Question

这似乎与this question非常相似。单击链接以按pass_count排序时，我得到：

无法将关键字u'pass_count'解析为字段。选择是：build，build_no

models.py：

class Run(DashboardBaseModel):
class Meta:
    db_table = 'runs'

build_no   = models.CharField(max_length=200, db_index = True)
release    = models.CharField(max_length=200, db_index = True)
extra_fields = [
        'pass_count',
        ]

@property
def pass_count(self):
    passes = 0
    for build in self.build_set.all():
        passes += build.pass_count
    return passes

tables.py：

class CombineTable(tables.Table):
build_no = tables.LinkColumn('run', args=[A('release'), A('id')], verbose_name="Build")
pass_count = tables.Column(verbose_name="Passed", attrs={"td": {"class": "num"}, "th": {"class": "num"}})

有办法解决这个问题吗？

Answer 1

您需要将数据转换为字典，以允许非db字段进行排序。

runs = Run.objects.all()

table_data = []
for run in runs:
    passes = 0
    for build in run.build_set.all():
        passes += build.pass_count
    table_data.append(dict(
        build_no=run.build_no,
        release=run.release,
        pass_Count=passes,
    ))

table = CombineTable(table_data)

注意：这将为每次运行查询数据库一次。解决方案需要在构建上使用QuerySet，并在所有构建上循环，根据运行将每个构建的pass_count添加到字典中。

非queryset数据排序

1 个答案: