We Keep Coding

How to use python to create url shortcut on mac?

There are answers to the window version. but not mac.
I have searched google but no appropriate result
let say I want to create a shortcut to google.com on desktop, then we could run the python function
createShortcut("www.google.com","~/Desktop")
what could be the function body?

You can create a macos .url file inside the function. It is formatted as follows:
[InternetShortcut]
URL=http://www.yourweb.com/
IconIndex=0
Here is a sample implementation:
import os
def createShortcut(url, destination):
# get home directory if ~ in destination
if '~' in destination:
destination = destination.replace('~', os.path.expanduser("~"))
# macos .url file format
text = '[InternetShortcut]\nURL=https://{}\nIconIndex=0'.format(url)
# write .url file to destination
with open(destination + 'my_shortcut.url', 'w') as fw:
fw.write(text)
return
createShortcut("www.google.com", '~/Desktop/')

How to get absolute path of the file selected as input file in python?

I want the absolute path of the file selected as input file (from file browser in the form) using the python code below:
for attr, document in request.files.iteritems():
orig_filename = document.filename
print os.path.abspath(orig_filename)
mhash = get_hash_for_doc(orig_filename)
This prints the path of current working directory along(where the python script is executing) with the 'orig_filename' appended to it, which is the wrong path. I am using python 2.7, flask 0.12 under linux OS.
The requirement is to find the hash value of the file before uploading it to the server to check deduplication. So I need to use the algorithm by passing the file selected for hashing to another function as:
def get_hash_for_doc(orig_filename):
mhash = None
hash = sha1()#md5()
with open(mfile, "rb") as f:
for chunk in iter(lambda: f.read(128 * hash.block_size), b""):
hash.update(chunk)
mhash = hash.hexdigest()
return mhash
In this function I want to read file from absolute path of the orig_filename before uploading. Avoided all other code checks here.

First you need to create a temp file to simulate this required file then make your process on it
import tempfile, os
try:
fd, tmp = tempfile.mkstemp()
with os.fdopen(fd, 'w') as out:
out.write(file.read())
mhash = get_hash_for_doc(tmp)
finally:
os.unlink(tmp)

If you want to find a folder/file.ext, for an input file, simply use 'os.path.abspath' like:
savefile = os.path.abspath(Myinputfile)
when "Myinputfile" is a variable that contains the relative path and file name. For instance, derived from an argument define by the user.
But if you prefer to have absolute address of the folder, without file name try this:
saveloc = os.path.dirname(os.path.realpath(Myinputfile))

You can use pathlib to find the absolute path of the selected file.

Python: Creating and writing to a file error

I am having trouble creating and writing to a text file in Python. I am running Python 3.5.1 and have the following code to try and create and write to a file:
from os import *
custom_path = "MyDirectory/"
if not path.exists(custom_path)
mkdir(custom_path)
text_path = custom_path + "MyTextFile.txt"
text_file = open(text_path, "w")
text_file.write("my text")
But I get a TypeError saying an integer is required (got type str) at the line text_file = open(text_path, "w").
I don't know what I'm doing wrong as my code is just about identical to that of several tutorial sites showing how to create and write to files.
Also, does the above code create the text file if it doesn't exist, and if not how do I create it?

Please don't import everything from os module:
from os import path, mkdir
custom_path = "MyDirectory/"
if not path.exists(custom_path):
mkdir(custom_path)
text_path = custom_path + "MyTextFile.txt"
text_file = open(text_path, 'w')
text_file.write("my text")
Because there also a "open" method in os module which will overwrite the native file "open" method.

