如何防止 UI 在 python 烧瓶中等待线程完成？

Question

我有一个程序可以获取音频文件并对其进行处理。

@app.route('/', methods=['GET', 'POST'])
def hello_world():
    if request.method == 'GET':
        return render_template('upload.html')
    else:
        file = request.files['file']
        path = os.getcwd() + '\\tempFilesAudio\\'
        if not os.path.exists(os.getcwd() + '\\' + 'tempFilesAudio'):
            os.mkdir(os.getcwd() + '\\' + 'tempFilesAudio')
        if not os.path.exists(os.getcwd() + '\\' + 'tempFilesTransciption'):
            os.mkdir(os.getcwd() + '\\' + 'tempFilesTransciption')
        file.save(path + secure_filename(file.filename))
        file_path = path + file.filename
        conv_path = convert(file_path)
        print('converted:{}'.format(conv_path))
        #this is a thread
        Recogniser(conv_path)
        print('Deleting MP3')
        os.remove(file_path)
        print('Deleting WAV')
        os.remove(conv_path)
        return render_template('upload.html')

我希望在文件提交给线程以在后台处理后重新渲染我的 UI。但它仍在等待。

下面是我的帖子的代码：

class Recogniser:

    def __init__(self, file):
        self.executor = ThreadPoolExecutor(5)
        self.file = file
        thread = threading.Thread(target=self.run(), daemon=True, args=file)
        thread.start()

    def run(self):
       #something

Answer 1

您可以使用多处理。试试这样：

from multiprocessing import Process

class Recogniser:

    def __init__(self, file):
        self.file = file
        thread = Process(target=self.run(), args=(file,))
        thread.start()

    def run(self):
       #something