django中多个字段的计数(不同)

Ice*_*man 8 django django-queryset django-filter

这里我尝试将 Django rawsql 查询转换为 Django 查询集。

但问题是——如何在两列中实现不同?

SQL查询:

count(distinct ra, category) as order_type
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Django 查询集 - order_type=Count('ra', 'category',distinct=True) - 这会导致错误。

queryset = Model.objects.raw("select id,category,count(category) 
as total_orders, count(distinct ra, category) as order_type, 
SUM(CASE WHEN service_status = 'success' THEN 1 ELSE 0 END) as total_success_order, 
SUM(CASE WHEN service_status = 'failed' THEN 1 ELSE 0 END) as total_failed_order 
from table 
group by ra;")


queryset = Model.objects.values('ra').annotate(category=F('category'),\
order_type=Count('ra', 'category', distinct=True),total_orders=Count('category'),\
total_success_order=Count('service_status', filter=Q(service_status='success')),\
total_failed_order=Count('service_status', filter=Q(service_status='failed'))).order_by()
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正确的语法应该是什么?

数据-

category   ra   service_status
cat1       11      success
cat1       12      success
cat2       11      success
cat2       11      success
cat1       15      success
cat3       5       failed
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ra 11 应返回为order_type = 2(即,如果相同的 ra 具有两个不同的类别)

O/p-

Category  order type   total order   successful  failed
 cat1        1            10              5         5
 cat2        2            14              10       4
 cat3        1            9               9        0
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小智 7

有点晚了,希望这对某人有帮助。

您可以使用 Django 数据库函数Concat

from django.db.models.functions import Concat

queryset = Model.objects.values('ra').annotate(order_type=Count(Concat('ra', 'category'), distinct=True))
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