如何使用 Python 图像库 (PIL) 将上传的图像保存到 FastAPI？

Question

I am using image compression to reduce the image size. When submitting the post request, I am not getting any error, but can't figure out why the images do not get saved. Here is my code:

@app.post("/post_ads")
async def create_upload_files(title: str = Form(),body: str = Form(), 
    db: Session = Depends(get_db), files: list[UploadFile] = File(description="Multiple files as UploadFile")):
    for file in files:
        im = Image.open(file.file)
        im = im.convert("RGB")
        im_io = BytesIO()
        im = im.save(im_io, 'JPEG', quality=50) 

Answer 1

PIL.Image.open() takes as fp argumnet the following:

fp – A filename (string), pathlib.Path object or a file object. The file object must implement file.read() , file.seek() , and file.tell() methods, and be opened in binary mode.

Using a BytesIO stream, you would need to do it like this: Image.open(io.BytesIO(await file.read())) , as shown in this answer (see client side ). However, you don't really have to use an in-memory bytes buffer, as you can get the the actual file object using the .file attribute ofUploadFile . As per thedocumentation :

file : A SpooledTemporaryFile (a file-like object). This is the actual Python file that you can pass directly to other functions or libraries that expect a "file-like" object.

Example:

# ...
from PIL import Image

@app.post("/upload")
def upload(file: UploadFile = File()):
    try:        
        im = Image.open(file.file)
        if im.mode in ("RGBA", "P"): 
            im = im.convert("RGB")
        im.save('out.jpg', 'JPEG', quality=50) 
    except Exception:
        # ...
    finally:
        file.file.close()
        im.close()

For more details and code examples on how to upload files/images using FastAPI, please have a look at this answer and this answer . Also, please have a look at this answer for more information on defining your endpoint with def or async def .

Answer 2

I assume you are writing to a BytesIO to get an "in memory" JPEG without slowing yourself down by writing to disk and cluttering your filesystem.

If so, you want:

im = Image.open(file.file)
im = im.convert("RGB")
im_io = BytesIO()
# create in-memory JPEG in RAM (not disk)
im.save(im_io, 'JPEG', quality=50)

# get the JPEG image in a variable called JPEG
JPEG = im_io.get_value()