Create a Simple API Using Django REST Framework in Python

WHAT IS AN API

API stands for application programming interface. API basically helps one web application to communicate with another application.

Let's assume you are developing an android application which has feature to detect the name of a famous person in an image.

Introduce to achieve this you have 2 options:

option 1:

Option 1 is to collect the images of all the famous personalities around the world, build a machine learning/ deep learning or whatever model it is and use it in your application.

option 2:

Just use someone elses model using api to add this feature in your application.

Large companies like Google, they have their own personalities. So if we use their Api, we would not know what logic/code whey have writting inside and how they have trained the model. You will only be given an api(or an url). It works like a black box where you send your request(in our case its the image), and you get the response(which is the name of the person in that image)

Here is an example:

Django REST框架创建一个简单的Api实例讲解

PREREQUISITES

conda install jango
conda install -c conda-forge djangorestframework

Step 1

Create the django project, open the command prompt therre and enter the following command:

django-admin startproject SampleProject

Step 2

Navigate the project folder and create a web app using the command line.

python manage.py startapp MyApp

Step 3

open the setting.py and add the below lines into of code in the INSTALLED_APPS section:

'rest_framework',
'MyApp'

Step 4

Open the views.py file inside MyApp folder and add the below lines of code:

from django.shortcuts import render
from django.http import Http404
from rest_framework.views import APIView
from rest_framework.decorators import api_view
from rest_framework.response import Response
from rest_framework import status
from django.http import JsonResponse
from django.core import serializers
from django.conf import settings
import json
# Create your views here.
@api_view(["POST"])
def IdealWeight(heightdata):
 try:
  height=json.loads(heightdata.body)
  weight=str(height*10)
  return JsonResponse("Ideal weight should be:"+weight+" kg",safe=False)
 except ValueError as e:
  return Response(e.args[0],status.HTTP_400_BAD_REQUEST)

Step 5

Open urls.py file and add the below lines of code:

from django.conf.urls import url
from django.contrib import admin
from MyApp import views
urlpatterns = [
 url(r'^admin/', admin.site.urls),
 url(r'^idealweight/',views.IdealWeight)
]

Step 6

We can start the api with below commands in command prompt:

python manage.py runserver

Finally open the url:

http://127.0.0.1:8000/idealweight/

Django REST框架创建一个简单的Api实例讲解

References:

Create a Simple API Using Django REST Framework in Python

以上就是本次介绍的关于Django REST框架创建一个简单的Api实例讲解内容,感谢大家的学习和对的支持。

华山资源网 Design By www.eoogi.com
广告合作:本站广告合作请联系QQ:858582 申请时备注:广告合作(否则不回)
免责声明:本站资源来自互联网收集,仅供用于学习和交流,请遵循相关法律法规,本站一切资源不代表本站立场,如有侵权、后门、不妥请联系本站删除!
华山资源网 Design By www.eoogi.com

《魔兽世界》大逃杀!60人新游玩模式《强袭风暴》3月21日上线

暴雪近日发布了《魔兽世界》10.2.6 更新内容,新游玩模式《强袭风暴》即将于3月21 日在亚服上线,届时玩家将前往阿拉希高地展开一场 60 人大逃杀对战。

艾泽拉斯的冒险者已经征服了艾泽拉斯的大地及遥远的彼岸。他们在对抗世界上最致命的敌人时展现出过人的手腕,并且成功阻止终结宇宙等级的威胁。当他们在为即将于《魔兽世界》资料片《地心之战》中来袭的萨拉塔斯势力做战斗准备时,他们还需要在熟悉的阿拉希高地面对一个全新的敌人──那就是彼此。在《巨龙崛起》10.2.6 更新的《强袭风暴》中,玩家将会进入一个全新的海盗主题大逃杀式限时活动,其中包含极高的风险和史诗级的奖励。

《强袭风暴》不是普通的战场,作为一个独立于主游戏之外的活动,玩家可以用大逃杀的风格来体验《魔兽世界》,不分职业、不分装备(除了你在赛局中捡到的),光是技巧和战略的强弱之分就能决定出谁才是能坚持到最后的赢家。本次活动将会开放单人和双人模式,玩家在加入海盗主题的预赛大厅区域前,可以从强袭风暴角色画面新增好友。游玩游戏将可以累计名望轨迹,《巨龙崛起》和《魔兽世界:巫妖王之怒 经典版》的玩家都可以获得奖励。