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筛选有选项的字段

django-filter django-rest-framework django

0

blueFast · 技术社区 · 7 年前

operation = models.CharField(max_length=10, choices=OPERATIONS)

使用此过滤器工作:

class OperationFilter(django_filters.Filter):

    def filter(self, qs, value):
        try:
            qs = qs.filter(operation=value.upper())
        except:
            pass
        return qs

带url:

/api/v1/operation/?operation=CREATE

但是有默认的过滤器(没有额外的 OperationFilter )失败原因:

{
    "operation": [
        "Select a valid choice. %(value)s is not one of the available choices."
    ]
}

对于其他非选项字段,默认筛选器工作正常:

/api/v1/operation/?recipient=recipient-19

编辑

OPERATIONS :

from enum import Enum

def enum_as_choices(enum_class):
    """From an enum class, generate choices for a django field"""
    return ((entry, entry.value) for entry in enum_class)

class OperationType(Enum):
    CREATE = 'CREATE'
    STATUS = 'STATUS'
    EXPAND = 'EXPAND'
    DELETE = 'DELETE'

OPERATIONS = enum_as_choices(OperationType)

3 回复 | 直到 7 年前

1

Mohsen Mahmoodi 7 年前

您编写的选项将转换为此pythonic表示:

(
    ('OperationType.CREATE', 'CREATE'), 
    ('OperationType.STATUS', 'STATUS'), 
    ('OperationType.EXPAND', 'EXPAND'), 
    ('OperationType.DELETE', 'DELETE')
)

'OperationType.CREATE' 等等。所以你应该把你的选择改成普通的常量选择,或者你应该用类似 '操作类型.创建' 这不是一个好的选择。

你也可以改变你的想法 enum_as_choices

def enum_as_choices(enum_class):
    """From an enum class, generate choices for a django field"""
    return ((entry.name, entry.value) for entry in enum_class)

2

Gabriel Muj 7 年前

您正在使用 django_filters 包,我建议您阅读文档,因为您已经有了对此的支持

https://django-filter.readthedocs.io/en/master/ref/filters.html#choicefilter

只要指出你的观点 choices

3

0

Daniel Holmes 7 年前

OPERATIONS . 为此,添加如下内容:

OPERATIONS = (
    ('', 'NONE'),
    # the rest of your choices here...
)

operation = models.CharField(max_length=10, choices=OPERATIONS, default='NONE')