Python/Django

[Django] API 만들기, Request 보내기

오늘의개발부 2019. 7. 31. 12:11
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* Request 보내기

## python

>>> import requests, json
>>> tmpData = {
  "cond" : {
    "VIENNA_CODE" : "0101"
  }
}
>>> jsonData = json.dumps(tmpData)
>>> r = requests.post("http://localhost:8888/mongodb/", data=jsonData)

 

## 오류일경우

>>> r.status_code
500
>>> with open("d:\\err.html", "w", encoding="utf-8") as f:
   f.write(r.text)

 

## 정상일경우

>>> r.status_code 
200 

 

 

* Django API 만들기

## urls.py

from django.conf.urls import url

from django.contrib import admin

from tripletBuilder import views

from tripletBuilder import mongo

urlpatterns = [

    url(r'^admin/', admin.site.urls),

    url(r'^mongodb/', mongo.getMongo),

]

 

## mongo.py

@csrf_exempt

def getMongo(request):

    ''' AJAX 조회 '''

    logger.info('## getMongo ########################################################')

    cond_json = json.loads(request.body.decode('utf-8'))

    cond = cond_json["cond"]

    print(cond)

    ```

    { 
      "VIENNA_CODE" : "0101" 
    } 

    ```

    result_json = cond

    return JsonResponse(result_json)

 

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