I am trying to return file path on click of Submit button in HTML. File path has to be sent to python variable. Requirement is user can select file and if this file location is different than pre-configured file location then new file location should be updated in python. But only file name is being returned.
<label for="excel">Choose file location: </label>
<input type="file" id="excel" name="excel" accept=".xlsx, .xlx, .doc, .pdf">
example.py
file = request.form.get("excel")
if 'Update File Location' in action:
if file:
c.update_file_location(file)
As of now, only selected file name is returned in 'file'. But I want to get file path also along with filename.
Is there any way to achieve it?
Thanks in advance!
Related
The code below is supposed to navigate through a macro enabled file delete the rows and save the file again. all the steps are running and the file is saved. However, when I try to open the file I get this error:
"Excel cannot open the file "Matrix_2022xlsm" because the file format or the file extension is not valid. Verify that the file has not been corrupted and that the file extension matches the format of the file.
warnings.simplefilter(action='ignore', category=UserWarning)
path = "path/Matrix_2022.xlsm" #file path
book = xl.load_workbook(path) #load file
sheet = book.worksheets[2]
print("Maximum rows before moving:", sheet.max_row)
sheet.delete_rows(1, sheet.max_row-1) # specify from which row to which row we are deleting
print("Maximum rows after removing:", sheet.max_row)
# save the file to the path
path = "path/Matrix_2022.xlsm"
book.save(path)
I am created a basic ocr system. When user uploads a PDF file to system ocr read it and created a xlsm file with the same name. I am using remote control with Azure (Windows). I created a download button. When I click it, downloads the file with same name, so it should work but appears a error.
How can I fix this problem?
<a href= "{% static path %}" class="btn btn-outline-primary btn-sm" download>Download Report File</a>
in my views:
name = pdf.pdf.name.replace(".xlsm", "").replace(".pdf", "")
path = "C:/user/ocr/" + name + ".xlsm"
Note: Uploaded and created files are stored outside of the project file.
please get more error info from django server. and list more code from my views.
Do you want to change file filename?
I am trying to upload a file in to a folder in the box using the below code:
folder_id = '22222'
new_file = client.folder(folder_id).upload('/home/me/document.pdf')
print('File "{0}" uploaded to Box with file ID {1}'.format(new_file.name, new_file.id))
This code is not replacing the existing document.pdf in the box folder, rather it is keeping the older version of the file. I would like to remove the file in the target and keep the latest file. How to achieve this?
Since your goal is to replace the original file, you can try to overwrite its existing content. Here is an example. You will need to check for the filename if it is already present in the BOX folder though
folder_id = '22222'
file_path = '/home/me/document.pdf'
results = client.search().query(query='document', limit=1, ancestor_folder_ids=[folder_id], type='file', file_extensions=['pdf'])
file_id = None
for item in results:
file_id = item.id
if file_id:
updated_file = client.file(file_id).update_contents(file_path)
print('File "{0}" has been updated'.format(updated_file.name))
else:
new_file = client.folder(folder_id).upload(file_path)
print('File "{0}" uploaded to Box with file ID {1}'.format(new_file.name, new_file.id))
Its not replacing it because every time you upload new file it assign it a new id so the old file will never be replaced.
This is what I found in official docs.
Try to give it a name and then try that.
upload[source]
Upload a file to the folder. The contents are taken from the given file path, and it will have the given name. If file_name is not specified, the uploaded file will take its name from file_path.
Parameters:
file_path (unicode) – The file path of the file to upload to Box.
file_name (unicode) – The name to give the file on Box. If None, then use the leaf name of file_path
preflight_check (bool) – If specified, preflight check will be performed before actually uploading the file.
preflight_expected_size (int) – The size of the file to be uploaded in bytes, which is used for preflight check. The default value is ‘0’, which means the file size is unknown.
upload_using_accelerator (bool) –
If specified, the upload will try to use Box Accelerator to speed up the uploads for big files. It will make an extra API call before the actual upload to get the Accelerator upload url, and then make a POST request to that url instead of the default Box upload url. It falls back to normal upload endpoint, if cannot get the Accelerator upload url.
