强制Django FileField对象的深层副本(主要是新文件本身)

时间:2015-03-26 03:51:35

标签: django file model deep-copy

我有这个型号:

class DocVersion(Commentable):

    name = models.CharField(
        'Version',
        max_length=100,
    )

    docfile = models.FileField(
        'File',
        upload_to='content/%Y/%m/%d/%H/%M/%S',
    )

    created = models.DateTimeField(
        'Created',
        auto_now_add=True,
    )

    creater = models.ForeignKey(
        User,
    )

class Document(DocumentBase):

    #..... blah .....

    versions = models.ManyToManyField(
        DocVersion,
    )

在视图中的一个函数中,我有这个函数来执行DocVersion的深层复制,如下所示:

def cp_document(transfer_object_id, parent_folder_id):
    document = Document.objects.get(pk=transfer_object_id)
    versions = []
    for version in document.versions.all():
        version.pk = None
        version.save()
        versions.append(version)
    document.pk = None
    document.parent = Folder.objects.get(pk=parent_folder_id)
    document.save()
    for version in versions:
        document.versions.add(version)
    document.save()

当我为cp_document()中的版本复制时,它会创建一个新字段,但仍然保留版本文件字段的相同路径。是否可以触发DocVersion创建具有新文件和路径的新版本(理想情况下使用与我在upload_to参数中指定的目录结构相同的时间格式)?或者还有其他方法可以做到这一点吗?

1 个答案:

答案 0 :(得分:1)

想出了这个:

from django.core.files.base import ContentFile

...

def cp_document(user, transfer_object_id, parent_folder_id):

    document = Document.objects.get(pk=transfer_object_id)
    versions = []

    for version in document.versions.all():

        dv = DocVersion()
        dv.name = version.name
        dv.docfile = ContentFile(version.docfile.read())
        dv.docfile.name = version.docfile.name
        dv.creater = user
        dv.save()
        versions.append(dv)

    document.pk = None
    document.parent = Folder.objects.get(pk=parent_folder_id)
    document.save()

    for version in versions:
        document.versions.add(version)
    document.save()

希望这有助于某人。