在 Django 中创建模板时迭代模型属性 [英] Iterating over model attributes when creating a template in Django

问题描述

我在 Google App Engine 中使用 Django.如果我有课

I'm using Django in Google App Engine. If I have the class

class Person():
    first_name = StringProperty()
    last_name = StringProperty()

我有一个实例，其中 Person.first_name = Bob 和 Person.last_name = Vance，我可以创建一个模板来迭代 Person 属性以生成:

and I have an instance where Person.first_name = Bob and Person.last_name = Vance, can I create a template that iterates over the Person attributes to produce:

<tr>
<td>First</td>
<td>Bob</td>
</tr>
<tr>
<td>Last</td>
<td>Vance</td>
</tr>

也许更简洁，是否有一个 model.as_table() 方法可以打印出我的实例 Bob Vance 的属性?

Perhaps more succinctly, is there a model.as_table() method that will print out the attributes of my instance, Bob Vance?

推荐答案

在模板中，您无法访问 __underscored__ 属性或函数.我建议您在模型/类中创建一个函数:

In template you cannot access __underscored__ attributes or functions. I suggest instead you create a function in your model/class:

class Person(models.Model):
  first_name = models.CharField(max_length=256)
  last_name = models.CharField(max_length=256)

  def attrs(self):
     for attr, value in self.__dict__.iteritems():
        yield attr, value

 def sorted_attrs(self):
     # Silly example of sorting
     return [(key, self.__dict__[key]) for key in sorted(self.__dict__)]

在模板中它只是:

 <tr>
 {% for name, value in person.attrs %}
   <td>{{name}}</td> 
   <td>{{value}}</td>
 {% endfor %}
 </tr>

现在这会给你first_name"而不是First"，但你明白了.您可以将该方法扩展为 mixin，或者存在于父类中等.同样，如果您有几个想要迭代的人员对象，则可以使用它:

Now this will give you "first_name" instead of "First", but you get the idea. You can extend the method to be a mixin, or be present in a parent-class etc.. Similarly you can use this if you have a few person objects you want to iterate over:

{% for person in persons %}
 <tr>
 {% for name, value in person.attrs %}
   <td>{{name}}</td> 
   <td>{{value}}</td>
 {% endfor %}
 </tr>
{% endfor %}

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