Please notice that this is a premium feature, which might not be available to your app.
Returns:
The newly uploaded file.
Return type:
File
I am creating a tool where either
A new XLSX file is generated for the user to download
The user can upload an XLSX file they have, I will read the contents of that file, aand use them to generate a new file for the user to download.
I would like to make use of Pandas to read the XLSX file into a dataframe, so I can work with it easily. However, I can't get it working. Can you help me?
Example extract from CGI file:
import pandas as pd
import cgi
from mako.template import Template
from mako.lookup import TemplateLookup
import http.cookies as Cookie
import os
import tempfile
import shutil
import sys
cookie = Cookie.SimpleCookie(os.environ.get("HTTP_COOKIE"))
method = os.environ.get("REQUEST_METHOD", "GET")
templates = TemplateLookup(directories = ['templates'], output_encoding='utf-8')
if method == "GET": # This is for getting the page
template = templates.get_template("my.html")
sys.stdout.flush()
sys.stdout.buffer.write(b"Content-Type: text/html\n\n")
sys.stdout.buffer.write(
template.render())
if method == "POST":
form = cgi.FieldStorage()
print("Content-Type: application/vnd.ms-excel")
print("Content-Disposition: attachment; filename=NewFile.xlsx\n")
output_path = "/tmp/" + next(tempfile._get_candidate_names()) + '.xlsx'
data = *some pandas dataframe previously created*
if "editfile" in form:
myfilename = form['myfile'].filename
with open(myfilename, 'wb') as f:
f.write(form['myfile'].file.read())
data = pd.read_excel(myfilename)
data.to_excel(output_path)
with open(path, "rb") as f:
sys.stdout.flush()
shutil.copyfileobj(f, sys.stdout.buffer)
Example extract from HTML file:
<p>Press the button below to generate a new version of the xlsx file</p>
<form method=post>
<p><input type=submit value='Generate new version of file' name='newfile'>
<div class="wrapper">
</div>
</form>
<br>
<p>Or upload a file.</p>
<p>In this case, a new file will be created using the contents of this file.</p>
<form method="post" enctype="multipart/form-data">
<input id="fileupload" name="myfile" type="file" />
<input value="Upload and create new file" name='editfile' type="submit" />
</form>
This works without the if "editfile" in form: bit so I know something is going wrong when I am trying to access the file that the user has uploaded.
The problem is that whilst a file is created, the created file has a file size of 0 KB and will not open in Excel. Crucially, the file that the user has uploaded can not be found in the location that I have written it out.
You've passed myfilename to pandas; however that file doesn't exist on the server yet. You'll have to save the file somewhere locally first before using it.
The following will download the file to the current directory (same directory as the CGI script). Of course, you're welcome to save it to some more suitable directory, depending on your setup.
form = cgi.FieldStorage()
myfilename = form['myfile'].filename
with open(myfilename, 'wb') as f: # Save the file locally
f.write(form['myfile'].file.read())
data = pd.read_excel(myfilename)
I am using the web2py framework.
I have uploaded txt a file via SQLFORM and the file is stored in the "upload folder", now I need to read this txt file from the controller, what is the file path I should use in the function defined in the default.py ?
def readthefile(uploaded_file):
file = open(uploaded_file, "rb")
file.read()
....
You can do join of application directory and upload folder to build path to file.
Do something like this:
import os
filepath = os.path.join(request.folder, 'uploads', uploaded_file_name)
file = open(filepath, "rb")
request.folder: the application directory. For example if the
application is "welcome", request.folder is set to the absolute path
"/path/to/welcome". In your programs, you should always use this
variable and the os.path.join function to build paths to the files you
need to access.
Read request.folder
The transformed name of the uploaded file is stored in the upload field of your database table, so you need a way to query the specific record that was inserted via the SQLFORM submission in order to get the name of the stored file. Here is how it would look assuming you know the record ID:
stored_filename = db.mytable(record_id).my_upload_field
original_filename, stream = db.mytable.my_upload_field.retrieve(stored_filename)
stream.read()
When you pass a filename to the .retrieve method of an upload field, it will return a tuple containing the original filename as well as the open file object (called stream in the code above).