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Unable to append data into Azure storage using Python

I am using the below code to append data to Azure blob using python.
from azure.storage.blob import AppendBlobService
append_blob_service = AppendBlobService(account_name='myaccount', account_key='mykey')
# The same containers can hold all types of blobs
append_blob_service.create_container('mycontainer')
# Append blobs must be created before they are appended to
append_blob_service.create_blob('mycontainer', 'myappendblob')
append_blob_service.append_blob_from_text('mycontainer', 'myappendblob', u'Hello, world!')
append_blob = append_blob_service.get_blob_to_text('mycontainer', 'myappendblob')
The above code works fine, but when I tried to insert new data, the old data gets overwritten.
Is there any way I can append data to 'myappendblob'

Considering you are calling the same code to append the data, the issue is with the following line of code:
append_blob_service.create_blob('mycontainer', 'myappendblob')
If you read the documentation for create_blob method, you will notice the following:
Creates a blob or overrides an existing blob. Use if_none_match=* to
prevent overriding an existing blob.
So essentially you are overriding the blob every time you call your code.
You should call this method with if_none_match="*" as the documentation suggests. If the blob exists, your code will throw an exception which you will need to handle.

Try this code which is taken from the Document and it is given by #Harsh Jain ,
from azure.storage.blob import AppendBlobService
def append_data_to_blob(data):
service = AppendBlobService(account_name="<Storage acc name>",
account_key="<Storage acc key>")
try:
service.append_blob_from_text(container_name="<name of Conatiner >", blob_name="<The name of file>", text = data)
except:
service.create_blob(container_name="<name of Conatiner >", blob_name="<the name of file>")
service.append_blob_from_text(container_name="<name of Conatiner>", blob_name="<the name of file>", text = data)
print('Data got Appended ')
append_data_to_blob('Hi blob')
Taken References from:
https://www.educative.io/answers/how-to-append-data-in-blob-storage-in-azure-using-python

Upload file to mediafire without api python script

I want to upload file to mediafire without API Delevoper(because when users use their accounts)
But I see mediafire api used Developer so i don't want to use it
I want to somethings like it:
import mediafire_uploader as mdf
mdf.create_login("USER_NAME","PASSWORD")
mdf.upload("file.txt")
And how can I do it?
Thanks for help!

You can't upload file to MediaFire without api,but i think it is source you want
from mediafire import MediaFireApi
from mediafire import MediaFireUploader
api = MediaFireApi()
uploader = MediaFireUploader(api)
session = api.user_get_session_token(
email='YOUR_EMAIL',
password='YOUR_PASSWORD',
app_id='42511')
api.session = session
response = api.user_get_info()
fd = open('PATH_FILE', 'rb')
result = uploader.upload(fd, 'OUTPUT_FILENAME',
folder_key='FOLDER KEY')
Default app_id is 42511, i think you don't know it

Get elasticloadbalancers names with boto3

When I try to print the load balancers from aws I get a huge dictionary with a lot of keys, but when I'm trying to print only the 'LoadBalancerName' value I get: None, I want to print all the load balancers names in our environment how I can do it? thanks!
What I tried:
import boto3
client = boto3.client('elbv2')
elb = client.describe_load_balancers()
Name = elb.get('LoadBalancerName')
print(Name)

The way in which you were handling the response object was incorrect, and you'll need to put it in a loop if you want all the Names and not just one. What you'll you'll need is this :
import boto3
client = boto3.client('elbv2')
elb = client.describe_load_balancers()
for i in elb['LoadBalancers']:
print(i['LoadBalancerArn'])
print(i['LoadBalancerName'])
However if your still getting none as a value it would be worth double checking what region the load balancers are in as well as if you need to pass in the use of a profile too.

How to send a file through Soap in python?

I want to send a zip file through SOAP (from a SOAP client to a SOAP server) in python.
Following the reading of this SO question, I choose to use suds as my python client.
But according to this, suds do not support sending attachment. A method is given to circumvent the problem but I've not been able to make it work. I'm puzzled over what I'm supposed to give as parameters.
Anyone know how to send a file through Soap in python ?
If needed I'll switch to another SOAP client library.

Download the provided wrapper, and then where you would normally say something like...
client.service.fooMethod(fooParam1,fooParam2,...)
...instead do...
soap_attachments.with_soap_attachment(client.service.fooMethod,binaryParam,fooParam1,fooParam2,...)
Where binaryParam is of the type expected by soap_attachements.py. For example, if you wanted to send a png image I think (never done this) you would do:
imageFile = open('imageFile.png','rb')
imageData = imageFile.read()
mimeType = 'image/png'
binaryParam = (imageData, uuid.uuid4(), mimeType)

Attachments are best way to send binary file through SOAP. If you can't use any other method but only SOAP, just encode your binaries with Base64 and paste it into SOAP method as a parameter. It isn't pure, but works great with small attachments. Large binaries? Use FTP, WebDAV and all others native ways for sending files between hosts.

I made the following changes to soap_attachments.py under suds to get my own uploads to work. You may not need some of the changes that I've made to this, but hopefully it'll at least give you a start.
--- /home/craig/Downloads/soap_attachments.py 2011-07-08 20:38:55.708038918 -0400
+++ soap_attachments.py 2011-06-21 10:29:50.090243052 -0400
## -1,4 +1,8 ##
+import uuid
+import re
def with_soap_attachment(suds_method, attachment_data, *args, **kwargs):
+ HUD_ARM_SERVICE_URL = suds_method.client.wsdl.url
+ HUD_ARM_SERVICE_URL = HUD_ARM_SERVICE_URL.replace('wsdl','xsd')
""" Add an attachment to a suds soap request.
attachment_data is assumed to contain a list:
## -16,7 +20,9 ##
soap_method = suds_method.method
if len(attachment_data) == 3:
+ print "here"
data, attachment_id, attachment_mimetype = attachment_data
+ attachment_id = uuid.uuid4()
elif len(attachment_data) == 2:
data, attachment_id = attachment_data
attachment_mimetype = MIME_DEFAULT
## -55,7 +61,7 ##
])
# Build the full request
- request_text = '\n'.join([
+ request_text = '\r\n'.join([
'',
'--%s' % boundary_id,
soap_headers,
I then use:
f = open(dir_path + infile,'rb')
data_file = f.read()
data_file_type = mimetypes.guess_type(infile)[0]
(filename,ext) = infile.split('.')
...
clientargs = [...]
identifier = with_soap_attachment(client.service.fooThing, [data_file, '1', data_file_type], credentials['foo'],credentials['bar'], morefoo)
You might not need all of these changes, but it's what got me going.
Hope this helps!

Link generator using django or any python module

I want to generate for my users temporary download link.
Is that ok if i use django to generate link using url patterns?
Could it be correct way to do that. Because can happen that I don't understand some processes how it works. And it will overflow my memory or something else. Some kind of example or tools will be appreciated. Some nginx, apache modules probably?
So, what i wanna to achieve is to make url pattern which depend on user and time. Decript it end return in view a file.

A simple scheme might be to use a hash digest of username and timestamp:
from datetime import datetime
from hashlib import sha1
user = 'bob'
time = datetime.now().isoformat()
plain = user + '\0' + time
token = sha1(plain)
print token.hexdigest()
"1e2c5078bd0de12a79d1a49255a9bff9737aa4a4"
Next you store that token in a memcache with an expiration time. This way any of your webservers can reach it and the token will auto-expire. Finally add a Django url handler for '^download/.+' where the controller just looks up that token in the memcache to determine if the token is valid. You can even store the filename to be downloaded as the token's value in memcache.

Yes it would be ok to allow django to generate the urls. This being exclusive from handling the urls, with urls.py. Typically you don't want django to handle the serving of files see the static file docs[1] about this, so get the notion of using url patterns out of your head.
What you might want to do is generate a random key using a hash, like md5/sha1. Store the file and the key, datetime it's added in the database, create the download directory in a root directory that's available from your webserver like apache or nginx... suggest nginx), Since it's temporary, you'll want to add a cron job that checks if the time since the url was generated has expired, cleans up the file and removes the db entry. This should be a django command for manage.py
Please note this is example code written just for this and not tested! It may not work the way you were planning on achieving this goal, but it works. If you want the dl to be pw protected also, then look into httpbasic auth. you can generate and remove entries on the fly in a httpd.auth file using htpasswd and the subprocess module when you create the link or at registration time.
import hashlib, random, datetime, os, shutil
# model to hold link info. has these fields: key (charfield), filepath (filepathfield)
# datetime (datetimefield), url (charfield), orgpath (filepathfield of the orignal path
# or a foreignkey to the files model.
from models import MyDlLink
# settings.py for the app
from myapp import settings as myapp_settings
# full path and name of file to dl.
def genUrl(filepath):
# create a onetime salt for randomness
salt = ''.join(['{0}'.format(random.randrange(10) for i in range(10)])
key = hashlib('{0}{1}'.format(salt, filepath).hexdigest()
newpath = os.path.join(myapp_settings.DL_ROOT, key)
shutil.copy2(fname, newpath)
newlink = MyDlink()
newlink.key = key
newlink.date = datetime.datetime.now()
newlink.orgpath = filepath
newlink.newpath = newpath
newlink.url = "{0}/{1}/{2}".format(myapp_settings.DL_URL, key, os.path.basename(fname))
newlink.save()
return newlink
# in commands
def check_url_expired():
maxage = datetime.timedelta(days=7)
now = datetime.datetime.now()
for link in MyDlink.objects.all():
if(now - link.date) > maxage:
os.path.remove(link.newpath)
link.delete()
[1] http://docs.djangoproject.com/en/1.2/howto/static-files/

It sounds like you are suggesting using some kind of dynamic url conf.
Why not forget your concerns by simplifying and setting up a single url that captures a large encoded string that depends on user/time?
(r'^download/(?P<encrypted_id>(.*)/$', 'download_file'), # use your own regexp
def download_file(request, encrypted_id):
decrypted = decrypt(encrypted_id)
_file = get_file(decrypted)
return _file
A lot of sites just use a get param too.
www.example.com/download_file/?09248903483o8a908423028a0df8032
If you are concerned about performance, look at the answers in this post: Having Django serve downloadable files
Where the use of the apache x-sendfile module is highlighted.
Another alternative is to simply redirect to the static file served by whatever means from django.