如何将 dataframe 转换为 csv 并将其保存以返回？

Question

I'm reading a.xlsx file and converting it to.csv but I need to return it as a.csv because I don't have access to the path where it should be stored. I've only been able to return it as a dataframe. Any help will be really appreciated. Thanks.

import pandas as pd


def excel_to_csv(myfile):
    print(myfile)
    data_xls = pd.read_excel(myfile,  index_col=None)
    data_xls.to_csv(encoding='utf-8', index=False)
    return(data_xls)

myfile = request.FILES['file']
if myfile.name.endswith('.xlsx'):
    myfile=excel_to_csv(myfile)

Answer 1

if you just want to read an Excel file and get the same data as csv printed to your screen, the following might be useful.

def excel_to_csv(file):
    df = pd.read_excel(file)
    csv_data = df.to_csv(encoding='utf-8', index=False)
    return csv_data

If you really want to save the file I think I would need more information to the systems your are using and what exactly are you trying to accomplish.

Answer 2

If you're using Django, just save it into /media/ root, then you can access it via url

First you need to know the config of MEDIA_ROOT & MEDIA_URL inside your settings.py .

If it's not exist, then you can create one

[ settings.py ]

MEDIA_ROOT = os.path.join(BASE_DIR, 'myproject/static')
MEDIA_URL = '/media/'

Then, make sure you're already add the media config into your urls.py

from django.conf.urls.static import static

urlpatterns = [
    # Project url patterns...
]

if settings.DEBUG:
    urlpatterns += static(settings.MEDIA_URL, document_root=settings.MEDIA_ROOT)

Since you're mentioned this code myfile = request.FILES['file'] , then I assume you're in views.py . Last but not least we need FileSystemStorage API.

There you go

[ views.py ]

from django.core.files.storage import FileSystemStorage

def upload(request):
    if request.method == "POST":
        csv_file = request.FILE['file']
        fs = FileSystemStorage()
        name = fs.save(csv_file.name, csv_file)
        print(f"file_name = {fs.url(name)}"
    return render(request, 'somepage.html')

Access the file via url, www.yourhost.com/media/file_name