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Implement similar module of pathlib in python 2

I have created a function where it will dynamically create a json file.
Currently i am using pathlib module which can be found in python 3 . However the server that i am deploying my script will only have python 2.7 and below.
The current approach im doing now work perfectly fine with pathlib module. When this was triggered in a higher environment, error im receiving is pathlib module not found. Hence i would go without using pathlib module when locating the base path to inject json content to publish_workbook.json
import json
import os
from os import listdir
from os.path import isfile, join
from pathlib import Path
def createJson():
print("Creating Json File ....... ")
..
..
event_dict = json.loads(_json)
_jsonFile = json.dumps(event_dict, indent = 4, sort_keys=True)
#Convert path to python 2.7 library orusing os.path
base = Path('/home/reporting/job/')
_saveFile = base / ('publish_workbook' + '.json')
print("Saving json file on : " + str(_saveFile))
_saveFile.write_text(_jsonFile)
createJson()
What is the best approach i can go with this ?

Install pathlib2 on the server, which is a backported version of pathlib. It's no longer supported but then neither is Python 2.7.
You'll then only have to change the import line to
from pathlib2 import Path

Just do it as strings. It's actually fewer characters to type. Pathlib has some cool concepts, but I'm not convinced it's a net win. And, of course, for Python 2 you don't have the print function.
base = '/home/reporting/job/'
_saveFile = base + 'publish_workbook.json'
print "Saving json file on : " + _saveFile
_saveFile.write_text(_jsonFile)

python: robustely import module from working directory

I define parameters for my programs in a python module for simplicity. These parameters are then loaded using import. Therefore, I have to make sure to always load from the working directory, and nowhere else (independent form the location of the executed script or available modules in python path).
I found two solutions. First modifying the path:
import sys
from os import getcwd
import_path = sys.path
sys.path = [str(getcwd(), ]
import xxx
sys.path = import_path
or using importlib
from pathlib import Path
from importlib.util import module_from_spec, spec_from_file_location
spec = spec_from_file_location('xxx', str(Path('.').expanduser()/'xxx.py'))
xxx = module_from_spec(spec)
spec.loader.exec_module(xxx)
Of course this can be wrapped into a context manager or a function respectively.
What would the pythonic way be to do this? Do those two approaches have advantages and disadvantages?
I checked How can I import a Python library located in the current working directory?
as well as Import python package from local directory into interpreter, but they lack the focus on robustness.

Local imports can be achieved by modifying the path. A context manager is a suitable solution:
import sys
from pathlib import Path
from contextlib import contextmanager
#contextmanager
def local_import(dir_=None):
"""Only import modules within `dir_` (default: cwd)."""
if dir_ is None:
dir_ = Path.cwd()
else:
dir_ = Path(dir_).absolute().resolve(strict=True)
import_path0 = sys.path[0]
sys.path[0] = str(dir_)
try:
yield
finally:
sys.path[0] = import_path0
Then the local import can be done using the standard import syntax
with local_import():
import xxx
This solution relys on the order, in which the paths are scanned, thus we temporarily replace sys.path[0]. We replace it, instead of prepending to avoid import conflicts with the script directory.
Note
You have to be careful to avoid name conflicts, as the import statement is used, modules with identical names will be imported only once. Thus if different modules with the same name exist in the working directory and in the original sys.path[0], only one of them will be imported. Thus, local_import should only be used for scripts, that only use the standard library or installed third party libraries, but not for scripts that import other scripts from the directory. For the unlikely case that you want to import different files with the same name, the following function can be used:
import uuid
from importlib.util import module_from_spec, spec_from_file_location
def import_file(file, content=None):
"""Try importing `file` as module avoiding name clashes.
If `content` is given `content = import_file('file.py', 'content')`
roughly corresponds to `from file import content`
else `file = import_file('file.py')`
roughly corresponds to `import file`.
Parameters
----------
file : str or Path
The Python file corresponding to the module.
content : str, optional
What to import from the module (optional).
"""
file = Path(file).expanduser().resolve(strict=True)
print(file)
spec = spec_from_file_location(file.stem + str(uuid.uuid4()), str(file))
module = module_from_spec(spec)
spec.loader.exec_module(module)
if content:
print(module)
return getattr(module, content)
else:
return module

Python - Get custom modules from current file

I am trying to get a list of the imported custom modules (modules I created by myself) in my current file but I could not find a proper way to achieve it.
For example:
<test.py>
import sys
import foo
import bar
I would like to get a list of [foo, bar] excluding the sys module.

