如何摆脱由Django Form的RadioSelect生成的虚假选择
发布于 2021-01-29 15:59:30
我在Django 1.3上使用ModelForm。
models.py:
class UserProfile(models.Model):
...
gender = models.CharField(max_length=1, blank=True, choices=(('M', 'Male'), ('F', 'Female'), ('Unspecified', '')), default='M')
...
forms.py:
class UserProfileForm(ModelForm):
class Meta:
model = UserProfile
fields = ('gender')
widgets = {
'gender': forms.RadioSelect(),
}
当此小部件呈现为HTML时,我得到了
<ul>
<li><label for="id_gender_0"><input type="radio" id="id_gender_0" value="" name="gender" />---------</label></li>
<li><label for="id_gender_1"><input checked="checked" type="radio" id="id_gender_1" value="M" name="gender" /> Male</label></li>
<li><label for="id_gender_2"><input type="radio" id="id_gender_2" value="F" name="gender" />Female</label></li>
<li><label for="id_gender_3"><input type="radio" id="id_gender_3" value="" name="gender" /> Unspecified</label></li>
</ul>
问题: 如何摆脱伪造的“ --------”?
几个月前,另一个stackoverflow用户也提出了相同的问题(Here)。我已经在那里尝试了可接受的解决方案(如您所见),但这对我没有用。
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1 个回答
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即使没有blank = True,它也会显示额外的输入。我创建了一个新的小部件:
from itertools import chain from django.forms import RadioSelect from django.utils.encoding import force_unicode class RadioSelectNotNull(RadioSelect): def get_renderer(self, name, value, attrs=None, choices=()): """Returns an instance of the renderer.""" if value is None: value = '' str_value = force_unicode(value) # Normalize to string. final_attrs = self.build_attrs(attrs) choices = list(chain(self.choices, choices)) if choices[0][0] == '': choices.pop(0) return self.renderer(name, str_value, final_attrs, choices)