The following code is successfully uploading an image file using the Bottle framework.
upload = bottle.request.files.get("filPhoto01")
if upload is not None:
name, ext = os.path.splitext(upload.filename)
if ext not in ('.png','.jpg','.jpeg'):
return "File extension not allowed."
save_path = "/tmp/abc".format(category=category)
if not os.path.exists(save_path):
os.makedirs(save_path)
file_path = "{path}/{file}".format(path=save_path, file=upload.filename)
with open(file_path, 'w') as open_file:
open_file.write(upload.file.read())
However, when I try to open this file manually after upload, I can't open the file. I can see the icon of the uploaded file with the correct size (implying the whole image was uploaded), but I cannot view it in any application like MS paint, etc.
I also tried referencing the file in my web application, but it does not render there either. What could possibly be wrong?
Just a guess, but since it sounds like you're on Windows, you'll want to write the file in binary mode:
with open(file_path, 'wb') as open_file:
(Also, you didn't mention your Python version, but FYI in Python 3 you'd need to use binary mode even on Linux.)
Related
Exactly as the title says, I have this code
from shareplum import Site
from shareplum import Office365
from shareplum.site import Version
authcookie = Office365('https://mysite.sharepoint.com/', username='username', password='password').GetCookies()
site = Site('https://mysite.sharepoint.com/sites/mysite/', version=Version.v2016, authcookie=authcookie)
folder = site.Folder('Shared Documents/Beta Testing')
file = folder.get_file('practice.xlsx')
with open("practice.xlsx", "wb") as fh:
fh.write(file)
print('---')
folder.upload_file('xlsx', 'practice.xlsx')
Currently it downloads the file just fine which is fantastic, however I do not know how to reverse what I did with opening and downloading the file. Basically I need to be able to upload the file with the exact same name as the one I downloaded in the exact same format (in this case xlsx) as to overwrite the one in the sharepoint with the updated document.
Your post indicates that you want to modify the file so you will need some file handling for the downloaded file once it is saved after modification. Once the file modification has been done you need to open the file in 'rb' and then read that to a variable which will be the content when calling folder_obj.upload_file(content, name).
#this is your step to modify the file.
with open("practice.xlsx", "wb") as fh:
#file modification stuff... pyxlsx?
fh.write(file)
#open the file and read it into a variable as binary
with open("practice.xlsx", "rb") as file_obj:
file_as_string = file_obj.read()
#upload the file including the file name and the variable (file_as_string)
folder.upload_file(file_as_string, 'practice.xlsx')
This has been working for me. If you want to change the name of the file to include a version, delete the old file by calling folder.delete_file("practice.xlsx").
Can you try the below and see if it works?
with open("practice.xlsx", "wb") as fh:
file_content = fh.write(file)
folder.upload_file(file_content,'practice.xlsx')
I am trying to create a simple GUI with streamlit and python for my aspect-based sentiment analysis project, the user should be able to upload a .txt file so that I can run the model on that file. I already created the widget for uploading a file. My question is:
The uploaded file should be added to a specific folder, how can I specify an exact location for the uploaded file to be saved?
uploaded_file = st.file_uploader('FILE UPLOAD')
(This is the code for the upload widget)
The file_uploader function does not save the file to disk, it writes to a BytesIO buffer.
The UploadedFile class is a subclass of BytesIO, and therefore it is “file-like”. This means you can pass them anywhere where a file is expected.
https://docs.streamlit.io/en/stable/api.html?highlight=file_uploader#streamlit.file_uploader
If you want to save the result as a file, use the standard Python file io capabilities:
with open(filename, "wb") as f:
f.write(buf.getbuffer())
To add to what #RandyZwitch said you can use this function to save to a directory of your choice (directory/folder "tempDir")
def save_uploaded_file(uploadedfile):
with open(os.path.join("tempDir",uploadedfile.name),"wb") as f:
f.write(uploadedfile.getbuffer())
return st.success("Saved file :{} in tempDir".format(uploadedfile.name))
And apply the function below your uploaded file like below
datafile = st.file_uploader("Upload CSV",type=['csv'])
if datafile is not None:
file_details = {"FileName":datafile.name,"FileType":datafile.type}
df = pd.read_csv(datafile)
st.dataframe(df)
# Apply Function here
save_uploaded_file(datafile)
You can define path like this:
from pathlib import Path
path = "C:/Projects/ex1/your_file"
file_path = Path(path)
uploaded_file = st.file_uploader(file_path)
I am using the following code to upload a SQLITE3 Database file. For some reason, the script does not completely upload the file (the uploaded filesize is less than the original)
FTP = ftplib.FTP('HOST','USERNAME','PASSWORD')
FTP.cwd('/public_html/')
FILE = 'Database.db';
FTP.storbinary("STOR " + FILE, open(FILE, 'r'))
FTP.quit()
When I go to open the uploaded file in SQLite Browser, it says it is an invalid file.
What am I doing incorrectly?
In the open() call, you need to specify that the file is a binary file, like so:
FTP.storbinary("STOR " + FILE, open(FILE, 'rb'))
Django 1.7, Python 3.4, windows apache 2.4.12 + wsgi
In my program, I generate some csv files and put them in a zip file. I would like to let the user download the zip file, either force download (after file is generated and render response) or a click of button (after display result, user has to click a button to download.)
Currently I am forcing them to download once the zip file is generated on the server.
I have referenced the following links and come up with my code below. But it is always give me this error "'charmap' codec can't decode byte 0x8d in position 80: character maps to " I tried to set it with utf-8 and ascii, and similar errors will be given.
referenced links:
django download file from server to user's machine,or read online
how to serve downloadable zip file in django
https://djangosnippets.org/snippets/365/
Generating file to download with Django
Anyone know why am I getting this error and how to get this to work?
Thank you very much!
zip_filename = time.strftime("%Y%m%d") + ".zip"
with zipfile.ZipFile(zip_filename, mode='a', compression=zipfile.ZIP_DEFLATED) as zf:
for s in sql_request:
// generate csv files
zf.write(csv_file)
// close zf
zip_file = open(zip_filename, 'r')
response = HttpResponse(zip_file, content_type='application/force-download')
response['Content-Disposition'] = 'attachment; filename="%s"' % 'foo.zip'
return response
If you're on Windows, you might need to change your open line to include the b flag.
zip_file = open(zip_filename, 'rb')
I'm using Dropbox API with Python. I don't have problems with Dropbox API, I make all the authentification steps without problems.
When I use this code:
pdf_dropbox = client.get_file('/Example.pdf')
new_file = open('/home/test.pdf','w')
new_file.write(pdf_dropbox.read())
I generate a file in the path /home/test.pdf, it's a PDF file and the content is displayed same as original.
But when I try same code with an .odt file, it fails generating the new file:
odt_dropbox = client.get_file('/Example.odt')
new_file = open('/home/test_odt.odt','w')
new_file.write(odt_dropbox.read())
This new file test_odt.odt has errors and I can't see it's content.
# With this instruction I have the content of the odt file inside odt_dropbox
odt_dropbox = client.get_file('/Example.odt')
Wich is the best way to save the content of an odt file ?
Is there a better way to write LibreOffice files ?
I'd appreciate any helpfull information,
Thanks
Solved, I forgot 2 things:
Open the file for binary writing wb instead of w
new_file = open('/home/test_odt.odt','wb')
Close the file after creation: new_file.close() to make the flush
Full Code:
odt_dropbox = client.get_file('/Example.odt')
new_file = open('/home/test_odt.odt','wb')
new_file.write(odt_dropbox.read())
new_file.close()