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我得到以下错误:Django的说:表X有没有名为Y.我不同意列
sqlite3.OperationalError: table gallery_image has no column named filename
这里是我的模型:
from django.db import models
class Image(models.Model):
filename= models.Field(max_length=40);
gallery = models.ForeignKey('Gallery')
def __unicode__(self):
return u'%s(%s)' % (self.filename,self.gallery)
class Gallery(models.Model):
title = models.CharField(max_length=100)
url = models.CharField(max_length=50, unique=True)
def __unicode__(self):
return u'title: %s, url: %s' % (self.title, self.url)
和这个脚本显然有在一个错误最后一行
from django.core.management.base import BaseCommand
from dev.gallery.models import Gallery, Image
import os
class Command(BaseCommand):
def handle(self, *args, **options):
importfrom='/srv/django/dev/gallery/import'
exportto='/var/www/dev-media/'
title=args[0]
url=args[1]
print(os.listdir(importfrom))
if not len(args)==2:
print 'wrong number of arguments'
return ;
#Create the gallery in the database
g=Gallery(title=title,url=title)
g.save();
files=os.listdir('/srv/django/dev/gallery/import');
for f in files:
Image(filename=f,gallery=g).save();
,如果这对你有意义的,请向我解释为什么:)
谢谢...应该是CharField,因为我想存储的所有文件的实际名称不是位置:) – niklasfi 2010-07-16 22:23:04
很高兴我可以帮助:-) – Lexo 2010-07-16 23:03:04