预期的字符串或缓冲区django图像上传

时间:2015-04-16 09:18:18

标签: python django

我用Django建立了我的第一个网站,我遇到了一个问题。当我尝试更改models.py中的内容时,会抛出错误:

    Applying Cooking1.0004_recipe_pub_date...Traceback (most recent call last):
  File "manage.py", line 10, in <module>
    execute_from_command_line(sys.argv)
  File "/usr/local/lib/python2.7/dist-packages/django/core/management/__init__.py", line 385, in execute_from_command_line
    utility.execute()
  File "/usr/local/lib/python2.7/dist-packages/django/core/management/__init__.py", line 377, in execute
    self.fetch_command(subcommand).run_from_argv(self.argv)
  File "/usr/local/lib/python2.7/dist-packages/django/core/management/base.py", line 288, in run_from_argv
    self.execute(*args, **options.__dict__)
  File "/usr/local/lib/python2.7/dist-packages/django/core/management/base.py", line 338, in execute
    output = self.handle(*args, **options)
  File "/usr/local/lib/python2.7/dist-packages/django/core/management/base.py", line 533, in handle
    return self.handle_noargs(**options)
  File "/usr/local/lib/python2.7/dist-packages/django/core/management/commands/syncdb.py", line 27, in handle_noargs
    call_command("migrate", **options)
  File "/usr/local/lib/python2.7/dist-packages/django/core/management/__init__.py", line 115, in call_command
    return klass.execute(*args, **defaults)
  File "/usr/local/lib/python2.7/dist-packages/django/core/management/base.py", line 338, in execute
    output = self.handle(*args, **options)
  File "/usr/local/lib/python2.7/dist-packages/django/core/management/commands/migrate.py", line 161, in handle
    executor.migrate(targets, plan, fake=options.get("fake", False))
  File "/usr/local/lib/python2.7/dist-packages/django/db/migrations/executor.py", line 68, in migrate
    self.apply_migration(migration, fake=fake)
  File "/usr/local/lib/python2.7/dist-packages/django/db/migrations/executor.py", line 102, in apply_migration
    migration.apply(project_state, schema_editor)
  File "/usr/local/lib/python2.7/dist-packages/django/db/migrations/migration.py", line 108, in apply
    operation.database_forwards(self.app_label, schema_editor, project_state, new_state)
  File "/usr/local/lib/python2.7/dist-packages/django/db/migrations/operations/fields.py", line 37, in database_forwards
    field,
  File "/usr/local/lib/python2.7/dist-packages/django/db/backends/sqlite3/schema.py", line 176, in add_field
    self._remake_table(model, create_fields=[field])
  File "/usr/local/lib/python2.7/dist-packages/django/db/backends/sqlite3/schema.py", line 74, in _remake_table
    self.effective_default(field)
  File "/usr/local/lib/python2.7/dist-packages/django/db/backends/schema.py", line 187, in effective_default
    default = field.get_db_prep_save(default, self.connection)
  File "/usr/local/lib/python2.7/dist-packages/django/db/models/fields/__init__.py", line 627, in get_db_prep_save
    prepared=False)
  File "/usr/local/lib/python2.7/dist-packages/django/db/models/fields/__init__.py", line 1290, in get_db_prep_value
    value = self.get_prep_value(value)
  File "/usr/local/lib/python2.7/dist-packages/django/db/models/fields/__init__.py", line 1269, in get_prep_value
    value = super(DateTimeField, self).get_prep_value(value)
  File "/usr/local/lib/python2.7/dist-packages/django/db/models/fields/__init__.py", line 1171, in get_prep_value
    return self.to_python(value)
  File "/usr/local/lib/python2.7/dist-packages/django/db/models/fields/__init__.py", line 1228, in to_python
    parsed = parse_datetime(value)
  File "/usr/local/lib/python2.7/dist-packages/django/utils/dateparse.py", line 70, in parse_datetime
    match = datetime_re.match(value)
TypeError: expected string or buffer

具体而言,当我尝试在image模型中插入Recipe字段时会出现此问题。

到目前为止,这是我的代码:

models.py:

class Recipe(models.Model):
    title = models.CharField(max_length=200)
    ingredients = models.CharField(max_length=1024)
    cooking_process = models.CharField(max_length=5024)
    image = models.ImageField(blank=True, null= True)
    # pub_date = models.DateTimeField(blank=True, null=True)
    def __unicode__(self):
        return u'{0}'.format(self.title)

admin.py

class RecipeAdmin(admin.ModelAdmin):
    list_display = ('title', 'ingredients', 'cooking_process', 'image')
admin.site.register(Recipe, RecipeAdmin)

0 个答案:

没有答案