Question

我的目标是对最后修改的查询集进行排序。我搜索了一个解决方案，并找到了“ xmin”属性（1）。我找不到如何访问此“伪”列的信息，这使我不得不考虑使用原始SQL从查询集中“注释”每个xmin，因为以下SQL起作用了：

SELECT xmin, * FROM media ORDER BY xmin::text::bigint;

我的尝试是以下三个：

将原始SQL应用于查询集我知道做错了，因为SQL不仅适用于当前查询集，而且适用于整个表：

mediaQS = Media.objects.filter(id__in=mediaIDList)
for media in mediaQS.annotate(xmin=RawSQL("SELECT xmin FROM media ;")):
     print "media.xmin",media.xmin`

哪个会引发TypeError：__init __（）至少接受3个参数（给定2个参数）

我的第二次和第三次尝试均基于this issue（Django 2.1）

for media in mediaQS.extra(select={'my_xmin': 'xmin'}):
    print "media.my_xmin",media.my_xmin

哪个给

django.db.utils.ProgrammingError: column "xmin" does not exist
LINE 1: SELECT (xmin) AS "my_xmin", "media"."id", "media"."path_id",...
            ^
HINT:  There is a column named "xmin" in table "media", but it cannot be referenced from this part of the query.

和

    from django.db.models import Expression    
    class XMin(Expression): 
        output_field = models.PositiveIntegerField()

        def as_postgresql(self, compiler, connection):
            return 'xmin', ()

    for media in mediaQS.annotate(my_xmin=XMin()):
         print "media.my_xmin",media.my_xmin

哪个给出类似的错误：

django.db.utils.ProgrammingError: column "xmin" does not exist
LINE 1: ..."path" AS "media_path", T4."path" AS "thumbnail", xmin AS "m...
                                                         ^
HINT:  There is a column named "xmin" in table "media", but it cannot be referenced from this part of the query.

如果有人可以为Django 1.11解决方案提供帮助，我表示感谢。

Django 1.11-通过PostgreSQL xmin列对Queryset进行排序

0 个答案: