問題描述

我用的是flask框架，圖片處理用的是pillow。

一般上傳都是在循環files，然后逐個file.save()我希望在save完成后，執行pillow的壓縮邏輯。

但是似乎save是一個I/O操作，存在延遲性，如果直接在file.save()下面直接調用pillow的Image.open，會出錯，因為圖片數據還沒有寫入圖片。

咋辦？

問題解答

def save(self, dst, buffer_size=16384):'''Save the file to a destination path or file object. If thedestination is a file object you have to close it yourself after thecall. The buffer size is the number of bytes held in memory duringthe copy process. It defaults to 16KB.For secure file saving also have a look at :func:`secure_filename`.:param dst: a filename or open file object the uploaded file is saved to.:param buffer_size: the size of the buffer. This works the same as the `length` parameter of :func:`shutil.copyfileobj`.'''from shutil import copyfileobjclose_dst = Falseif isinstance(dst, string_types): dst = open(dst, ’wb’) close_dst = Truetry: copyfileobj(self.stream, dst, buffer_size)finally: if close_dst:dst.close()

你看save操作不是異步的吖

更新

copyfileobj是個阻塞操作

https://github.com/pallets/we...

其實這類圖片處理，直接使用阿里云的OSS或者七牛等類似的存儲功能更好，直接將圖片上傳到OOS中，然后調用特別的后綴進行指定的圖片處理，未來也訪問OSS上處理后的地址。這樣既可以規避用自己服務器處理圖片的負荷，而且也降低了訪問的壓力，對于降低程序的復雜度也是大有好處的。

樓主看看Image.open 的fp參數，也可以A filename (string), pathlib.Path object or a file object PIL.Image.open(fp, mode=’r’)

你直接傳file給Image.open(file)就可以了吧！

PIL.Image.open(fp, mode=’r’)Opens and identifies the given image file.This is a lazy operation; this function identifies the file, but the file remains open and the actual image data is not read from the file until you try to process the data (or call the load() method). See new().Parameters: fp – A filename (string), pathlib.Path object or a file object. The file object must implement read(), seek(), and tell() methods, and be opened in binary mode.mode – The mode. If given, this argument must be “r”.Returns: An Image object.Raises: IOError – If the file cannot be found, or the image cannot be opened and identified.