How to compress image size after uploading in python

Question

I use the flask framework and pillow for image processing. Generally, uploading is done by looping files, and then file.save() one by one. I hope that after the save is completed, the pillow's compression logic will be executed. But it seems that save is an I/O operation, and there is delay. If it is directly under file.save()...

習慣沉默 · Answer

    def save(self, dst, buffer_size=16384):
        """Save the file to a destination path or file object.  If the
        destination is a file object you have to close it yourself after the
        call.  The buffer size is the number of bytes held in memory during
        the 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 copyfileobj
        close_dst = False
        if isinstance(dst, string_types):
            dst = open(dst, 'wb')
            close_dst = True
        try:
            copyfileobj(self.stream, dst, buffer_size)
        finally:
            if close_dst:
                dst.close()

You see the save operation is not asynchronous

Update

copyfileobj is a blocking operation

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

阿神 · Answer

In fact, for this type of image processing, it is better to directly use Alibaba Cloud's OSS or Qiniu and other similar storage functions. Directly upload the image to OOS, and then call a special suffix for specified image processing. In the future, you will also access OSS for processing. the address of. This can not only avoid the load of using your own server to process images, but also reduce the access pressure, which is also very beneficial to reducing the complexity of the program.

某草草 · Answer

Please take a look at the fp parameter of Image.open. You can also use A filename (string), pathlib.Path object or a file object PIL.Image.open(fp, mode='r')

You can just pass file to Image.open(file) directly!

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.