So here is my issue:
def get_datasheets(self, build_id):
catalog = self._get_catalogue(build_id)
paths = []
for key, value in catalog.items():
if "data_sheets" in value:
if value["data_sheets"] is not None:
paths.append(value["data_sheets"]["data_sheet"])
for path in paths:
media_url = "http://some_url.com:8000" + path
print(media_url)
# Parsing the url and extracting only the datasheet name
parsed_url = parse.urlsplit(media_url).path
datasheet_name = parsed_url.split("/")[-1]
response = requests.get(media_url)
datasheet_path = os.path.join(self._output_path, datasheet_name)
with open(datasheet_path, 'wb') as file:
file.write(response.content)
I am trying to get the datasheet to be in a subfolder called datasheets, but the current implementation only allows me to get the datasheets in the ._output_path, which is the folder that I select.
What I really want is to be able to have something like: output_path_folder/datasheets_folder/datasheet_name.
UPDATE: so basically this is a method that extracts datasheets from a website and saves them locally, some sort of download. I gets stored with some other files so I want to have a subdirectory inside of the main directory that would allow me to store the datasheets. Hope it makes sense now
Any help would be greatly appreciated
I would suggest the standard python library pathlib for working with paths.
from pathlib import Path
datasheet_path = Path(self._output_path) / "datasheets" / datasheet_name
datasheet_path.parent.mkdir(parents=True, exist_ok=True)
To fit your edit it would look something like this:
from pathlib import Path
def get_datasheets(self, build_id):
catalog = self._get_catalogue(build_id)
paths = []
for key, value in catalog.items():
if "data_sheets" in value:
if value["data_sheets"] is not None:
paths.append(value["data_sheets"]["data_sheet"])
for path in paths:
media_url = "http://some_url.com:8000" + path
print(media_url)
# Parsing the url and extracting only the datasheet name
parsed_url = parse.urlsplit(media_url).path
datasheet_name = parsed_url.split("/")[-1]
response = requests.get(media_url)
datasheet_path = Path(self._output_path) / "datasheets" / datasheet_name
# takes the parent from the hole path and creates all directorys that are missing
datasheet_path.parent.mkdir(parents=True, exist_ok=True)
with open(datasheet_path, 'wb') as file:
file.write(response.content)
Related
I need to create an script that helps me pick the URL from a Box file that has a path in my computer.
I've the Box desktop application installed.
Like: C:\Users\Thiago.Oliveira\Box\
I've created this script:
# Providing the folder path
origin = r'C:\\Users\\Thiago.Oliveira\\Box\\XXXXXXX.xlsx'
target = f'C:\\Users\\Thiago.Oliveira\\Box\\XXXXXXX{datetime.date.today()}.xlsx'
# Fetching the list of all the files
shutil.copy(origin, target)
print("Files are copied successfully")
This helps me out to copy and rename the file for this box folder. But I also want to pick up the URL from the newly created file so I can send it over in an e-mail.
I haven't found anything that would help me with this.
Is this possible?
Yes, you can, see the example below.
This example is using the box python sdk, and this is a JWT auth application script.
After uploading the file, the box sdk returns a file object, which has many properties.
I'm not sure what you mean by "pick up the URL", it could be the direct download URL of the file or a shared link.
The example is for the direct download URL.
To get the destination FOLDER_ID, you can look at the browser URL when you open the folder on the box.com app.
e.g.:image of demo folder url in browser
from datetime import date
import os
from boxsdk import JWTAuth, Client
def main():
auth = JWTAuth.from_settings_file(".jwt.config.json")
auth.authenticate_instance()
client = Client(auth)
folder_id = "163422716106"
user = client.user().get()
print(f"User: {user.id}:{user.name}")
folder = client.folder(folder_id).get()
print(f"Folder: {folder.id}:{folder.name}")
with open("./main.py", "rb") as file:
basename = os.path.basename(file.name)
file_name = os.path.splitext(basename)[0]
file_ext = os.path.splitext(basename)[1]
file_time = date.today().strftime("%Y%m%d")
box_file_name = file_name + "_" + file_time + file_ext
print(f"Uploading {box_file_name} to {folder.name}")
box_file = folder.upload_stream(file, box_file_name)
box_file.get()
print(f"File: {box_file.id}:{box_file.name}")
print(f"Download URL: {box_file.get_download_url()}")
if __name__ == "__main__":
main()
print("Done")
Resulting in:
User: 20344589936:UI-Elements-Sample
Folder: 163422716106:Box UI Elements Demo
Uploading main_20230203.py to Box UI Elements Demo
File: 1130939599686:main_20230203.py
Download URL: https://public.boxcloud.com/d/1/b1!5CgJ...fLc./download
I'm trying to copy all pictures from one directory (also including subdirectories) to another target directory. Whenever the exact picture name is found in one of the xml files the tool should grap all information (attributes in the parent and child nodes) and create subdirectories based on those node informations, also it should rename the picture file.
