Question

我想用数据库中的数据填充表单下拉列表。这些数据不是直接来自模型，而是来自原始查询。

当数据库可用且已经生成迁移时，此方法有效。否则，生成迁移（python manage.py makemigrations myapp）会失败，因为Django评估无法找到适当表的_all_departments()。

def _all_departments() -> List[Tuple[str, str]]:
    from django.db import connection
    with connection.cursor() as cursor:
        cursor.execute("select distinct department from crm_mytable order by department")
        return [(row[0], row[0]) for row in cursor.fetchall()]


class MyForm(forms.Form):
    department = forms.CharField(
        widget=forms.SelectMultiple(choices=_all_departments()))

我天真地尝试手动更新__init__上的选择，但没有成功（选择始终为空）：

class MyForm(forms.Form):
    def __init__(self, *args, **kwargs):
        super(MyForm, self).__init__(*args, **kwargs)
        self.fields['department'].widget.choices = _all_departments()

    department = forms.CharField(
        widget=forms.SelectMultiple(choices=[]))

正确选择按需选择的方式是什么？

Answer 1

您不应不要将choices传递给 widget ，而应传递给 field 。您可能还想使用MultipleChoiceField [Django-doc]，它使用SelectMultiple [Django-doc]作为默认小部件：

class MyForm(forms.Form):

    department = forms.MultipleChoiceField(choices=[])

    def __init__(self, *args, **kwargs):
        super(MyForm, self).__init__(*args, **kwargs)
        self.fields['department'].choices = _all_departments()

使用数据库（而不是模型）中的选项填充Django表单字段

1 个答案: