django文件字段仅在模板中返回文件名

I've got a field in my model of type FileField. This gives me an object of type type File, which has the following method:

我的FileField类型模型中有一个字段。这给了我一个类型类型文件的对象，它有以下方法:

File.name: The name of the file including the relative path from MEDIA_ROOT.

文件名:文件的名称，包括来自MEDIA_ROOT的相对路径。

What I want is something like .filename that will only give me the filename and not the path as well

我想要的是像.filename这样的东西，它只给我文件名，而不是路径

something like:

喜欢的东西:

{% for download in downloads %}
    <div class="download">
        <div class="title">{{download.file.filename}}</div>
    </div>
{% endfor %}

which would give something like myfile.jpg

哪个会给出类似myfile.jpg的东西

thanks

谢谢

3 个解决方案

#1

100

In your model definition:

在您的模型中定义:

import os

class File(models.Model):
    file = models.FileField()
    ...

    def filename(self):
        return os.path.basename(self.file.name)

#2

You can do this by creating a template filter:

您可以通过创建模板过滤器来实现这一点:

In myapp/templatetags/filename.py:

在myapp / templatetags / filename.py:

import os

from django import template


register = template.Library()

@register.filter
def filename(value):
    return os.path.basename(value.file.name)

And then in your template:

然后在模板中:

{% load filename %}

{# ... #}

{% for download in downloads %}
  <div class="download">
      <div class="title">{{download.file|filename}}</div>
  </div>
{% endfor %}

#3

You can access the filename from the file field object with the name property.

您可以使用name属性从file field对象访问文件名。

class CsvJob(Models.model):

    file = models.FileField()

then you can get the particular objects filename using.

然后您可以使用特定的对象文件名。

obj = CsvJob.objects.get()
obj.file.name property

#1

100