烧瓶静息分页

Question

我需要在很短的期限内将一个非常简单的API组合在一起。烧瓶安宁似乎很理想，除了以下几点：我在分页文档中找不到任何内容。给定一个像这样的简单端点：

from flask import Flask, request 
from flask_restful import Resource, Api 
from sqlalchemy import create_engine
import json 

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

class Employees(Resource):
    def get(self):
        return json.dumps([{'employees': 'hello world'} for i in range(1000)])

api.add_resource(Employees, '/employees')

if __name__ == '__main__':
    app.run(port='5002')

有什么办法可以使flask_restful对端点进行分页，以便每页仅接收100个这样的字典，并具有“下一个”和“上一个”的URL？如果不是，是否有可能在Flask中以其他方式创建分页？谢谢。

Answer 1

You can either use:

• Pagination provided by flask_sqlalchemy (API documentation can be found here)
• Custom method to paginate existing data as shown in this tutorial by Avi Aryan.

As I am not sure if you are using flask_sqlalchemy or any model information, I am showing the custom pagination technique.

I have modified the data to show the employee id. And I also used jsonify from Flask.

from flask import Flask, request, jsonify, abort
from flask_restful import Resource, Api 


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

data = [{'employee_id': i+1} for i in range(1000)]

def get_paginated_list(results, url, start, limit):
    start = int(start)
    limit = int(limit)
    count = len(results)
    if count < start or limit < 0:
        abort(404)
    # make response
    obj = {}
    obj['start'] = start
    obj['limit'] = limit
    obj['count'] = count
    # make URLs
    # make previous url
    if start == 1:
        obj['previous'] = ''
    else:
        start_copy = max(1, start - limit)
        limit_copy = start - 1
        obj['previous'] = url + '?start=%d&limit=%d' % (start_copy, limit_copy)
    # make next url
    if start + limit > count:
        obj['next'] = ''
    else:
        start_copy = start + limit
        obj['next'] = url + '?start=%d&limit=%d' % (start_copy, limit)
    # finally extract result according to bounds
    obj['results'] = results[(start - 1):(start - 1 + limit)]
    return obj

class Employees(Resource):
    def get(self):
        return jsonify(get_paginated_list(
        data, 
        '/employees', 
        start=request.args.get('start', 1), 
        limit=request.args.get('limit', 20)
    ))

api.add_resource(Employees, '/employees')

if __name__ == '__main__':
    app.run(port='5002', debug=True)

Output:

Footnote:

• API can be called with parameter or without parameter. Example of valid API calls:
  • http://127.0.0.1:5002/employees
  • http://127.0.0.1:5002/employees?start=41&limit=20
  • http://127.0.0.1:5002/employees?limit=5
  • http://127.0.0.1:5002/employees?start=100
• Default value for start is 1 and limit is 20.
• If the start value is greater than data length or limit is negative then the API will return HTTP 404 error with an error message: