如何从临时文件对象获取文件名？

Question

My goal is to create a temporary file that's written by csv.writer which will be used in an upload to a database. Once the upload completes, I want to delete the file.

The upload and deletion portions are not included here as I haven't fully built it out but I need to get past this part first.

import csv, io, tempfile

filename = tempfile.NamedTemporaryFile(suffix='.csv', delete=False)

file = io.StringIO(report_downloader.DownloadReportAsString(report, skip_report_header=False, skip_column_header=False, skip_report_summary=True))
reader = csv.reader(file)

with open(filename, 'w', encoding='utf8', newline='') as f:
    writer = csv.writer(f, delimiter=',')
    for row in reader:
        writer.writerows([row])     

The resulting error from doing this is:

Traceback (most recent call last):
File "aw-ad-performance-tempfile.py", line 99, in <module> get_api_report()
File "aw-ad-performance-tempfile.py", line 92, in get_api_report
with open(filename, 'w', encoding='utf8', newline='') as f:

OSError: [Errno 22] Invalid argument:'<tempfile._TemporaryFileWrapper object at 0x00000274FBED0FD0>'

I've tried a few different ways of addressing the problem and I've identified that it expects a string where filename is here:

with open(filename, 'w', encoding='utf8', newline='') as f:

Is it possible to reference the temporary file as a string using the tempfile module so csv.writer recognizes it? What's the syntax to do so if it is?

Answer 1

You're on the right track. You need reference your file with the .name attribute of your tempfile.NamedTemporaryFile object. See the documentation here.

with open(filename.name, 'w', encoding='utf8', newline='') as f:
    writer = csv.writer(f, delimiter=',')
        for row in reader:
            writer.writerows([row])