Hi is it possible for a user to upload a file using flask;
user would select it from there computer, select submit, which would be downloaded to a ZIP file folder on webserver(local host) and unzip that file, and search for a certain file within that unzip file directory
I have the functionality of the form down to upload it can’t figuire out how to unzip the file and save its content in a folder
This may work for your problem:
You'd used this first then,
response = make_response(log_file.text)
this for the second
handle.write(response.content)
.content is "Content of the response, in bytes." .text is "Content of the response, in unicode."
If you want a byte stream, use .content.
for further comprehension go to:
Can't unzip file retrieved with Requests in Flask app
or to:
Unzipping files in Python
One of them should work
its pure python unzip file
import zipfile
with zipfile.ZipFile(path_to_zip_file, 'r') as zip_ref:
zip_ref.extractall(directory_to_extract_to)
then you can use it in your flask app
import zipfile
#app.route("/",methods=["POST"])
def page_name_post():
file = request.files['data_zip_file']
file_like_object = file.stream._file
zipfile_ob = zipfile.ZipFile(file_like_object)
file_names = zipfile_ob.namelist()
# Filter names to only include the filetype that you want:
file_names = [file_name for file_name in file_names if file_name.endswith(".txt")]
files = [(zipfile_ob.open(name).read(),name) for name in file_names]
return str(files)
Related
I have a function that opens a zip file, finds a pdf with a given filename, then reads the first page of the pdf to get some specific text. My issue is that after I locate the correct file, I can't open it to read it. I have tried to use a relative path within the zip folder and a absolute path in my downloads folder and I keep getting the error:
no such file: 'Deliverables_Rev B\Plans_Rev B.pdf'
no such file: 'C:\Users\MyProfile\Downloads\Deliverables_Rev B\Plans_Rev B.pdf'
I have been commenting out the os.path.join line to change between the relative and absolute path as self.prefs['download_path'] returns my download folder.
I'm not sure what the issue with with the relative path is, any insight would be helpful, as I think it has to do with trying to read out of a zipped folder.
import zipfile as ZipFile
import fitz
def getjobcode(self, filename):
if '.zip' in filename:
with ZipFile(filename, 'r') as zipObj:
for document in zipObj.namelist():
if 'plans' in document.lower():
document = os.path.join(self.prefs['download_path'], document)
doc = fitz.open(document)
page1 = doc.load_page(0)
page1text = page1.get_text('text')
jobcode = page1text[page1text.index(
'PROJECT NUMBER'):page1text.index('PROJECT NUMBER') + 30][-12:]
return jobcode
I ended up extracting the zip folder into the downloads folder then parsing the pdf to get the data I needed. Afterwords I created a job folder where I wanted it and moved the extracted folder into it from the downloads folder.
I am trying to loop through all mp3 files in my directory in MacOS Monterrey and for every iteration get the file's more info attributes, like Title, Duration, Authors etc. I found a post saying use xattr, but when i create a variable with xattr it doesn't show any properties or attributes of the files. This is in Python 3.9 with xattr package
import os
import xattr
directory = os.getcwd()
for filename in os.listdir(directory):
f = os.path.join(directory, filename)
# checking if it is a file
if os.path.isfile(f):
print(f)
x = xattr.xattr(f)
xs = x.items()
xattr is not reading mp3 metadata or tags, it is for reading metadata that is stored for the particular file to the filesystem itself, not the metadata/tags thats stored inside the file.
In order to get the data you need, you need to read the mp3 file itself with some library that supports reading ID3 of the file, for example: eyed3.
Here's a small example:
from pathlib import Path
import eyed3
root_directory = Path(".")
for filename in root_directory.rglob("*.mp3"):
mp3data = eyed3.load(filename)
if mp3data.tag != None:
print(mp3data.tag.artist)
print(mp3data.info.time_secs)
I am generating multiple .csv files using a script which is plugged in to my Flask app. Below is my route file section where required inputs are taken and passing in to the script.
#app.route('/summary_report', methods= ['GET', 'POST'])
def summary_report():
"""
Showing page for generating daily report
:return:
"""
form = frm.SummaryReportForm()
if form.validate_on_submit():
from_date = form.from_date.data
is_active = form.is_active.data
reports = current_summary_report.fetch_report(from_date=from_date, status=is_active) # Go the script
return send_file(reports, as_attachment=True, attachment_filename="reports.zip") # Download attachment
return render_template('pages/reports/current/summary.html', form=form)
And my script file is running a loop and creating multiple csv files.
def fetch_report(from_date=from_date, status=is_active):
for record in records:
....
f = open(file_name + '-' + str(time) + '.csv','w+')
f.write(csv)
f.close()
What changes I should do here to make all csv files in to a zip file and making it downloadable.
Python allows you to quickly create zip/tar archives.
Following command will zip entire directory
shutil.make_archive(output_filename, 'zip', dir_name)
Following command gives you control on the files you want to archive
ZipFile.write(filename)
Code Explanation
Import make_archive class from module shutil.
Use the split function to split out the directory and the file name from the path to the location of the text file.
Then we call the module "shutil.make_archive("guru99 archive, "zip", root_dir)" to create archive file, which will be in zip format
I want to open a html file and that html file is in a zip file(both name is same) and i'm trying to open that html file.
old_file = input("DRAG:") #dir C:\Users\GG\PycharmProjects\pythonProject\f1dbef77-342b-4026-85d8-7f30fe691a63_f.zip
file_parts = old_file.split(".") #[C:\Users\GG\PycharmProjects\pythonProject\f1dbef77-342b-4026-85d8-7f30fe691a63_f] [zip]
first= file_parts[0]
direcs = first.split("\\")
file_itself = direcs[-1] # the file name that i need to use
last = file_parts[1]
file = open(f'{first}.zip\\{file_itself}.html', encoding="UTF-8").read()
You should first unzip the archive in a temporary folder, then you should open the file from there, and when everything is done, you may delete the folder in which you have extracted your data.
You may use python ZipFile as library and the extract() call to unzip your html.
See ZipFile Docs
I am newer to python. I was try open the pdf files and write its content into the
new text files. That the text files name are generate by the pdf name. I tried so far but it is not give what i expect. How can i achieve it
import glob, os
import pyPdf
os.chdir("pdf/")
for file in glob.glob("*.pdf"):
filena = file
filename = "c:/documents/"+filena+".txt"
target = open(filename,'w')
pdf = pyPdf.PdfFileReader(open(filena,"rb"))
for page in pdf.pages:
target.write (page.extractText())
target.close()
Results the Error
File "c:/documents/atpkinase.pdf.txt",line 7, in <module>
target = open(filename,'w')
IOError: [Errno 2] No such file or directory: "c:/documents/atpkinase.pdf.txt"
Looks like if the directory "c:/documents/" does not exist. To write file to it you must create directory first. To check directory existent (and create it if needed) you can use
dir = "c:/documents"
if not os.path.exists(dir):
os.makedirs(dir)
Also, filea contains file name with extension, and when you create filename you need only a file name of old file without extension.