[fn] = (cfn, None)
if file_exists(cfn): return Text, Html, Files, 1
save_file(cfn, m.get_payload(decode=True))
return Text, Html, Files, 1
# Not an attachment!
# See where this belongs. Text, Html or some other data:
cp = m.get_content_type()
if cp=="text/plain":
Text += str(m.get_payload(decode=True))
elif cp=="text/html":
Html += str(m.get_payload(decode=True))
else:
# Something else!
# Extract a message ID and a file name if there is one:
# This is some packed file and name is contained in content-type header
# instead of content-disposition header explicitly
cp = m.get("content-type")
try: id = disgra(m.get("content-id"))
except: id = None
# Find file name:
o = cp.find("name=")
if o==-1: return Text, Html, Files, 1
ox = cp.find(";", o)
if ox==-1: ox = None
o += 5; fn = cp[o:ox]
fn = disqo(fn)
cfn = construct_name(key, fn)
Files[fn] = (cfn, id)
if file_exists(cfn): return Text, Html, Files, 1
save_file(cfn, m.get_payload(decode=True))
return Text, Html, Files, 1
# This IS a multipart message.
# So, we iterate over it and call pullout() recursively for each part.
y = 0
while 1:
# If we cannot get the payload, it means we hit the end:
try:
pl = m.get_payload(y)
except: break
# pl is a new Message object which goes back to pullout
t, h, f, p = pullout(pl, key)
Text += t; Html += h; Files.update(f); Parts += p
y += 1
return Text, Html, Files, Parts
def extract (msgfile, key):
"""Extracts all data from e-mail, including From, To, etc., and returns it as a dictionary.
msgfile -- A file-like readable object
key -- Some ID string for that particular Message. Can be a file name or anything.
Returns dict()
Keys: from, to, subject, date, text, html, parts[, files]
Key files will be present only when message contained binary files.
For more see __doc__ for pullout() and caption() functions.
"""
m = email.message_from_file(msgfile)
From, To, Subject, Date = caption(m)
Text, Html, Files, Parts = pullout(m, key)
Text = Text.strip(); Html = Html.strip()
msg = {"subject": Subject, "from": From, "to": To, "date": Date,
"text": Text, "html": Html, "parts": Parts}
if Files: msg["files"] = Files
return msg
def caption (origin):
"""Extracts: To, From, Subject and Date from email.Message() or mailbox.Message()
origin -- Message() object
Returns tuple(From, To, Subject, Date)
If message doesn't contain one/more of them, the empty strings will be returned.
"""
Date = ""
if origin.__contains__("date"): Date = origin["date"].strip()
From = ""
if origin.__contains__("from"): From = origin["from"].strip()
To = ""
if origin.__contains__("to"): To = origin["to"].strip()
Subject = ""
if origin.__contains__("subject"): Subject = origin["subject"].strip()
return From, To, Subject, Date
Try this:
#!python3
# -*- coding: utf-8 -*-
import email
import os
SOURCE_DIR = 'email'
DEST_DIR = 'temp'
def extractattachements(fle,suffix=None):
message = email.message_from_file(open(fle))
filenames = []
if message.get_content_maintype() == 'multipart':
for part in message.walk():
if part.get_content_maintype() == 'multipart': continue
#if part.get('Content-Disposition') is None: continue
if part.get('Content-Type').find('application/octet-stream') == -1: continue
filename = part.get_filename()
if suffix:
filename = ''.join( [filename.split('.')[0], '_', suffix, '.', filename.split('.')[1]])
filename = os.path.join(DEST_DIR, filename)
fb = open(filename,'wb')
fb.write(part.get_payload(decode=True))
fb.close()
filenames.append(filename)
return filenames
def main():
onlyfiles = [f for f in os.listdir(SOURCE_DIR) if os.path.isfile(os.path.join(SOURCE_DIR, f))]
for file in onlyfiles:
#print path.join(SOURCE_DIR,file)
extractattachements(os.path.join(SOURCE_DIR,file))
return True
if __name__ == "__main__":
main()
Here I am simplifying things for you so that you can get a more clear data to process on .....
.eml will consist of 2 parts on broad level 1) Headers 2)Content/Body
(Note it will discard any attachements if they are there)
Moreover I've removed https links also from .eml file but I'll tell you what to do if you want them .
1) Header :
So I used eml-parser to get Header information you can install it using :
pip install eml-parser
View their documentation to get more info about how to get headers :
https://pypi.org/project/eml-parser/
2)Content/Body : Now here I modified some older scripts to get best result in output
from email import policy
from email.parser import BytesParser
import glob
import os
path = './' # set this to "./" if in current directory
eml_files = glob.glob(path + '*.eml') # get all .eml files in a list
for eml_file in eml_files:
with open(eml_file, 'rb') as fp: # select a specific email file from the list
name = fp.name # Get file name
msg = BytesParser(policy=policy.default).parse(fp)
text = msg.get_body(preferencelist=('plain')).get_content()
fp.close()
print (name) # Get name of eml file
# print (text) # Get list of all text in email
This is a part of code which was already available on many places and of which I don't take credit of......
Now I've added few conditions to print out the body in more pretty way these lines of code are mine and you can give me credit for that :
newText = ""
flag = 0
urlFlag = 0
for i in range(len(text)):
if(flag==1):
flag = 0
continue
if(text[i]=="\\"):
flag = 1
continue
if(text[i]=='<'): //to remove hyperlinks
urlFlag = 1
continue
if(text[i]=='>'): //to remove hyperlinks
urlFlag = 0
continue
if(urlFlag==0): //to remove hyperlinks
newText = newText+text[i]
print(newText)
Now this will remove all the break-lines , tab space and other stuff (\t,\r,\n)
Moreover if you want to have links (http,https links present in your .eml file) then just remove 3 conditions and new code will look like :
newText = ""
flag = 0
urlFlag = 0
for i in range(len(text)):
if(flag==1):
flag = 0
continue
if(text[i]=="\\"):
flag = 1
continue
newText = newText+text[i]
print(newText)
Final Code (with removing links) :
from email import policy
from email.parser import BytesParser
import glob
import os
path = './' # set this to "./" if in current directory
eml_files = glob.glob(path + '*.eml') # get all .eml files in a list
for eml_file in eml_files:
with open(eml_file, 'rb') as fp: # select a specific email file from the list
name = fp.name # Get file name
msg = BytesParser(policy=policy.default).parse(fp)
text = msg.get_body(preferencelist=('plain')).get_content()
fp.close()
print (name) # Get name of eml file
# print (text) # Get list of all text in email
newText = ""
flag = 0
urlFlag = 0
for i in range(len(text)):
if(flag==1):
flag = 0
continue
if(text[i]=="\\"):
flag = 1
continue
if(text[i]=='<'):
urlFlag = 1
continue
if(text[i]=='>'):
urlFlag = 0
continue
if(urlFlag==0):
newText = newText+text[i]
print(newText)
This is my 1st answer on StackOverflow hope this will help you guys !
My Python version is : 3.8.10