我想创建一个与我正在创建的对象同名的新文件。
我想在数据库中创建一个名为str的新“结构对象”,同时在指定的上传文件夹中创建一个名为str的新文件。当我这样做时,我得到'file' object has no attribute '_committed'
的AttributeError
这是我的代码:
views.py:
def create(request):
print request.POST
filename = request.POST['name']
f = open(filename, "w")
structure = Structure(name=request.POST['name'], file=f)
structure.save()
return redirect('/structures')
models.py:
class Structure(models.Model):
name = models.CharField(max_length=120)
file = models.FileField(upload_to='structures')
created_at = models.DateTimeField(auto_now_add = True)
updated_at = models.DateTimeField(auto_now = True)
def __str__(self):
return self.name
urls.py: url(r'^create$', views.create),
模板:
<div class="col-md-12">
<div class="panel panel-primary">
<div class="panel-heading">
<h3 class="panel-title">Créer une nouvelle structure</h3>
</div>
<div class="panel-body">
<form class="form-horizontal" action="/create", method="post">
{% csrf_token %}
<fieldset>
<div class="form-group">
<label for="name" class="col-lg-6 control-label">Nom de la structure</label>
<div class="col-lg-6">
<input type="text" name="name" id="name">
</div>
</div>
<div class="form-group">
<div class="col-lg-10 col-lg-offset-2" align="center">
<button type="submit" value="Create" class="btn btn-primary">Créer</button>
</div>
</div>
</fieldset>
</form>
</div>
</div>
</div>
代码工作正常,直到我在views.py
中添加了创建行的文件这是我得到的错误的屏幕截图:
Environment:
Request Method: POST
Request URL: http://localhost:8000/create
Django Version: 1.8
Python Version: 2.7.13
Installed Applications:
('apps.structure',
'django.contrib.admin',
'django.contrib.auth',
'django.contrib.contenttypes',
'django.contrib.sessions',
'django.contrib.messages',
'django.contrib.staticfiles')
Installed Middleware:
('django.contrib.sessions.middleware.SessionMiddleware',
'django.middleware.common.CommonMiddleware',
'django.middleware.csrf.CsrfViewMiddleware',
'django.contrib.auth.middleware.AuthenticationMiddleware',
'django.contrib.auth.middleware.SessionAuthenticationMiddleware',
'django.contrib.messages.middleware.MessageMiddleware',
'django.middleware.clickjacking.XFrameOptionsMiddleware',
'django.middleware.security.SecurityMiddleware')
Traceback:
File "/home/kaiss/.local/lib/python2.7/site-packages/django/core/handlers/base.py" in get_response
132. response = wrapped_callback(request, *callback_args, **callback_kwargs)
File "/home/kaiss/Documents/proj/apps/structure/views.py" in create
27. structure.save()
File "/home/kaiss/.local/lib/python2.7/site-packages/django/db/models/base.py" in save
710. force_update=force_update, update_fields=update_fields)
File "/home/kaiss/.local/lib/python2.7/site-packages/django/db/models/base.py" in save_base
738. updated = self._save_table(raw, cls, force_insert, force_update, using, update_fields)
File "/home/kaiss/.local/lib/python2.7/site-packages/django/db/models/base.py" in _save_table
822. result = self._do_insert(cls._base_manager, using, fields, update_pk, raw)
File "/home/kaiss/.local/lib/python2.7/site-packages/django/db/models/base.py" in _do_insert
861. using=using, raw=raw)
File "/home/kaiss/.local/lib/python2.7/site-packages/django/db/models/manager.py" in manager_method
127. return getattr(self.get_queryset(), name)(*args, **kwargs)
File "/home/kaiss/.local/lib/python2.7/site-packages/django/db/models/query.py" in _insert
920. return query.get_compiler(using=using).execute_sql(return_id)
File "/home/kaiss/.local/lib/python2.7/site-packages/django/db/models/sql/compiler.py" in execute_sql
962. for sql, params in self.as_sql():
File "/home/kaiss/.local/lib/python2.7/site-packages/django/db/models/sql/compiler.py" in as_sql
920. for obj in self.query.objs
File "/home/kaiss/.local/lib/python2.7/site-packages/django/db/models/fields/files.py" in pre_save
313. if file and not file._committed:
Exception Type: AttributeError at /create
Exception Value: 'file' object has no attribute '_committed'
编辑:我没有显示错误的完整痕迹:它告诉我在将对象保存到数据库时找到错误:structure.save()
答案 0 :(得分:1)
我认为,您需要在保存时为FileField
类型的相应字段传递File
对象。
从Django docs起保存/检索文件:
当您访问模型上的 FileField 时,您将获得 FieldFile 的实例作为访问基础文件的代理。 FieldFile 的API反映了文件。
的API
FieldFile.save(name, content, save=True)
请注意,content参数应该是 django.core.files.File 的实例,而不是Python的内置文件对象。您可以从现有的Python文件对象构建文件,如下所示:
from django.core.files import File f = open('/path/to/hello.world') myfile = File(f)
因此,您可能想尝试:
from django.core.files import File
def create(request):
print request.POST
filename = request.POST['name']
f = open(filename, "w")
structure = Structure(name=request.POST['name'], file=File(f))
structure.save()
return redirect('/structures')