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How to replace a part of a File Path in Python

import os.path
original = input(str("Filepath:"))
filename = os.path.basename(original)
print(filename)
target = r'C:\Users\Admin\Desktop\transfer\filename'
path = filename.replace('filename', filename)
print(path)
I have a problem with getting new target path... I need to copy original file and paste it to new directory, that is always the same and the name must stay the same as it was in previous directory, I was trying to do it by code on top but it doesn't work, only thing I need to know is how to replace name of the Path file at the end. (Example: r'C:\Users\Admin\Desktop\Directory2\***' and replace *** with filename of first file)

Considering your code, if you want to change C:\Users\Admin\Desktop\transfer\filename into C:\Users\Admin\Desktop\transfer\{new filename} you need to call replace() function on «target» variable, not on the «filename» variable.
So your code would look something like:
import os.path
original = input(str("Filepath:"))
filename = os.path.basename(original)
target = r'C:\Users\Admin\Desktop\transfer\filename'
path = target.replace('filename', filename)
On entering D:\Documents\program.py, the output is C:\Users\Admin\Desktop\transfer\program.py

python check if the folder content existed

The purpose of this code is:
Read a csv file which contains a column for a list of file names
here is the csv file:
https://drive.google.com/open?id=0B5bJvxM9TZkhVGI5dkdLVzAyNTA
Then check a specific folder to check if the files exist or not
If its found a file is not in the list delete it
here is the code:
import pandas as pd
import os.path
data = pd.read_csv('data.csv')
names = data['title']
path = "C:\\Users\\Sayed\\Desktop\\Economic Data"
for file in os.listdir(path):
os.path.exists(file)
print(file)
file = os.path.join(path, file)
fileName = os.path.splitext(file)
if fileName not in names:
print('error')
os.remove(file)
I modified the first code, and this is the new code and I got no error but the simply delete all the files in the directory

os.chdir does not return anything, so assigning the result to path means that path has None, which causes the error.
Since you're using pandas, here's a little trick to speed this up using pd.Series.isin.
root = "C:\Users\Sayed\Desktop\Economic Data"
files = os.listdir(root)
for f in data.loc[~data['title'].isin(files), 'title'].tolist():
try:
os.remove(os.path.join(root, f))
except OSError:
pass
Added a try-except check in accordance with EAFP (since I'm not doing an os.path.exists check here). Alternatively, you could add a filter based on existence using pd.Series.apply:
m = ~data['title'].isin(files) & data['title'].apply(os.path.exists)
for f in data.loc[m, 'title'].tolist():
os.remove(os.path.join(root, f))

Your path is the return value of the os.chdir() call. Which is obviously None.
You want to set path to the string representing the path ... leave the chdir out.

Cant change string value to variable with string value

I upload file to dropbox api, but it post on dropbox all directories from my computer since root folder. I mean you have folder of your project inside folder home, than user until you go to file sours folder. If I cut that structure library can't see that it is file, not string and give mistake message.
My code is:
def upload_file(project_id, filename, dropbox_token):
dbx = dropbox.Dropbox(dropbox_token)
file_path = os.path.abspath(filename)
with open(filename, "rb") as f:
dbx.files_upload(f.read(), file_path, mute=True)
link = dbx.files_get_temporary_link(path=file_path).link
return link
It works, but I need something like:
file_path = os.path.abspath(filename)
chunks = file_path.split("/")
name, dir = chunks[-1], chunks[-2]
which gives me mistake like:
dropbox.exceptions.ApiError: ApiError('433249b1617c031b29c3a7f4f3bf3847', GetTemporaryLinkError('path', LookupError('not_found', None)))
How could I make only parent folder and filename in the path?
For example if I have
/home/user/project/file.txt
I need
/project/file.txt

you have /home/user/project/file.txt and you need /project/file.txt
I would split according to os default separator (so it would work with windows paths as well), then reformat only the 2 last parts with the proper format (sep+path) and join that.
import os
#os.sep = "/" # if you want to test that on Windows
s = "/home/user/project/file.txt"
path_end = "".join(["{}{}".format(os.sep,x) for x in s.split(os.sep)[-2:]])
result:
/project/file.txt

I assume the following code should works:
def upload_file(project_id, filename, dropbox_token):
dbx = dropbox.Dropbox(dropbox_token)
abs_path = os.path.abspath(filename)
directory, file = os.path.split(abs_path)
_, directory = os.path.split(directory)
dropbox_path = os.path.join(directory, file)
with open(abs_path, "rb") as f:
dbx.files_upload(f.read(), dropbox_path, mute=True)
link = dbx.files_get_temporary_link(path=dropbox_path).link
return link

Python converting url into directory

I am trying to convert a url like "www.example.com/images/dog.png" into directories from the current directory.
So I get a folder named "www.example.com", inside that "images" and finally inside that the file saved as "dog.png"
I've tried using urllib.url2pathname(path) but it keeps appending P:\ to the start of it.

You can use os.makedirs() to create the directory tree, but that will fail if the final directory already exists. So you can test if it exists before attempting to create the directory tree, or use try: ... except OSError:. In Python 3 you can supply an exist_ok parameter to over-ride this behaviour, see the Python docs of os.makedirs for further info.
#!/usr/bin/env python
import os
cwd = os.getcwd()
url = "www.example.com/images/dog.png"
fullname = os.path.join(cwd, url)
path, basename = os.path.split(fullname)
if not os.path.exists(path):
os.makedirs(path)
with open(fullname, 'w') as f:
f.write('test\n')
If your system doesn't support directory names containing periods you can translate them to another character, eg _, like this:
fullname = fullname.replace('.', '_')
(just insert this after the fullname = os.path.join(cwd, url) line).
And as jwilner mentions in the comments, it's more efficient to use
path = os.path.dirname
than path, basename = os.path.split(fullname) if you don't need the base component of the file name (in this example "dog.png").

How to use Popen in Windows to invoke an external .py script and wait for its completion

Have you ever tried this feedback calling an external zip.py script to work? My CGITB does not show any error messages. It simply did not invoke external .py script to work. It simply skipped over to gush. I should be grateful if you can assist me in making this zip.py callable in feedback.py.
Regards. David
#**********************************************************************
# Description:
# Zips the contents of a folder.
# Parameters:
# 0 - Input folder.
# 1 - Output zip file. It is assumed that the user added the .zip
# extension.
#**********************************************************************
# Import modules and create the geoprocessor
#
import sys, zipfile, arcgisscripting, os, traceback
gp = arcgisscripting.create()
# Function for zipping files. If keep is true, the folder, along with
# all its contents, will be written to the zip file. If false, only
# the contents of the input folder will be written to the zip file -
# the input folder name will not appear in the zip file.
#
def zipws(path, zip, keep):
path = os.path.normpath(path)
# os.walk visits every subdirectory, returning a 3-tuple
# of directory name, subdirectories in it, and filenames
# in it.
#
for (dirpath, dirnames, filenames) in os.walk(path):
# Iterate over every filename
#
for file in filenames:
# Ignore .lock files
#
if not file.endswith('.lock'):
gp.AddMessage("Adding %s..." % os.path.join(path, dirpath, file))
try:
if keep:
zip.write(os.path.join(dirpath, file),
os.path.join(os.path.basename(path),
os.path.join(dirpath, file)[len(path)+len(os.sep):]))
else:
zip.write(os.path.join(dirpath, file),
os.path.join(dirpath[len(path):], file))
except Exception, e:
gp.AddWarning(" Error adding %s: %s" % (file, e))
return None
if __name__ == '__main__':
try:
# Get the tool parameter values
#
infolder = gp.GetParameterAsText(0)
outfile = gp.GetParameterAsText(1)
# Create the zip file for writing compressed data. In some rare
# instances, the ZIP_DEFLATED constant may be unavailable and
# the ZIP_STORED constant is used instead. When ZIP_STORED is
# used, the zip file does not contain compressed data, resulting
# in large zip files.
#
try:
zip = zipfile.ZipFile(outfile, 'w', zipfile.ZIP_DEFLATED)
zipws(infolder, zip, True)
zip.close()
except RuntimeError:
# Delete zip file if exists
#
if os.path.exists(outfile):
os.unlink(outfile)
zip = zipfile.ZipFile(outfile, 'w', zipfile.ZIP_STORED)
zipws(infolder, zip, True)
zip.close()
gp.AddWarning(" Unable to compress zip file contents.")
gp.AddMessage("Zip file created successfully")
except:
# Return any python specific errors as well as any errors from the geoprocessor
#
tb = sys.exc_info()[2]
tbinfo = traceback.format_tb(tb)[0]
pymsg = "PYTHON ERRORS:\nTraceback Info:\n" + tbinfo +
"\nError Info:\n " + str(sys.exc_type) +
": " + str(sys.exc_value) + "\n"
gp.AddError(pymsg)
msgs = "GP ERRORS:\n" + gp.GetMessages(2) + "\n"
gp.AddError(msgs)

zip() is a built-in function in Python. Therefore it is a bad practice to use zip as a variable name. zip_ can be used instead of.
execfile() function reads and executes a Python script.
It is probably that you actually need just import zip_ in feedback.py instead of execfile().

Yay ArcGIS.
Just to clarify how are you trying to call this script using popen, can you post some code?
If your invoking this script via another script in the ArcGIS environment, then the thing is, when you use Popen the script wont be invoked within the ArcGIS environment, instead it will be invoked within windows. So you will loose all real control over it.
Also just another ArcGIS comment you never initalize a license for the geoprocessor.
My suggestion refactor your code, into a module function that simply attempts to zip the files, if it fails print the message out to ArcGIS.
If you want post how you are calling it, and how this is being run.