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django 按照时间降序

django 按照时间降序

作者: 叶叶阿姨 | 来源:发表于2020-01-09 15:54 被阅读0次

    比较简单所以就不写注释了 自己看看吧

    def api_back_descending_news(request):
      # 降序
      if request.POST.get('DescendingOrder'):
          cursor = connection.cursor()
          page = int(request.POST.get('page', 1))
          limit = int(request.POST.get('limit', 10))
          data = find_data(cursor,
                           f'select * from (select id from quick_poll_alerts order by alerts_releasetime asc limit '
                           f'{(page - 1) * limit},{limit}) a left join quick_poll_alerts b on a.id=b.id')
          cursor.execute(f'select count(*) from quick_poll_alerts')
          count = cursor.fetchall()
          cursor.close()
          return JsonResponse({'code': 200, 'data': data, 'count': count})
    

    find_data

    def find_data(cursor, sql):
        # 格式化sql输出
        cursor.execute(sql)
        res = cursor.fetchall()
        # label 多对多
        list = []
        for r in res:
            data1 = {}
            cursor.execute(f'select quick_poll_alerts.id, quick_poll_label.* '
                           f'from quick_poll_alerts, quick_poll_label, quick_poll_alerts_label '
                           f'where quick_poll_alerts.id=quick_poll_alerts_label.alerts_id '
                           f'and quick_poll_label.id=quick_poll_alerts_label.label_id '
                           f'and quick_poll_alerts.id={r[0]};')
            list1 = cursor.fetchall()
            labels = [l[2] for l in list1]
            data1['id'] = r[0]
            data1['img'] = r[4]
            data1['labels'] = labels
            data1['title'] = r[13]
            data1['regional'] = Regional.objects.filter(id=r[8]).first().regional_name
            data1['source'] = Source.objects.filter(id=r[11]).first().source_name
            data1['total_views'] = r[10]
            data1['time'] = r[12].strftime("%Y-%m-%d")
            list.append(data1)
        return list
    
    

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