Creating an on-the-fly zip file from string content for AWS Lambda in Python

I have a Python script that creates a Lambda script in AWS along with all the policies and triggers. I use python boto3 library for that. I create the zip file for the lambda as on-the-fly rather than uploading a static zip file from the hard drive. I use this simple code from below to create my zip file. It creates the zip file without any problems and my python code uploads this zip file as a lambda script and I can view my lambda script in the AWS without any problems. But when I run my lambda script it gives me the module not found error even though I can clearly see that both the module name and the file name does exist and is view-able.
Unable to import module 'xxxx': No module named xxxx
In the file system I double click that zip file that was created by this code and see that the content is created and everything looks normal.
If I bypass zipping on the fly and create the zip statically using WinZip and let the rest of the Python & boto3 script upload this file then it works just fine.
def CreateLambdaZip(self, fileName, fileContent):
with zipfile.ZipFile('Lambda/' + fileName + '.zip', 'w') as myzipc:
myzipc.writestr( fileName + '.py', fileContent)
myzipc.close()
It kinda looks like for the zip file I'm skipping some special headers that is needed by Aws Lambda. Is there such thing? Because in the file system the zip file that is created by Python code and the other one that is created by WinZip are exactly the same. So I know there's nothing wrong with the lambda script.
Update: I'm uploading the zip file using the below code that reads the zip file which was created using the above snippet.
with open('Lambda/'+ fileName +'.zip', 'rb') as zipFile:
func = boto3.client("Lambda").create_function(
FunctionName=lambdaFunction,
Runtime='python2.7',
Role=role['Role']['Arn'],
Handler= fileName + "." + functionName,
Description=description,
Timeout=10,
MemorySize=256,
Publish=True,
Code={'ZipFile': zipFile.read()},
)
When I use zipFile.read() I get 2 different headers for the same content when I zip it using WinZip and when I zip it using Python's module.
Zip file that's created programmatically using Python
b'PK\x03\x04\x14\x00\x00\x00\x00\x00\xe4~\x01IO\x96J=Z\x07\x00\x00Z\x07\x00\x00\x19\x00\x00\x00schedule-ec2-snapshots.pyimport json\nimport boto3\nimport time\nfrom datetime import date, timedelta\n\nprint(\'Loading scheduled EC2 backup actions\')\n\ndef create_snapshots(event, context):\n """\n Lambda function that executes daily snapshots for the instances that
and zipfile created by WinZip
b'PK\x03\x04\x14\x00\x02\x00\x08\x004X\xfcH\x88\x1f\xce\xb5&\x03\x00\x00b\x07\x00\x00\x19\x00\x00\x00schedule-ec2-snapshots.py\x8dU]k\xdb#\x10|7\xf4?,\nA\x12qL\xda\x06B\r~I\x93Bh\x9b\x87&\xf4E\x15\xe1\xac[\xdb\xd7HwBw2\t\xc1\xff\xbd{+\xeb\xcb.\xb4\n\xc4\xba\xdb\xd1\xec\xce\xdc\xae\xa4\x8a\xd2T\x0e~[\xa3\'\xaa\xb9_\x1ag>\xb6\x0b\xa7\n\x9c\xac*S\x80\x14\x0e\xfd\n\xf6\x11\xbf\x9er\\b\xee\xc4dRVJ\xbb(\xfcf\x84Tz\r6\xdb\xa0\xacs\x94p\xfb\xf9\x03,E\xf6\\\x97

With the info above I was able to start the in-memory solution. The deployment of that zip file worked but I could not use the resulting function. Got error:
Unable to import module '<function-name>': No module named <function-name>
I got it to work by specifying the file permissions.
I then use the in-mem-zip to create an AWS lambda function.
Setup:
file_map is a dictionary of full_path->file_bytes.
files is a list of full_paths
def create_lambda_function(function_name, desc, role, handler, file_map, files)
zip_contents = create_in_mem_zip_archive(file_map, files)
result = lambda_code.create_function(
FunctionName=function_name,
Runtime="python2.7",
Description=desc,
Role=role,
Handler=handler,
Code={'ZipFile': zip_contents},
)
return result
def create_in_mem_zip_archive(file_map, files):
buf = io.BytesIO()
logger.info("Building zip file: " + str(files))
with zipfile.ZipFile(buf, 'w', zipfile.ZIP_DEFLATED) as zfh:
for file_name in files:
file_blob = file_map.get(file_name)
if file_blob is None:
logger.error("Missing file {} from files".format(file_name))
continue
try:
info = zipfile.ZipInfo(file_name)
info.date_time = time.localtime()
info.compress_type = zipfile.ZIP_DEFLATED
info.external_attr = 0777 << 16L # give full access
# info.external_attr = 0644 << 16L # -r-wr--r--
# info.external_attr = 0755 << 16L # -rwxr-xr-x
zfh.writestr(info, file_blob)
except Exception as ex:
logger.info("Error reading file: " + file_name + ", error: " + ex.message)
buf.seek(0)
return buf.read()

I have experienced the exactly same problem you have. My solution is do NOT use on the fly zip file. Create a real zip file and add real file into it, and it just works. You can do that even in the lambda environment, by create filepath like "/tmp/yourfile.txt" you can create temp real file when lambda execute.

IOError: [Errno 2] No such file or directory, but file does exist

I keep getting an IO Error stating my directory doesn't exist. What am I doing wrong?
I have this in a separate file named pirate.py:
with open("/images/image.jpg", "rb") as fin:
image_data = fin.read()
with open("pirate.py","wb") as fout:
fout.write("image_data="+repr(image_data))
And then the code in my main file
from pirate import image_data
# Content-type declaration
print('Content-type: text/html\n')
def main():
print('<!doctype html><head><meta charset="utf-8">')
print('<style>html {background:url (data:image/gif;base64,' + pirate.image_data + ')
I'm trying to encode an image in base64 and then use it as a background in a .cgi, I'm sure the rest of my code is working, what am I doing wrong here?

try to use absolute file paths. or use root prefix ROOT_PATH = os.path.dirname(os.path.realpath(file))

repr is not what you want, you are saying the image data is base64 ... so make it base 64
with open("pirate.py","wb") as fout:
fout.write("image_data="+base64.b64encode(image_data))
then in addition you say
from pirate import image_data
then you reference it as
pirate.image_data
when you should just reference it as
image_data
theres probably some other probelems since the image image path you show is not the image path from your comment

Your code says
open("/images/image.jpg", "rb")
I suspect the "/" at the beginning is the problem. Or is your images folder in your root?