Lets say, that if file located near curent file, or in some subfolder, when it's "our" module, not system one.
For this example, I've created three files: main.py, other.py and another.py, and placed whem into one folder.
The code of main.py is:
# os and sys are needed to work
import sys
import os
import shutil
import io
import datetime
import other
import another
def get_my_modules():
# Get list of modules loaded
modules = list(sys.modules.keys())
mymodules = []
# Get current dir
curdir = os.path.realpath(os.path.dirname(__file__))
for m in modules:
try:
# if some module's file path located in current folder or in some subfolder
# lets sey, that it's our self-made module
path = sys.modules[m].__file__
if path.startswith(curdir):
mymodules.append(m)
except Exception:
# Exception could happen if module doesn't have any __file__ property
pass
# Return list of our moudles
return mymodules
print(get_my_modules())
And this code actually outputs ["other", "another"]
This approach has a problem: If you import module, that somehow located in upper folder, it woudln't be detected.

python import a module with source code in temporary file

import tempfile
tmp = tempfile.NamedTemporaryFile(delete=True)
try:
# do stuff with temp
tmp.write(b'def fun():\n\tprint("hello world!")\n')
if __name__ == '__main__':
func = __import__(tmp.name)
func.fun()
finally:
tmp.close() # deletes the file
So I want to create a temporary file, add some source code to it and then import the module and call the function, but I always run into this error:
ModuleNotFoundError: No module named '/var/folders/3w/yyp887lx4018h9s5sr0bwhkw0000gn/T/tmp84bk0bic'
It doesn't seem to find the module of the temporary file. How do I solve this?

There are a few problems with your code:
Your filename does not end with .py, but Python modules are expected to. You can fix this by setting suffix='.py' in NamedTemporaryFile().
__import__() is not the right way to load a module from a full path. See here: How to import a module given the full path?
You do not flush after writing and before importing, so even if Python does find the file, it may well be empty. Add tmp.flush() after writing to fix this.

Importing can only be done from certain directories which are part of the PYTHON_PATH. You can extend that. Then you will have to use __import__() with a module name (not a path in the file system). You will have to deal with the suffix for the temp file.
I implemented a simple version using the local directory for the temp module file and a version using a proper tempfile:
#!/usr/bin/env python3
import sys
import os
import tempfile
SCRIPT = '''\
def fun():
print("hello world!")
'''
# simple version using the local directory:
with open('bla.py', 'w') as tmp_module_file:
tmp_module_file.write(SCRIPT)
import bla
bla.fun()
# version using the tempfile module:
tmpfile = tempfile.NamedTemporaryFile(suffix='.py', delete=True)
try:
tmpfile.write(SCRIPT.encode('utf8'))
tmpfile.flush()
tmpmodule_path, tmpmodule_file_name = os.path.split(tmpfile.name)
tmpmodule_name = tmpmodule_file_name[:-3] # strip off the '.py'
sys.path.append(tmpmodule_path)
tmpmodule = __import__(tmpmodule_name)
finally:
tmpfile.close()
tmpmodule.fun()

How to get relative file path on Python 2.5?

I was using python 2.6 for a script, but for requirement limitations I have to downgrade my script to python 2.5, how can I get the relative path using python 2.5?
I was previously using:
os.path.relpath(path[, start])
But since this is new from 2.6 I can't use it anymore.
Thanks and regards!

Try Relative paths in Python it should have valuable information to you.
>>> import sys
>>> import os.path
>>> sys.path[0]
'C:\\Python25\\Lib\\idlelib'
>>> os.path.relpath(sys.path[0], "path_to_libs") # if you have python 2.6
>>> os.path.join(sys.path[0], "path_to_libs")
'C:\\Python25\\Lib\\idlelib\\path_to_libs'
EDIT: Found something more http://www.saltycrane.com/blog/2010/03/ospathrelpath-source-code-python-25/
It is a re-implementation by James Gardner:
from posixpath import curdir, sep, pardir, join
def relpath(path, start=curdir):
"""Return a relative version of a path"""
if not path:
raise ValueError("no path specified")
start_list = posixpath.abspath(start).split(sep)
path_list = posixpath.abspath(path).split(sep)
# Work out how much of the filepath is shared by start and path.
i = len(posixpath.commonprefix([start_list, path_list]))
rel_list = [pardir] * (len(start_list)-i) + path_list[i:]
if not rel_list:
return curdir
return join(*rel_list)