如何从最近48小时获取数据-Django [英] How to get data from last 48 hours - Django

问题描述

我正在建立一个新闻网站。我需要显示48小时内观看次数最多的新闻。
因此，我需要首先获取48小时新闻，然后获取其pv。
当前，我正在使用一种非常复杂的方法，该方法来自教程：

I'm building a news website.I need display 48 hours most viewed news. So I need first to get the 48 hours news, and then get its pv. Currently I'm using a very complicated method which is from a tutorial:

def get_two_days_read_data(content_type):
    today = timezone.now().date()
    dates = []
    read_nums = []

    for i in range(2, 0, -1):
        date = today - datetime.timedelta(days=i)
        dates.append(date.strftime('%m/%d'))
        read_details = ReadDetail.objects.filter(content_type=content_type, date=date)
        result = read_details.aggregate(read_num_sum=Sum('read_num'))
        read_nums.append(result['read_num_sum'] or 0)
    return dates, read_nums

我的问题是，有没有更简单的方法？

My question is, is there any easier way?

例如这样的东西：

def newsDetailView(request, news_pk):
    news = get_object_or_404(News, id=news_pk)
    News.objects.filter(id=news_pk).update(pv=F('pv') + 1)
    48_hours_hot_news = news.objects.filter(**48_housrs**).order_by('-pv')

    return render(request, "news_detail.html", {
        'news': news,
        '48_hours_hot_news' : 48_hours_hot_news
    })

有谁可以帮忙？非常感谢！

Any friends can help?Thank you so much!

推荐答案

您可以使用过滤器进行操作，从他的<$ c中减去48小时$ c>创建日期，如果其结果大于 48小时或相等，则您会收到最新消息

You can do it with filter, your substract 48 hours from his date created, in case its result is greater than 48 hours or equal, you got recent news

from datetime import datetime, timedelta

thetime = datetime.now() - timedelta(hours=48)
results = news.objects.filter(date_created__gte=thetime)

注意变量名不能以 digit 开始： 48_hours_hot_news ：错误

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