Python 如何从Django管理员上传多个文件?
我想在Django管理员中上传多个文件,而不必放置多个文件字段。用户能够以简单的方式管理文件;删除或更改每个上载的文件,但同时上载多个 我认为可行的解决方案是使用几个文件字段,但问题是,我不知道用户将上载多少文件Python 如何从Django管理员上传多个文件?,python,django,django-models,django-forms,django-file-upload,Python,Django,Django Models,Django Forms,Django File Upload,我想在Django管理员中上传多个文件,而不必放置多个文件字段。用户能够以简单的方式管理文件;删除或更改每个上载的文件,但同时上载多个 我认为可行的解决方案是使用几个文件字段,但问题是,我不知道用户将上载多少文件 def case_upload_location(instance, filename): case_name = instance.name.lower().replace(" ", "-") file_name = filename.lower().replace(
def case_upload_location(instance, filename):
case_name = instance.name.lower().replace(" ", "-")
file_name = filename.lower().replace(" ", "-")
return "casos/{}/{}".format(case_name, file_name)
class Case(models.Model):
name = models.CharField(max_length=250)
observations = models.TextField(null = True, blank = True)
number_folder = models.CharField('Folder', max_length=250)
file1 = models.FileField('file 1', upload_to=case_upload_location, null = True, blank = True)
file2 = models.FileField('file 2', upload_to=case_upload_location, null = True, blank = True)
file3 = models.FileField('file 3', upload_to=case_upload_location, null = True, blank = True)
file4 = models.FileField('file 4', upload_to=case_upload_location, null = True, blank = True)
最终目标
要上载多个文件(用户需要逐个删除或更改,但一次上载所有文件)。看起来您需要一个多外键关系,从“案例文件”模型到您定义的“案例”模型 型号.py
from django.db import models
def case_upload_location(instance, filename):
case_name = instance.name.lower().replace(" ", "-")
file_name = filename.lower().replace(" ", "-")
return "casos/{}/{}".format(case_name, file_name)
class Case(models.Model):
# datos del caso
name = models.CharField('Nombre', max_length=250)
observations = models.TextField('Observaciones', null = True, blank = True)
number_folder = models.CharField('Numero de Carpeta', max_length=250)
class CaseFile(models.Model):
case = models.ForeignKey(Case, on_delete=models.CASCADE) # When a Case is deleted, upload models are also deleted
file = models.FileField(upload_to=case_upload_location, null = True, blank = True)
from django.contrib import admin
from .models import Case, CaseFile
class CaseFileAdmin(admin.StackedInline):
model = CaseFile
@admin.register(Case)
class CaseAdmin(admin.ModelAdmin):
inlines = [CaseFileAdmin]
@admin.register(CaseFile)
class CaseFileAdmin(admin.ModelAdmin):
pass
然后,您可以添加StackedLine管理表单,将案例文件直接添加到给定案例中
admin.py
from django.db import models
def case_upload_location(instance, filename):
case_name = instance.name.lower().replace(" ", "-")
file_name = filename.lower().replace(" ", "-")
return "casos/{}/{}".format(case_name, file_name)
class Case(models.Model):
# datos del caso
name = models.CharField('Nombre', max_length=250)
observations = models.TextField('Observaciones', null = True, blank = True)
number_folder = models.CharField('Numero de Carpeta', max_length=250)
class CaseFile(models.Model):
case = models.ForeignKey(Case, on_delete=models.CASCADE) # When a Case is deleted, upload models are also deleted
file = models.FileField(upload_to=case_upload_location, null = True, blank = True)
from django.contrib import admin
from .models import Case, CaseFile
class CaseFileAdmin(admin.StackedInline):
model = CaseFile
@admin.register(Case)
class CaseAdmin(admin.ModelAdmin):
inlines = [CaseFileAdmin]
@admin.register(CaseFile)
class CaseFileAdmin(admin.ModelAdmin):
pass
为什么不建立一个单独的模型,它只包含一个文件字段,可以用FK链接到
案例
&您可以通过admin添加内联线来添加文件