Python 和 Flask - 试图让函数返回文件内容

Question

我正在努力将文件内容返回给用户。有一个从用户接收 txt 文件的 Flask 代码，然后调用 Python 函数 transform() 来解析 infile，这两个代码都在做这项工作。

当我尝试将新文件（输出文件）发送（返回）给用户时，问题正在发生，Flask 代码也可以正常工作。 但我不知道如何让这个 Python 转换函数（）“返回”该文件内容，已经测试了几个选项。

以下详细信息：

def transform(filename):

    with open(os.path.join(app.config['UPLOAD_FOLDER'],filename), "r") as infile:
        with open(os.path.join(app.config['UPLOAD_FOLDER'], 'file_parsed_1st.txt'), "w") as file_parsed_1st:

            p = CiscoConfParse(infile)
            ''' 
            parsing the file uploaded by the user and 
            generating the result in a new file(file_parsed_1st.txt)  
            that is working OK
            '''

    with open (os.path.join(app.config['UPLOAD_FOLDER'], 'file_parsed_1st.txt'), "r") as file_parsed_2nd:
        with open(os.path.join(app.config['UPLOAD_FOLDER'], 'file_parsed_2nd.txt'), "w") as outfile:
            '''
            file_parsed_1st.txt is a temp file, then it creates a new file (file_parsed_2nd.txt)
            That part is also working OK, the new file (file_parsed_2nd.txt) 
            has the results I want after all the parsing;
            Now I want this new file(file_parsed_2nd.txt) to "return" to the user
            '''

    #Editing -  
    #Here is where I was having a hard time, and that now is Working OK
    #using the follwing line:

        return send_file(os.path.join(app.config['UPLOAD_FOLDER'], 'file_parsed_2nd.txt')) 

Answer 1

您确实需要使用flask.send_file() callable 来产生正确的响应，但需要传入尚未关闭或即将关闭的文件名或文件对象。所以传入完整路径就可以了：

return send_file(os.path.join(app.config['UPLOAD_FOLDER'], 'file_parsed_2nd.txt'))

当你传入一个文件对象时，你不能使用with 语句，因为它会在你从视图返回的那一刻关闭文件对象；只有当响应对象被处理为 WSGI 响应时，它才会被实际读取，在您的视图函数之外。

如果您想向浏览器建议文件名以将文件另存为，您可能需要传入attachment_filename 参数；它还有助于确定 mimetype。您可能还想使用 mimetype 参数显式指定 mimetype。

您也可以使用flask.send_from_directory() function;它的作用相同，但需要一个文件名和一个目录：

return send_from_directory(app.config['UPLOAD_FOLDER'], 'file_parsed_2nd.txt')

关于 mimetype 的警告同样适用；对于.txt，默认mimetype 为text/plain。该函数实质上连接了目录和文件名（flask.safe_join() 应用额外的安全检查以防止使用.. 结构破坏目录）并将其传递给flask.send_file()。