有什么办法可以停止 pycurl perform() 的执行吗？

Question

我目前正在使用 pycurl 和 python 开发一个简单的多线程下载器，现在我可以根据需要暂停下载，但这只会占用太多内存，如果进程被终止，我不能直接恢复吧。所以我想出了一个解决方法（尚未实施）来停止下载，将字节位置（我可以通过进度函数获得）保存到 XML 文件，然后通过指定所述字节从那里继续下载位置。

我卡在的地方是我需要在执行 perform() 时关闭连接，因为 perform() 正在阻塞我想我可以从另一个线程调用 close() 但这只会抛出一个例外。我似乎无法在 pycurl 或 libcurl 文档中找到任何可以帮助我完成我想要的东西。

至于为什么我在更简单的请求上使用 pycurl，我已经使用 pycurl 进行基本自动化已有一段时间了，我已经习惯了，如果这里没有其他方法可以做我想做的事情的话pycurl 然后我猜 requests 可能是最后的手段。

class Downloader:
    def __init__(self, url, parts):
        self.url = url
        self.fileName = re.search(r"(?:[^/][\d\w.]+)+$", self.url, flags=re.IGNORECASE).group(0)
        self.parts = parts
        self.pause = False
        self.fileSize = round(self._getSize())
        self.partSize = round(self.fileSize / self.parts)
        self.threads = list()
        self.curlObjs = list()

    # Get file size by only downloading the HEADER and then calling getinfo for the length.
    def _getSize(self):
        curl = pycurl.Curl()
        curl.setopt(curl.URL, self.url)
        curl.setopt(curl.FOLLOWLOCATION, True)
        curl.setopt(curl.NOBODY, True)
        curl.perform()
        fileSize = curl.getinfo(curl.CONTENT_LENGTH_DOWNLOAD)
        curl.close()
        return fileSize

    # Track individual file part download progress.
    def _trackProgress(self, totalDown, currentDown, totalUp, currentUp):
        pass  # TODO
        # if currentDown != 0 and currentDown == totalDown:
            # print(f"Download Completed!\n{currentDown}/{totalDown}")

    # Calculate the part size, execute _downloadRange in separate threads, merge file parts on download completion.
    def download(self):
        partStart = 0
        partEnd = self.partSize

        for part in range(1, self.parts + 1):
            t = threading.Thread(target=self._downloadRange, args=(partStart, partEnd, part))
            self.threads.append(t)
            t.start()
            partStart += self.partSize + 1 if part == 1 else self.partSize
            partEnd += self.partSize

        for t in self.threads:
            t.join()

        self._mergeFiles(self.fileName)

    # Download the specified range and write it to a file part.
    def _downloadRange(self, startRange, endRange, fileNo):
        with open(f"{self.fileName}{fileNo}.part", "wb") as f:
            curl = pycurl.Curl()
            self.curlObjs.append(curl)
            curl = curlObj.curl
            curl.setopt(curl.URL, self.url)
            curl.setopt(curl.FOLLOWLOCATION, True)
            curl.setopt(curl.RANGE, f"{startRange}-{endRange}")
            curl.setopt(curl.WRITEDATA, f)
            curl.setopt(curl.NOPROGRESS, False)
            curl.setopt(curl.XFERINFOFUNCTION, self._trackProgress)
            curl.perform()
            curl.close()

    # Merge the file parts into one and delete the parts.
    def _mergeFiles(self, fileName):
        with open(fileName, "wb") as o:
            for part in range(1, self.parts + 1):
                with open(f"{self.fileName}{part}.part", "rb") as p:
                    o.write(p.read())
                os.remove(f"{self.fileName}{part}.part")```

Answer 1

https://curl.haxx.se/libcurl/c/CURLOPT_PROGRESSFUNCTION.html 有一个中止回调执行的规定。

有什么办法可以停止 pycurl perform() 的执行吗？

Is there any way to stop pycurl perform() execution?

python

pycurl