Django 如果模板中的值为“无”,则显示字符串

Django 如果模板中的值为“无”,则显示字符串,django,django-models,django-templates,Django,Django Models,Django Templates,如果值为None,我希望在表中显示-,而不是呈现None 但是,我得到以下错误: Could not parse the remainder: ' or '-'' from 'employee.full_name or '-'' 如果可能的话,我想避免做大量的if-else语句 这是我的密码: {% for employee in employees %} <tr> <td>{{employee.full_name or '-'}}</td>

如果值为
None
,我希望在表中显示
-
,而不是呈现
None

但是,我得到以下错误:

Could not parse the remainder: ' or '-'' from 'employee.full_name or '-''
如果可能的话,我想避免做大量的if-else语句

这是我的密码:

  {% for employee in employees %}
  <tr>
    <td>{{employee.full_name or '-'}}</td>
    <td>{{employee.position or '-'}}}</td>
    <td>{{employee.dob or '-'}}}</td>
    <td>{{employee.phone or '-'}}}</td>
    <td>{{employee.email or '-'}}}</td>
    <td>{{employee.address or '-'}}}</td>
    <td>{{employee.joined or '-'}}}</td>
  </tr>
  {% endfor %}
您可以使用:

{employees%中的员工百分比]
{{employee.full{u name}如果没有默认值:'-'}
{{employee.position}默认值{u如果没有:'-'}
{{employee.dob}默认值{u如果没有:'-'}
{{employee.phone |默认值_如果没有:'-'}
{{employee.email |默认值_如果没有:'-'}
{{employee.address}默认值{u如果没有:'-'}
{{employee.joined}默认值{u如果没有:'-'}

{%endfor%}
您可以使用
template\u标记
dog = None
print(dog or '-')
{% for employee in employees %}
  <tr>
    <td>{{ employee.full_name|default_if_none:'-' }}</td>
    <td>{{ employee.position|default_if_none:'-' }}</td>
    <td>{{ employee.dob|default_if_none:'-' }}</td>
    <td>{{ employee.phone|default_if_none:'-' }}</td>
    <td>{{ employee.email|default_if_none:'-' }}</td>
    <td>{{ employee.address|default_if_none:'-' }}</td>
    <td>{{ employee.joined|default_if_none:'-' }}</td>
  </tr>
{% endfor %}