Python Flask-RESTful - 上传图片
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Flask-RESTful - Upload image
提问by Sigils
I was wondering on how do you upload files by creating an API service?
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'}
Route
路线
api.add_resource(UploadImage, '/api/uploadimage/<string:fname>')
And then the HTML
然后是 HTML
<input type="file" name="file">
I have enabled CORS on the server side
我在服务器端启用了 CORS
I am using angular.js as front-end and ng-uploadif that matters, but can use CURL statements too!
如果重要的话，我使用 angular.js 作为前端和ng-upload，但也可以使用 CURL 语句！
采纳答案by Sibelius Seraphini
class UploadWavAPI(Resource):
def post(self):
parse = reqparse.RequestParser()
parse.add_argument('audio', type=werkzeug.FileStorage, location='files')
args = parse.parse_args()
stream = args['audio'].stream
wav_file = wave.open(stream, 'rb')
signal = wav_file.readframes(-1)
signal = np.fromstring(signal, 'Int16')
fs = wav_file.getframerate()
wav_file.close()
You should process the stream, if it was a wav, the code above works. For an image, you should store on the database or upload to AWS S3 or Google Storage
您应该处理流，如果它是 wav，则上面的代码有效。对于图像，您应该存储在数据库中或上传到 AWS S3 或 Google Storage
回答by iJade
Something on the lines of the following code should help.
以下代码行中的某些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})
回答by Ron Harlev
The following is enough to save the uploaded file
以下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")
回答by Jethro Guce
you can use the request from flask
您可以使用来自烧瓶的请求
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'}
回答by Mudidi Jimmy
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)
Here is a link
这是一个链接
