Python - Flask - 传递图像 URL 而不是文件

Question

我使用 Python 3.6.5 - Flask 0.12.2 获取了此代码，公开服务并接收图像文件:

@app.route('/image', methods=['POST'])
def image():
    try:
        image_file = request.files['image']  # get the image

        # Set an image confidence threshold value to limit returned data
        threshold = request.form.get('threshold')
        if threshold is None:
            threshold = 0.5
        else:
            threshold = float(threshold)

        # finally run the image through tensor flow object detection`
        image_object = Image.open(image_file)
        objects = od_ws_api.get_objects(image_object, threshold)
        return objects

    except Exception as e:
        print(e)

当我使用此 CURL 命令运行程序时，一切都按预期正常工作:

curl -F "image=@xxx.jpg" http://localhost:5000/image

我的目标是使用相同的 POST 方法传递图像 URL 而不是本地文件，例如:

curl -F "image=https://i.ytimg.com/vi/aeLgjgoy_kE/maxresdefault.jpg" http://localhost:5000/image

如果我这样做，我会收到以下错误消息:

400 Bad Request: The browser (or proxy) sent a request that this server could not understand.
127.0.0.1 - - [10/Apr/2018 08:20:52] "POST /image HTTP/1.1" 200

我应该使用其他方法或库吗？ 谢谢 法规

<小时/>

我确实修改为:

@app.route('/image_url', methods=['POST'])
def image_url():
    try:
        image_url = request.value['image_url']  # get the image URL
        local_filename='c:/tensorflow/temp.jpg'
        local_filename, headers = urllib.request.urlretrieve(image_url)
        # Set an image confidence threshold value to limit returned data
        threshold = request.form.get('threshold')
        if threshold is None:
            threshold = 0.5
        else:
            threshold = float(threshold)

        # finally run the image through tensor flow object detection`
        image_object = Image.open(image_file)
        objects = od_ws_api.get_objects(image_object, threshold)
        return objects

    except Exception as e:
        print(e)
        return 'error'

尝试 CURL 后:

curl -F "image_url=@https://i.ytimg.com/vi/aeLgjgoy_kE/maxresdefault.jpg" http://localhost:5000/image_url

我收到以下错误:

Warning: setting file https://i.ytimg.com/vi/aeLgjgoy_kE/maxresdefault.jpg
Warning: failed!
curl: (26) read function returned funny value

Answer 1

如果您使用带文件名的curl，它不会存储在request.files中，而是存储在request.values中。因此，要获取图像网址，您需要调用

image_url = request.value['image']

现在您需要下载图像，例如通过使用 urlretrieve :

import urllib.request
local_filename, headers = urllib.request.urlretrieve(image_url)

图像现在存储在临时文件中，您可以通过 local_filename 访问该文件。