如何在烧瓶/Python上解压缩文件

Question

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

Answer 1

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

Answer 2

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)