Django中實現簡單的CRUD
阿新 • • 發佈:2021-11-03
# models.py
class User(models.Model):
id = models.IntegerField(primary_key=True)
username = models.CharField(max_length=32, unique=True)
# views.py @require_GET def get_user_list(request): users = User.objects.all() data = [{'id': user.id, 'username': user.username} for user in users] return JsonResponse({'code': 200, 'data': data, 'message': '查詢成功'}) @require_GET def add_user(request): request.encoding = 'utf-8' if request.GET: username = request.GET['username'] id = request.GET['id'] if username and id: User.objects.create(id=id, username=username) return JsonResponse({'code': 200, 'msg': '新增成功'}) else: return HttpResponse('請傳入username和id') else: return HttpResponse('請求方式錯誤') @require_GET def delete_user(request): request.encoding = 'utf-8' if request.GET: id = request.GET['id'] if username: User.objects.filter(id=id).delete() return JsonResponse({'code': 200, 'msg': '刪除成功'}) else: return HttpResponse('請傳入id') else: return HttpResponse('請求方式錯誤') @require_GET def update_use(request): request.encoding = 'utf-8' if request.GET: id = request.GET['id'] username = request.GET['username'] if username and id: User.objects.filter(id=id).update(username=username) return JsonResponse({'code': 200, 'msg': '修改成功'}) else: return HttpResponse('請傳入username和id') else: return HttpResponse('請求方式錯誤')
本文來自部落格園,作者:胡圖人,轉載請註明原文連結:https://www.cnblogs.com/huturen/p/15504767.html