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How to run python file from subfolder so current folder file path automatically detect and edit data?

Y.py python file location "D:\\Folder\\Sub Folder\\Y.py"
1.html file location "D:\\Folder\\1.html"
every time Folder name change python code outFileName has to edit.There is any way Folder path catch automatically.
python code is
import os
import glob
import re
html = ""
strPath = os.path.realpath(__file__)
print( f"Full Path :{strPath}" )
nmFolders = strPath.split( os.path.sep )
print( f"Folder Name :{nmFolders[-2]}" )
for X in {nmFolders[-2]}:
for file in sorted(glob.glob( os.path.join('*.html') )):
for Y in {'*.webp', '*.png', '*.jpg'}:
for image in sorted(glob.glob( os.path.join(Y))[:1]):
for X in {nmFolders[-2]}:
html += f'''<div class="card"><img class="card__image" src="{os.path.realpath(image)}" width="200px alt=""><div class="card__content"><p1>{X}</p1></div><div class="card__info"></div></div>'''
for X in {nmFolders[-2]}:
outFileName = "D:\\Folder\\1.html"
# Read the contents of the HTML file into a string
with open(outFileName, "r", encoding="utf-8") as f:
existing_html = f.read()
if html not in existing_html:
# Write the new HTML to the file
with open(outFileName, "a", encoding="utf-8") as f:
f.write(html + '\n')

so what is changing? The parent folder that contains
Y.py and 1.html?
what is the "Folder Path" you are trying to catch automatically? Sharing your code doesn't clarify what you are trying to do in this situation.
Assuming you are trying to catch D:\\Folder, you can traverse the current directory you're in using the .. in your file path.
so if you current directory (aka current folder) is
D:\\Folder\\Sub Folder\\Y.py
you can get to
D:\\Folder\\1.html
by using
..\\1.html
so basically two dots: .., means "go up one folder from where I currently am"

How to open and read text files in a folder python

I have a folder which has a text files in it. I want to be able to put in a path to this file and have python go through the folder, open each file and append its content to a list.
import os
folderpath = "/Users/myname/Downloads/files/"
inputlst = [os.listdir(folderpath)]
filenamelist = []
for filename in os.listdir(folderpath):
if filename.endswith(".txt"):
filenamelist.append(filename)
print(filename list)
So far this outputs:
['test1.txt', 'test2.txt', 'test3.txt', 'test4.txt', 'test5.txt', 'test6.txt', 'test7.txt', 'test8.txt', 'test9.txt', 'test10.txt']
I want to have the code take each of these files, open them and put all of its content into a single huge list not just print the file name. Is there any way to do this?

You should use file open for this.
Read here a documentation about its advanced options
Anyway, here is one way how you can do it:
import os
folderpath = r"yourfolderpath"
inputlst = [os.listdir(folderpath)]
filenamecontent = []
for filename in os.listdir(folderpath):
if filename.endswith(".txt"):
f = open(os.path.join(folderpath,filename), 'r')
filenamecontent.append(f.read())
print(filenamecontent)

If you are using Python3, you can use :
for filename in filename_list :
with open(filename,"r") as file_handler :
data = file_handler.read()
Please do mind that you will need the full (either relative or absolute) path to your file in filename
This way, your file handler will be automatically closed when you get out of the with scope.
More information around here : https://docs.python.org/fr/3/library/functions.html#open
On a side note, in order to list files, you might want to have a look to glob and use :
filename_list = glob.glob("/path/to/files/*.txt")

You can use fileinput
Code:
import fileinput
folderpath = "your_path_to_directory_where_files_are_stored"
file_list = [a for a in os.listdir(folderpath) if a.endswith(".txt")]
# This will return all the files which are in .txt format
get_all_files = fileinput.input(file_list)
with open("alldata.txt", 'ab+') as writefile:
for line in get_all_files:
writefile.write(line+'\n')
The above code will read all the data from .txt from a specified directory(folderpath) and store it in alldata.txt So, you wanted to have that long list, that list is now stored in .txt file if you want, else you can remove the write process.
Links:
https://docs.python.org/3/library/fileinput.html
https://docs.python.org/3/library/functions.html#open

Create folder by book id , download file from batch file, move files into folder

I'm using Python 3.6.5 on Windows 10
First, I'll create folder by book id, it works :
And then, Download file from batch file, one URL per line in file.
URL:
http://url.com/page-1.jpg
http://url.com/a.mp3
http://url.com/b.mp3
Here are code and works :
import os
import wget
book_id = ["5151","5152","5153"]
for id in book_id:
directory ="new/"+ str(id)
if not os.path.exists(directory):
os.makedirs(directory)
with open ("%s_url.txt" % id, encoding='utf-8', mode = 'r') as f:
for url in f.readlines():
filename = wget.download(url.strip(), out=directory)
print (filename)
5151_url.txt file is not in 5151 folder. I need to move download file into 5151 folder.
I know it could be use wget.download(url, fullfilename), and fullfilename = os.path.join(directory, filename)
filename = page-1.jpg ..etc.
Solution: add
out=directory)
But this will change file name, I don't wanna change file name. How to?

Try this.
import os
import wget
import time
book_id = ["5151","5152","5153"]
for id in book_id:
directory ="new/"+ str(id)
if not os.path.exists(directory):
os.makedirs(directory)
with open ("new" + "/" + id + "/%s_url.txt" % id, encoding='utf-8', mode = 'r') as f:
for url in f.readlines():
time.sleep(4)
filename = wget.download(url.strip())
print (filename)
Second part is bit modified, and could probably be done litle bit better, but you get an idea.
And this code reports no problems to me.
I also added time.sleep(4) just in case if you have blocking mechanism in place.
And I assume that you have 5151.txt inside the 5151 folder, and others in their respective locations.

how to zip all pdf files under a static folder? django

I have a folder named pdfs under static folder.
I am trying to have a returned zip which contains all the pdf files in the pdfs folder.
I have tried a few threads and used their codes, but I tried to workout things but then couldn't solve the last part that I get a message saying no file / directory
I know static folders are a bit different than usual folders.
can someone please give me a hand and see what I have missed?
Thanks in advance
from StringIO import StringIO
import zipfile
pdf_list = os.listdir(pdf_path)
print('###pdf list################################')
print(pdf_path) # this does show me the whole path up to the pdfs folder
print(pdf_list) # returns ['abc.pdf', 'efd.pdf']
zip_subdir = "somefiles"
zip_filename = "%s.zip" % zip_subdir
# Open StringIO to grab in-memory ZIP contents
s = StringIO()
# Grab ZIP file from in-memory, make response with correct MIME-type
resp = HttpResponse(content_type='application/zip')
# ..and correct content-disposition
resp['Content-Disposition'] = 'attachment; filename=%s' % zip_filename
# The zip compressor
zf = zipfile.ZipFile(s, "w")
for pdf_file in pdf_list:
print(pdf_file)
zf.write(pdf_file, pdf_path + pdf_file)
zf.writestr('file_name.zip', pdf_file.getvalue())
zf.close()
return resp
here I am getting errors for not able to find file / directory for 'abc.pdf'
P.S. I don't really need any sub folders zipped into the zip file. As long as all files are inside the zip, it'll be all good. (There won't be any sub folders in the pdfs folder)

I solved it myself and made it into a function with comments.
complicated things myself earlier
# two params
# 1. the directory where files want to be zipped
# e.g. of file directory is /et/ubuntu/vanfruits/vanfruits/static/pdfs/
# 2. filename of the zip file
def render_respond_zip(self, file_directory, zip_file_name):
response = HttpResponse(content_type='application/zip')
response['Content-Disposition'] = 'attachment; filename=' + zip_file_name
# open a file, writable
zip = ZipFile(response, 'w')
# loop through the directory provided
for single_file in os.listdir(file_directory):
# open the file, full path to the file including file name and extension is needed as first param
f = open(file_directory + single_file, 'r')
# write the file into the zip with
# first param is the name of the file inside the zip
# second param is read the file
zip.writestr(single_file, f.read())
zip.close()
return response

Writing to a new directory in Python without changing directory

Currently, I have the following code...
file_name = content.split('=')[1].replace('"', '') #file, gotten previously
fileName = "/" + self.feed + "/" + self.address + "/" + file_name #add folders
output = open(file_name, 'wb')
output.write(url.read())
output.close()
My goal is to have python write the file (under file_name) to a file in the "address" folder in the "feed" folder in the current directory (IE, where the python script is saved)
I've looked into the os module, but I don't want to change my current directory and these directories do not already exist.

First, I'm not 100% confident I understand the question, so let me state my assumption:
1) You want to write to a file in a directory that doesn't exist yet.
2) The path is relative (to the current directory).
3) You don't want to change the current directory.
So, given that:
Check out these two functions: os.makedirs and os.path.join. Since you want to specify a relative path (with respect to the current directory) you don't want to add the initial "/".
dir_path = os.path.join(self.feed, self.address) # will return 'feed/address'
os.makedirs(dir_path) # create directory [current_path]/feed/address
output = open(os.path.join(dir_path, file_name), 'wb')

This will create the file feed/address/file.txt in the same directory as the current script:
import os
file_name = 'file.txt'
script_dir = os.path.dirname(os.path.abspath(__file__))
dest_dir = os.path.join(script_dir, 'feed', 'address')
try:
os.makedirs(dest_dir)
except OSError:
pass # already exists
path = os.path.join(dest_dir, file_name)
with open(path, 'wb') as stream:
stream.write('foo\n')

Commands like os.mkdir don't actually require that you make the folder in your current directory; you can put a relative or absolute path.
os.mkdir('../new_dir')
os.mkdir('/home/you/Desktop/stuff')
I don't know of a way to both recursively create the folders and open the file besides writing such a function yourself - here's approximately the code in-line. os.makedirs will get you most of the way there; using the same mysterious self object you haven't shown us:
dir = "/" + self.feed + "/" + self.address + "/"
os.makedirs(dir)
output = open(os.path.join(dir, file_name), 'wb')