The part when it extracts all the information from the nodes is already done.
from bs4 import BeautifulSoup as bs
path_xml = r"path\file.xml"
content = []
with open(res, "r") as file:
content = file.readlines()
content = "".join(content)
def get_filename(_content):
bs_content = bs(_content, "html.parser")
# some code
picture_path = f'{pm_1}{pm_2}\{pm_3}\{pm_4}\{pm_5}_{pm_6}_{pm_7}\{pm_8}\{pm_9}.jpg'
get_filename(content)
So in the end I get a string value with the directory path and the file name I want.
Now I struggle with opening all xml files in one directory instead of just opening one file. I tryed this:
import os
dir_xml = r"path"
res = []
for path in os.listdir(dir_xml):
if os.path.isfile(os.path.join(dir_xml, path)):
res.append(path)
with open(res, "r") as file:
content = file.readlines()
but it gives me this error: TypeError: expected str, bytes or os.PathLike object, not list
How can i read through all xml files instead of just one? I have hundreds of xml files so that will take a wile :D
And another question: How can i create directories base on string?
Lets say the value of picture_path is AB\C\D\E_F_G\H\I.jpg
I would need another directory path for the destination of the created folders and a function that somehow creates folders based on that string. How can I do that?
To read all XML files in a directory, you can modify your code as follows:
import os
dir_xml = r"path"
for path in os.listdir(dir_xml):
if path.endswith(".xml"):
with open(os.path.join(dir_xml, path), "r") as file:
content = file.readlines()
content = "".join(content)
get_filename(content)
This code uses the os.listdir() function to get a list of all files in the directory specified by dir_xml. It then uses a for loop to iterate over the list of files, checking if each file ends with the .xml extension. If it does, it opens the file, reads its content, and passes it to the get_filename function.
To create directories based on a string, you can use the os.makedirs function. For example:
import os
picture_path = r'AB\C\D\E_F_G\H\I.jpg'
dest_path = r'path_to_destination'
os.makedirs(os.path.join(dest_path, os.path.dirname(picture_path)), exist_ok=True)
In this code, os.path.join is used to combine the dest_path and the directory portion of picture_path into a full path. os.path.dirname is used to extract the directory portion of picture_path. The os.makedirs function is then used to create the directories specified by the path, and the exist_ok argument is set to True to allow the function to succeed even if the directories already exist.
Finally, you can use the shutil library to copy the picture file to the destination and rename it, like this:
import shutil
src_file = os.path.join(src_path, picture_path)
dst_file = os.path.join(dest_path, picture_path)
shutil.copy(src_file, dst_file)
Here, src_file is the full path to the source picture file and dst_file is the full path to the destination. The shutil.copy function is then used to copy the file from the source to the destination.
You can use os.walk() for recursive search of files:
import os
dir_xml = r"path"
for root, dirs, files in os.walk(dir_xml): #topdown=False
for names in files:
if ".xml" in names:
print(f"file path: {root}\n XML-Files: {names}")
with open(names, 'r') as file:
content = file.readlines()
I'm in need fo specifying file paths to sort JSON files that gets retreived from an API. I have a class module that saves and loads the files
import os
import json
class Directory:
def __init__(self):
self.working_dir = os.path.dirname(__file__) #Get's the current working directory
def mkdir(self, *path):
"""Creates folder in the same level as the working directory folder
Args:
*args: path to folder that is to be created
"""
target_dir = os.path.join(self.working_dir, *path)
try:
if os.path.exists(target_dir) == True:
print(target_dir, 'exists')
else:
os.mkdir(os.path.join(self.working_dir, *path))
print(os.path.join(self.working_dir, *path), 'succesfully created')
except OSError as e:
print(folder,'coult not be created', e)
def check_if_file_exist(self, *path):
if os.path.exists(os.path.join(self.working_dir, *path)):
return True
else:
return False
def save_json(self, filename, content, *path):
"""save dictionarirys to .json files
Args:
file (str): The name of the file that is to be saved in .json format
filename (dict): The dictionary that is to be wrote to the .json file
folder (str): The folder name in the target directory
"""
target_dir = os.path.join(self.working_dir, *path)
file_dir = os.path.join(self.working_dir, target_dir, filename)
with open(file_dir + '.json', "w") as f:
#pretty prints and writes the same to the json file
f.write(json.dumps(content, indent=4, sort_keys=False))
But it often result in hideously long lines when a file-path has to be specified, for example.
