Python 和 Flask - 尝试使用函数 return 文件内容

Question

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

当我尝试将新文件 (outfile) 发送回 (return) 给用户时出现问题，该文件的 Flask 代码也工作正常。但是我不知道如何让这个 Python transform function() "return" 文件内容，已经测试了几个选项。

更多详情如下：

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。所以传递完整路径会做：

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

当您传入一个文件对象时，您不能使用 with 语句，因为它会在您从视图中 return 时关闭文件对象；只有当响应对象被处理为 WSGI 响应时，它才会被实际读取，在你的视图函数之外。

如果您想向浏览器建议一个文件名以将文件另存为，您可能需要传入一个 attachment_filename 参数；它还将有助于确定模仿类型。您可能还想使用 mimetype 参数明确指定 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，默认的 mimetype 将是 text/plain。该函数实质上连接了目录和文件名（使用 flask.safe_join() 应用额外的安全检查以防止使用 .. 构造突破目录）并将其传递给 flask.send_file().

Python 和 Flask - 尝试使用函数 return 文件内容

Python and Flask - Trying to have a function return a file content

python

return

file

flask