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Django实现列表页-----制作Json格式输出模板(一)

时间:2020-05-07 16:58:33      阅读:97      评论:0      收藏:0      [点我收藏+]
 1 # -*- coding: utf-8 -*-
 2 from dss.Serializer import serializer
 3 from django.http import HttpResponse
 4 
 5 
 6 def response_as_json(data, foreign_penetrate=False):
 7     jsonString = serializer(data=data, output_type="json", foreign=foreign_penetrate)
 8     response = HttpResponse(
 9         # json.dumps(data, cls=MyEncoder),
10         jsonString,
11         content_type="application/json; charset=utf-8",
12     )
13     response["Access-Control-Allow-Origin"] = "*"
14     return response
15 
16 
17 def json_response(data, code=200, foreign_penetrate=False, **kwargs):
18     data = {
19         "code": code,
20         "msg": "success",
21         "data": data,
22     }
23     data.update(**kwargs)
24     return response_as_json(data, foreign_penetrate=foreign_penetrate)
25 
26 
27 def json_error(error_string="", code=500, **kwargs):
28     data = {
29         "code": code,
30         "msg": error_string,
31         "data": {}
32     }
33     data.update(kwargs)
34     return response_as_json(data)
35 
36 
37 JsonResponse = json_response
38 JsonError = json_error

 

Django实现列表页-----制作Json格式输出模板(一)

原文:https://www.cnblogs.com/nieliangcai/p/12844047.html

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