Question

我想知道如何通过创建API服务来上传文件？

class UploadImage(Resource):
    def post(self, fname):
        file = request.files['file']
        if file:
            # save image
        else:
            # return error
            return {'False'}

路线

api.add_resource(UploadImage, '/api/uploadimage/<string:fname>')

然后是HTML

   <input type="file" name="file">

我在服务器端启用了CORS

我使用angular.js作为前端，ng-upload如果重要，但也可以使用CURL语句！

Answer 1

以下内容足以保存上传的文件

    from flask import Flask
    from flask_restful import Resource, Api, reqparse
    import werkzeug

    class UploadAudio(Resource):
      def post(self):
        parse = reqparse.RequestParser()
        parse.add_argument('file', type=werkzeug.datastructures.FileStorage, location='files')
        args = parse.parse_args()
        audioFile = args['file']
        audioFile.save("your_file_name.jpg")

Answer 2

以下代码的内容应该会有所帮助。

 @app.route('/upload', methods=['GET', 'POST'])
 def upload():
    if request.method == 'POST':
        file = request.files['file']
        extension = os.path.splitext(file.filename)[1]
        f_name = str(uuid.uuid4()) + extension
        file.save(os.path.join(app.config['UPLOAD_FOLDER'], f_name))
        return json.dumps({'filename':f_name})

Answer 3

您可以使用烧瓶中的请求

class UploadImage(Resource):
    def post(self, fname):
        file = request.files['file']
        if file and allowed_file(file.filename):
            # From flask uploading tutorial
            filename = secure_filename(file.filename)
            file.save(os.path.join(app.config['UPLOAD_FOLDER'], filename))
            return redirect(url_for('uploaded_file', filename=filename))
        else:
            # return error
            return {'False'}

http://flask.pocoo.org/docs/0.12/patterns/fileuploads/

Answer 4

from flask import Flask, url_for, send_from_directory, request
import logging, os
from werkzeug import secure_filename

app = Flask(__name__)
file_handler = logging.FileHandler('server.log')
app.logger.addHandler(file_handler)
app.logger.setLevel(logging.INFO)

PROJECT_HOME = os.path.dirname(os.path.realpath(__file__))
UPLOAD_FOLDER = '{}/uploads/'.format(PROJECT_HOME)
app.config['UPLOAD_FOLDER'] = UPLOAD_FOLDER


def create_new_folder(local_dir):
    newpath = local_dir
    if not os.path.exists(newpath):
        os.makedirs(newpath)
    return newpath

@app.route('/', methods = ['POST'])
def api_root():
    app.logger.info(PROJECT_HOME)
    if request.method == 'POST' and request.files['image']:
        app.logger.info(app.config['UPLOAD_FOLDER'])
        img = request.files['image']
        img_name = secure_filename(img.filename)
        create_new_folder(app.config['UPLOAD_FOLDER'])
        saved_path = os.path.join(app.config['UPLOAD_FOLDER'], img_name)
        app.logger.info("saving {}".format(saved_path))
        img.save(saved_path)
        return send_from_directory(app.config['UPLOAD_FOLDER'],img_name, as_attachment=True)
    else:
        return "Where is the image?"

if __name__ == '__main__':
    app.run(host='0.0.0.0', debug=False)

这里是a link

Answer 5

以下是一个使用 curl 上传图像并将其保存在本地的简单程序。

from flask import Flask
from flask_restful import Resource, Api, reqparse
import werkzeug
import cv2
import numpy as np

app = Flask(__name__)
api = Api(app)

parser = reqparse.RequestParser()
parser.add_argument('file',
                    type=werkzeug.datastructures.FileStorage,
                    location='files',
                    required=True,
                    help='provide a file')

class SaveImage(Resource):
    def post(self):
        args = parser.parse_args()
        # read like a stream
        stream = args['file'].read()
        # convert to numpy array
        npimg = np.fromstring(stream, np.uint8)
        # convert numpy array to image
        img = cv2.imdecode(npimg, cv2.IMREAD_UNCHANGED)
        cv2.imwrite("saved_file.jpg", img)

api.add_resource(SaveImage, '/image')

if __name__ == '__main__':
    app.run(debug=True)

你可以像这样使用 curl：

curl localhost:port/image -F file=@image_filename.jpg

Reference

Flask-RESTful - 上传图片

5 个答案: