How to change the file name of an uploaded file in

问题:

Is it possible to change the file name of an uploaded file in django? I searched, but couldn't find any answer.

My requirement is whenever a file is uploaded its file name should be changed in the following format.

format = userid + transaction_uuid + file_extension

Thank you very much...

回答1:

How are you uploading the file? I assume with the FileField.

The documentation for FileField.upload_to says that the upload_to field,

may also be a callable, such as a function, which will be called to obtain the upload path, including the filename. This callable must be able to accept two arguments, and return a Unix-style path (with forward slashes) to be passed along to the storage system. The two arguments that will be passed are:

"instance": An instance of the model where the FileField is defined. More specifically, this is the particular instance where the current file is being attached.

"filename":The filename that was originally given to the file. This may or may not be taken into account when determining the final destination path.

So it looks like you just need to make a function to do your name handling and return the path.

def update_filename(instance, filename):
    path = "upload/path/"
    format = instance.userid + instance.transaction_uuid + instance.file_extension
    return os.path.join(path, format)

回答2:

You need to have a FileField with the upload_to that calls to a callback, see [1]

Your callback should call a wrapper method which gets an instance as one of the params and filename as the other. [2]

Change it the way you like and return the new path [3]

1. LOGIC

FileField(..., upload_to=method_call(params),....)

2. define method

def method_call(params):
    return u'abc'

3. Wrapper:

def wrapper(instance, filename):
    return method

this is the rapper method that you need for getting the instance.

def wrapper(instance, filename):
... Your logic
...
return wrapper

Complete Code

def path_and_rename(path, prefix):
    def wrapper(instance, filename):
        ext = filename.split('.')[-1]
        project = "pid_%s" % (instance.project.id,)
        # get filename
        if instance.pk:
            complaint_id = "cid_%s" % (instance.pk,)
            filename = '{}.{}.{}.{}'.format(prefix, project, complaint_id, ext)
        else:
            # set filename as random string
            random_id = "rid_%s" % (uuid4().hex,)
            filename = '{}.{}.{}.{}'.format(prefix, project, random_id, ext)
            # return the whole path to the file
        return os.path.join(path, filename)

    return wrapper

Call to Method

sales_attach = models.FileField("Attachment", upload_to=path_and_rename("complaint_files", 'sales'), max_length=500,
                                help_text="Browse a file")

Hope this helps. Thanks.

回答3:

if you want your function re-usable:

import hashlib
import datetime
import os
from functools import partial

def _update_filename(instance, filename, path):
    path = path

    filename = "..."

    return os.path.join(path, filename)

def upload_to(path):
    return partial(_update_filename, path=path)

You just have to use it this way:

document = models.FileField(upload_to=upload_to("my/path"))

回答4:

The basic way is

import os

os.rename('a.txt', 'b.html')

For your situation, it would probably look like

os.rename ("a.txt", "{id}{uuid}.{ext}".format(id=userid, uuid=transaction_uuid, ext=file_extension))