#EXAMPLE
filename = self.league + '_' + year + '_' + 'fixturestats'
self.dir.save_json(filename, stats, '..', 'json', 'params', 'stats')
path = '/'.join(('..', 'json', 'params'))
#OTHER EXAMPLE
league_season_info = self.dir.load_json('season_params.json', '..', 'json', 'params')
I'm wondering what optimal praxis is when having a repository that has rather static folders relative to the working directory. All my modules are now build to create the folders that are needed in the repository if they do not exist, so I can count on that the paths exist when loading or saving files.
A bit late to the party, but here's how I might do it. As you mention these folders are fairly static, they can be placed into a configuration object of sorts (e.g. as class members of the thing owning dir, or a stand-alone object). Since I do not know your class structure, I'll opt for a standalone object.
This might look like the following, using pathlib:*
from pathlib import Path
class StorageConfig:
STORAGE_BASE_DIR = Path("json")
PARAMS_DIR = STORAGE_BASE_DIR / "params"
STATS_DIR = PARAMS_DIR / "stats"
# Etc.
This can be used as follows:
self.dir.save_json(filename, stats, StorageConfig.STATS_DIR)
params = self.dir.load_json('season_params.json', StorageConfig.PARAMS_DIR)
Or, doing away with the directory argument altogether:
self.dir.save_json(StorageConfig.STATS_DIR / (filename + ".json"), stats)
params = self.dir.load_json(StorageConfig.PARAMS_DIR / 'season_params.json')
*Pathlib can simplify most of the code currently in your Directory class. Have a look!
I have a python script that creates a PDF and saves it in a subfoler of the folder where the script is saved. I have the following that saves the file to the subfolder:
outfilename = "Test" + ".pdf" #in real code there is a var that holds the name of the file
outfiledir = 'C:/Users/JohnDoe/Desktop/dev/PARENTFOLDER/SUBFOLDER/' #parent folder is where the script is - subfolder is where the PDFs get saved to
outfilepath = os.path.join(outfiledir, outfilename)
Is there a way I can save the PDFs to the subfolder without having to specify the full path? Lets say I wanted yto make this script an exe that multiple computers could use, how would I display the path so that the PDFs are just saved in the subfoler?
Thanks!
Try it:
import os
dir_name = os.path.dirname(os.path.abspath(__file__)) + '/subdir'
path = os.path.join(dir_name, 'filename')
I am looking for an simple file chooser dialog in sl4a. I have found a few native dialogs here but did not find the one I was looking for.
I wish I could save time by finding something readily available. A minimal code like filename = fileopendialog() would be a bonus.
Any ideas ?
I decided to write my own (see below for reference). This could probably be made better, any suggestions welcome.
import android, os, time
droid = android.Android()
# Specify root directory and make sure it exists.
base_dir = '/sdcard/sl4a/scripts/'
if not os.path.exists(base_dir): os.makedirs(base_dir)
def show_dir(path=base_dir):
"""Shows the contents of a directory in a list view."""
#The files and directories under "path"
nodes = os.listdir(path)
#Make a way to go up a level.
if path != base_dir:
nodes.insert(0, '..')
droid.dialogCreateAlert(os.path.basename(path).title())
droid.dialogSetItems(nodes)
droid.dialogShow()
#Get the selected file or directory .
result = droid.dialogGetResponse().result
droid.dialogDismiss()
if 'item' not in result:
return
target = nodes[result['item']]
target_path = os.path.join(path, target)
if target == '..': target_path = os.path.dirname(path)
#If a directory, show its contents .
if os.path.isdir(target_path):
show_dir(target_path)
#If an file display it.
else:
droid.dialogCreateAlert('Selected File','{}'.format(target_path))
droid.dialogSetPositiveButtonText('Ok')
droid.dialogShow()
droid.dialogGetResponse()
if __name__ == '__main__':
show_dir()