如何将 Django 中 Count() 的输出转换为 Django 1.8 中的 FloatField

Question

我正在使用 django 1.8。

我有两个模型：-

class Query(models.Model):
    user = models.ForeignKey(settings.AUTH_USER_MODEL)
    title = models.TextField()
    details = models.TextField()
    pub_date = models.DateTimeField('date published')
    post_score=models.FloatField(default=0.0)
    ....

class Tags(models.Model):
    """docstring for Tags"""
    tagname=models.TextField()
grouppostlist=models.ManyToManyField(Query,through='GroupPostsTag',through_fields=('tag','query'))

# and another model(it stores tags for a query)

class GroupPostsTag(models.Model):
    query=models.ForeignKey('Query')
    tag=models.models.ForeignKey('Tags')

现在我想根据 "number of tags" 和查询的 "post_score" 的总和对查询进行排序。我正在寻找这样的东西：-

tagged_user_posts = Query.objects.filter(Q(tags__id__in=tags_list)).annotate(num_tags=Cast(Count('tags'),models.FloatField())).annotate(post_scores=F('num_tags')+F('post_score')).order_by('-post_scores')

Cast 在 django 1.10 中提供。那么我可以使用什么替代方案？

Answer 1

正如我在评论中所说，您可以简单地复制 Cast class 的源代码并将其保存在您的项目中并使用它。

事实证明，事情并不那么简单。您还需要更改 Cast.as_sql as it is in v1.10。而且，无需对 Func.as_sql 进行 super 调用。

只需复制以下代码并将其保存在您的项目中。我已经测试过它并且有效。

from django.db.models import Func, fields


class Cast(Func):
    """
    Coerce an expression to a new field type.
    """
    function = 'CAST'
    template = '%(function)s(%(expressions)s AS %(db_type)s)'

    mysql_types = {
        fields.CharField: 'char',
        fields.IntegerField: 'signed integer',
        fields.FloatField: 'signed',
    }

    def __init__(self, expression, output_field):
        super(Cast, self).__init__(expression, output_field=output_field)

    def as_sql(self, compiler, connection, function=None, template=None, arg_joiner=None, **extra_context):
        if 'db_type' not in extra_context:
            extra_context['db_type'] = self._output_field.db_type(connection)
        connection.ops.check_expression_support(self)
        sql_parts = []
        params = []
        for arg in self.source_expressions:
            arg_sql, arg_params = compiler.compile(arg)
            sql_parts.append(arg_sql)
            params.extend(arg_params)
        data = self.extra.copy()
        data.update(**extra_context)
        # Use the first supplied value in this order: the parameter to this
        # method, a value supplied in __init__()'s **extra (the value in
        # `data`), or the value defined on the class.
        if function is not None:
            data['function'] = function
        else:
            data.setdefault('function', self.function)
        template = template or data.get('template', self.template)
        arg_joiner = arg_joiner or data.get('arg_joiner', self.arg_joiner)
        data['expressions'] = data['field'] = arg_joiner.join(sql_parts)
        return template % data, params


    def as_mysql(self, compiler, connection):
        extra_context = {}
        output_field_class = type(self._output_field)
        if output_field_class in self.mysql_types:
            extra_context['db_type'] = self.mysql_types[output_field_class]
        return self.as_sql(compiler, connection, **extra_context)

    def as_postgresql(self, compiler, connection):
        # CAST would be valid too, but the :: shortcut syntax is more readable.
        return self.as_sql(compiler, connection, template='%(expressions)s::%(db_type)s')

Answer 2

正如您在 django 1.10 的问题中提到的，您可以使用内置的 Cast 函数。对于旧版本，您可以使用 Func() expression

from django.db.models import Count, Func, F, ExpressionWrapper, FloatField
tagged_user_posts = Query.objects.filter(Q(tags__id__in=tags_list)).
    annotate(num_tags=Func(Count('tags'), 
                           template='%(function)s(%(expressions)s AS %(type)s)', 
                           function='Cast', 
                           type='float')
    ).
    annotate(post_scores=ExpressionWrapper(F('num_tags')+F('post_score'), output_field=FloatField())).
    order_by('-post_scores')

Answer 3

您可以将 Cast 定义为

def Cast(what, as_type):
    if isinstance(as_type, Field):
        as_type = as_type.db_type(connection)
    return Func(
        what,
        template='%(function)s(%(expressions)s AS %(type)s)',
        function='Cast',
        type=as_type,
    )

如何将 Django 中 Count() 的输出转换为 Django 1.8 中的 FloatField

how to cast the output of Count() in django to FloatField in django 1.8

django

django-models

django-orm