使用ftplib进行多线程上传

Question

我正在尝试进行多线程上传，但出现错误。 我猜想也许不可能在 ftplib 中使用多线程？

这是我的代码：

    class myThread (threading.Thread):
    def __init__(self, threadID, src, counter, image_name):
        self.threadID = threadID
        self.src = src
        self.counter = counter
        self.image_name = image_name
        threading.Thread.__init__(self)
    def run(self):
        uploadFile(self.src, self.image_name)

def uploadFile(src, image_name):
    f = open(src, "rb")            
    ftp.storbinary('STOR ' + image_name, f)
    f.close()

ftp = FTP('host')   # connect to host, default port
ftp.login()               # user anonymous, passwd anonymous@   
dirname = "/home/folder/"
i = 1   
threads = []

for image in os.listdir(dirname):
    if os.path.isfile(dirname + image):
        thread = myThread(i , dirname + image, i, image )   
        thread.start()
        threads.append( thread )        
        i += 1  

for t in threads:
    t.join()

获取一堆 ftplib 错误，例如

raise error_reply、resp error_reply: 200 Type set to I

如果我尝试一张一张上传，一切正常

Answer 1

您是否尝试过将连接代码放入线程内？

换句话说，让每个线程使用 FTP.host() 和 FTP.login() 进行自己单独的连接。服务器可能不喜欢在单个连接上同时进行多个上传，因为它可能一次解析一个命令，并且无法处理第二个上传或“STOR”命令。但是，如果您可以从同一 IP 地址进行多个连接，您将拥有单独的会话来发出“STOR”命令。

这是一个例子：

    class myThread (threading.Thread):
        def __init__(self, threadID, src, counter, image_name):
             ###############
             #Add ftp connection here!
             self.ftp = FTP('host')   # connect to host, default port
             self.ftp.login()               # user anonymous, passwd anonymous@   
             ################
             self.threadID = threadID
             self.src = src
             self.counter = counter
             self.image_name = image_name
             threading.Thread.__init__(self)
        def run(self):
             uploadFile(self.src, self.image_name)

    def uploadFile(src, image_name):
          f = open(src, "rb")            
          self.ftp.storbinary('STOR ' + image_name, f)
          f.close()

     dirname = "/home/folder/"
     i = 1   
     threads = []

     for image in os.listdir(dirname):
          if os.path.isfile(dirname + image):
             thread = myThread(i , dirname + image, i, image )   
             thread.start()
             threads.append( thread )        
             i += 1  

      for t in threads:
          t.join()

看看效果是否更好。

PS 不确定我的所有选项卡是否对齐。

Answer 2

我最终使用信号量将 FTP 连接的使用一次限制为一个线程。共享连接比为每个线程创建连接更快。在你的情况下，它看起来像：

from threading import Semaphore

ftp_semaphore = Semaphore(1)  # limiting connection to only one thread

class myThread (threading.Thread):
    def __init__(self, threadID, src, counter, image_name):
        self.threadID = threadID
        self.src = src
        self.counter = counter
        self.image_name = image_name
        threading.Thread.__init__(self)
    def run(self):
        uploadFile(self.src, self.image_name)

def uploadFile(src, image_name):
    f = open(src, "rb")
    with ftp_semaphore:     
        ftp.storbinary('STOR ' + image_name, f)
        f.close()

ftp = FTP('host')   # connect to host, default port
ftp.login()               # user anonymous, passwd anonymous@   
dirname = "/home/folder/"


i = 1   
threads = []

for image in os.listdir(dirname):
    if os.path.isfile(dirname + image):
        thread = myThread(i , dirname + image, i, image )   
        thread.start()
        threads.append( thread )        
        i += 1  

for t in threads:
    t.join()