我有以下数据,
此查询按topicid分组,然后在每个组中获取帖子的最大日期,频率并计算作者的贡献者数量,
info_model = InfoModel.objects.values('topicid')
.annotate( max=Max('date'), freq=Count('postid'),
contributors=Count('author', distinct=True))
此查询可以显示如下,
Q.1(已解决)如何从最近的日期开始按日期订购行?我在查询中添加了.order_by(' date'),这似乎是最明显的解决方案,但这会产生,
完全改变了频率'和#'贡献'。
编辑:可以通过附加.order_by(' -max')来实现订购
Q.2如何显示'帖子'那个日期?所以post列应该显示,
亚
见ya
侨
溜溜球
我认为以下内容适用于{{item.post}},但没有运气。
<table class='table table-striped table-hover'>
<thead>
<tr>
<th>freq</th>
<th>topicid</th>
<th>date</th>
<th>contributors</th>
<th>post</th>
</tr>
</thead>
<tbody>
{% for item in info %}
<tr>
<td>{{ item.freq }}</td>
<td>{{ item.topicid }}</td>
<td>{{ item.max }}</td>
<td>{{ item.contributors }}</td>
<td>{{ item.post }}</td>
</tr>
{% endfor %}
</tbody>
</table>
谢谢,
编辑:
我可以使用原始SQL获得正确的结果,但不能使用Django查询,
info_model = list(InfoModel.objects.raw('SELECT *,
max(date),
count(postid) AS freq,
count(DISTINCT author) AS contributors FROM
crudapp_infomodel GROUP BY topicid ORDER BY date DESC'))
我在此处简化并重新发布此问题Rewrite raw SQL as Django query
答案 0 :(得分:1)
以下视图合并了两个查询来解决问题,
def info(request):
info_model = InfoModel.objects.values('topic')
.annotate( max=Max('date'),
freq=Count('postid'),
contributors=Count('author', distinct=True))
.order_by('-max')
info2 = InfoModel.objects.all()
columnlist = []
for item in info2:
columnlist.append([item])
for item in info_model:
for i in range(len(columnlist)):
if item['max'] == columnlist[i][0].date:
item['author'] = columnlist[i][0].author
item['post'] = columnlist[i][0].post
print item['max']
paginator = Paginator(info_model, 20)
page = request.GET.get('page')
try:
info = paginator.page(page)
except PageNotAnInteger:
# If page is not an integer, deliver first page.
info = paginator.page(1)
except EmptyPage:
# If page is out of range (e.g. 9999), deliver last page of results.
info = paginator.page(paginator.num_pages)
return render(request, 'info.html', {'info': info})