将Excel文件加载到以用户上传的mod_wsgi apache运行的pandas dataframe flask应用程序中

Question

I have a flask app where a user uploads files to an upload folder. Then I want to take those files and read them into pandas dataframes for further processing. The process works fine using app.run() on my localhost. I am trying to get it to work on aws with mod_wsgi and apache.

    @app.route('/uploader', methods=['POST'])
    def upload_file():
        if request.method == 'POST':
            filenames=[]
            uploaded_files = request.files.getlist("file[]")
            file.save(os.path.join(app.root_path,app.config['UPLOAD_FOLDER'], filename))
            filenames.append(filename)
        plotfiles=parse_all(filenames)

    def parse_all(filenames):
        folder_path=os.path.join(app.root_path, app.config['UPLOAD_FOLDER'])
        for f in filenames:
            f=send_from_directory(folder_path,filename))
            excel_file=pandas.ExcelFile(f)
            #do more stuff

I get the error ValueError: Must explicitly set engine if not passing in buffer or path for io.

The file is uploaded to the upload folder correctly but obviously not fetched correctly into the f variable. The type of f is <class 'flask.wrappers.Response'> and f.__dict__ returns

{'_on_close': [], 'response': [], 'headers': Headers([('X-Sendfile', u'/var/www/html/cluster_app/data/filename.xlsx'), ('Content-Length', u'82668'), ('Content-Type', u'application/vnd.openxmlformats-officedocument.spreadsheetml.sheet'), ('Cache-Control', u'public, max-age=43200'), ('Expires', u'Tue, 07 Jun 2016 22:59:11 GMT'), ('ETag', u'"1465297151.54-82668-755509703"')]), '_status_code': 200, '_status': '200 OK', 'direct_passthrough': True}

When running on my localhost on my machine there was a .file attribute in the response, now response is empty. Printing folder_path gives /var/www/html/cluster_app/data which is the uploads folder.

I'm very green on flask/wsgi/apache. Would really appreciate some advice on how to access the file system in my code.

Answer 1

嗨，我建议您在此处查看关于上传文件的Flask文档，稍后再检查您的代码。

Answer 2

Instead of

f=send_from_directory(folder_path,filename))

I use

f = open(os.path.join(app.root_path, app.config['UPLOAD_FOLDER'], filename))

to open the file. I just assumed send_from_directory would work as it does when I used flask app.run() on my localhost. I'd still like to know why send_from_directory does not work.