问题
我正在为嵌入式设备编写代码.CRC-CCITT 16位计算的许多解决方案都需要库.
鉴于使用库几乎是不可能的并且耗尽其资源,需要一个函数.
可能解决方案
在线发现以下CRC计算.但是,它的实现是不正确的.
http://bytes.com/topic/python/insights/887357-python-check-crc-frame-crc-16-ccitt
def checkCRC(message):
#CRC-16-CITT poly, the CRC sheme used by ymodem protocol
poly = 0x11021
#16bit operation register, initialized to zeros
reg = 0xFFFF
#pad the end of the message with the size of the poly
message += '\x00\x00'
#for each bit in the message
for byte in message:
mask = 0x80
while(mask > 0):
#left shift by one
reg<<=1
#input the next bit from the message into the right hand side of the …我试图ModelForm从这里构建一个解决方案，但是我收到此错误：
'User' object has no attribute 'get'
这个想法是构建ModelForm一个表单，提交时登录的用户会更新条目。
models.py 是：
class UserDetailsForm(ModelForm):
class Meta:
model = UserProfile
fields = ['mobile_phone']
def __init__(self, *args, **kwargs):
self.request = kwargs.pop('request', None)
return super(UserDetailsForm, self).__init__(*args, **kwargs)
def save(self, *args, **kwargs):
kwargs['commit']=False
obj = super(UserDetailsForm, self)
if self.request:
obj.user = UserProfile.objects.get(user=self.request.user)
obj.save()
我在 models.py 中的模型是
class UserProfile(models.Model):
user = models.OneToOneField(User)
mobile_phone = models.CharField(max_length=30,help_text='Max 30 characters.',blank=True)
#have shortened this for simplicity
def __unicode__(self):
return self.mobile_phone
根据这里的请求，我们从views.py 中回溯了问题：
userprofile …我有一个从Django表单加载到内存中的文件.它给了我以下错误:
new-line character seen in unquoted field - do you need to open the file in universal-newline mode?
另一种解决方案使用此.
但是,我已经将文件存储在内存中,因此我需要将其更改为以"rU"格式打开.
这是我的相关代码 views.py
form = ResultsUploadForm(request.POST, request.FILES)
if form.is_valid():
form.save()
reader = csv.reader(request.FILES['file'])
for row in reader:
#etc...
有谁知道如何解决这个问题？从此声明中抛出错误for row in reader:
如果我使用模型表单,我的views.py看起来像:
def dog_image_upload(request):
if request.user.is_authenticated():
if request.method == 'POST':
form = DogImageForm(request.POST, request.FILES)
if form.is_valid():
form.save()
else:
form = DogImageForm(user)
return render_to_response("dog-image-upload.html", {'form': form}, context_instance=RequestContext(request))
else:
return HttpResponseRedirect('/')
在model.py中,我想这样做:
class DogImageForm(ModelForm):
dogs = forms.ModelChoiceField(queryset=Dog.objects.filter(user=request.user))
class Meta:
model = ResultsUpload
fields = ['dogs','image']
但是,我无法尝试将用户发送到model.py这方面的帮助将是非常棒的,值得点!