Django:显示上传文件的文件名 [英] Django: Display File Name for Uploaded File

问题描述

在我的模型中，我定义了一个 FileField ，我的模板将其显示为链接.我的问题是链接的文件将URL显示为名称.在html页面上显示的是:

上传的文件:./picture.jpg

我查看了 DjangoDocs 关于文件名的信息和以前的SO问题，但无法弄清楚./p>

我如何:

1. 让它显示不同的名称，而不是URL.
2. 允许上传文件的管理员给它起一个名字，然后可以在模板上查看它.

我的models.py:

  class model_name(models.Model):附件= models.FileField() 

我的views.py(如果存在条目，则显示它，如果不存在，则返回消息):

django.core.files中的

 导入文件从vendor_db.models导入model_namedef网页(请求，ID):尝试:变量= model_name.objects.get(id = id)除了model_name.DoesNotExist:引发Http404('该项目不存在')返回render(request，'page.html'，{'variable':变量，}) 

我的page.html:

 < p>上传的文件:< a href ="{{variable.attachment.url}}"> {{variable.attachment}}</a></p> 

解决方案

对于您的代码:

  class model_name(models.Model):附件= models.FileField() 

附件是一个 FileField 对象，它具有一个名为 filename 的属性，因此只需索取该属性即可.即

  foo = model_name.objects.create(attachment = some_file)foo.attachment.filename#以字符串形式返回文件名 

In my model, I have a defined a FileField which my template displays it as a link. My problem is that the linked file displays the url as the name. What shows on html page is:

Uploaded File: ./picture.jpg

I've looked on the DjangoDocs regarding file names and a previous S.O. question, but just can't figure it out.

How can I:

1. Have it display a different name, not a url.
2. Allow the admin who uploaded the file to give it a name, which would then be viewed on the template.

my models.py:

class model_name(models.Model):
    attachment = models.FileField()

my views.py (if entry exists, display it, if not, return message):

from django.core.files import File
from vendor_db.models import model_name

def webpage(request, id):
    try:
        variable = model_name.objects.get(id=id)
    except model_name.DoesNotExist:
        raise Http404('This item does not exist')
    return render(request, 'page.html', {
        'variable': variable,
    })

my page.html:

<p>Uploaded File: <a href="{{ variable.attachment.url }}">{{ variable.attachment }}</a></p>

解决方案

For your code:

class model_name(models.Model):
    attachment = models.FileField()

attachment is a FileField object, that has a property called filename, so just ask for that property. i.e.

foo = model_name.objects.create(attachment=some_file)
foo.attachment.filename # filename as a string is returned

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