Re: UserCreationForm decode str object has not decode attribute

[Aldian Fazrihady]

url_safe_base64_encode returns string, and strings don't have decode method.
However, bytes objects have decode method.

string.encode will change string to bytes.
bytes.decode will change bytes to string.

On Thu, Jul 18, 2019 at 8:28 PM Ahmet İnal  wrote:

>
>
> What's wrong taht code if i try register i seeing.
>
> AttributeError at /register/
>
> 'str' object has no attribute 'decode'
>
> 'uid':urlsafe_base64_encode(force_bytes(user.pk)).decode('utf-8'), whats
> problem on this line.
>
> class MyUserCreationForm(UserCreationForm):
> class Meta:
> # Yeni Model
> model = User
> # Yeni alanlar
> fields = { 'username', 'password1', 'password2', 'email',}
> exclude = ['Avatar','gender','birth_date',]
>
>
>
>
>
> def clean_username(self):
> username = self.cleaned_data['username'].lower()
> if username in blacklist:
> raise ValidationError("Lütfen başka bir kullanıcı adı seçin.")
> return username
>
> def save(self, commit=True):
> user = super(UserCreationForm, self).save(commit=True)
> current_site = Site.objects.get_current()
> mail_subject = 'Activate your blog account.'
> message = render_to_string('includes/emails/activation-mail.html', {
> 'user': user,
> 'domain': current_site.domain,
> 'uid':urlsafe_base64_encode(force_bytes(user.pk)).decode('utf-8'),
> 'token':account_activation_token.make_token(user),
> })
> to_email = form.cleaned_data.get('email')
> email = EmailMessage(
> mail_subject, message, to=[to_email]
> )
> email.send()
>
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> 
> .
> For more options, visit https://groups.google.com/d/optout.
>


-- 
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Aldian Fazrihady
http://aldianfazrihady.com

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Re: Using a 2-dimensional array in a template

[Yves de Champlain]

That looks just perfect ! I knew but I see I still need to learn more about 
data structures in Python.

Thanks a lot !

yves

Le jeudi 18 juillet 2019 15:29:48 UTC-4, Daniel Roseman a écrit :
>
> On Thursday, 18 July 2019 15:56:59 UTC+1, Yves de Champlain wrote:
>>
>> Indeed, it feels like I'm taking things the wrong way, maybe I coded too 
>> much PHP to be redeemable ...
>>
>> My main problem might be coming from the fact that I need to parse the 
>> data to generate my view and can't simply pull the data from the DB.
>>
>> Here are the models, template and view attached, as well as a screenshot 
>> of the result. 
>>
>> Please note that as far as the logic of the app is concerned, 
>> "Competency" means the same as "Skill". I need to fix that dual terminology 
>> in the code. 
>>
>> Thanks a lot for your answer.
>>
>> yves
>>
>
>  The key is to think about the structure you need for output when building 
> up your data in the view. In this case, what you want is a simple nested 
> list which includes *all* the data for each row - including the skill 
> object. (Also, there's no need to pad your lists with "X" and spaces; use 
> bools.) In other words, the list looks like:
>
> [
>   [skill1, [True, False, False...],
>   [skill2, [True, False, True...].
>   ...
> ]
>
> So, in the view:
>
> for s in skills:
> v_skill = []
> valeur = False
> for a in achievements:
> ...
> if...
> valeur = True
> v_skill.append(valeur)
> v_table.append(skill, v_skill)
>
> Now in the template you can simply do:
>
> {% for skill, achievements in skills %}
>  {% if skill.is_key %}
> style="font-weight:bold;"
>   {% endif %}
>   >
>   {{ skill.code }}
>   {{ skill.name }}
>
>   {% for achievement in achievements %}
> 
>   {{ achievement|yesno:"X, " }}
> 
>   {% endfor %}
>   
> {% endfor %}
>
> -- 
> DR.
>

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Re: Using a 2-dimensional array in a template

[Daniel Roseman]

On Thursday, 18 July 2019 15:56:59 UTC+1, Yves de Champlain wrote:
>
> Indeed, it feels like I'm taking things the wrong way, maybe I coded too 
> much PHP to be redeemable ...
>
> My main problem might be coming from the fact that I need to parse the 
> data to generate my view and can't simply pull the data from the DB.
>
> Here are the models, template and view attached, as well as a screenshot 
> of the result. 
>
> Please note that as far as the logic of the app is concerned, "Competency" 
> means the same as "Skill". I need to fix that dual terminology in the code. 
>
> Thanks a lot for your answer.
>
> yves
>

 The key is to think about the structure you need for output when building 
up your data in the view. In this case, what you want is a simple nested 
list which includes *all* the data for each row - including the skill 
object. (Also, there's no need to pad your lists with "X" and spaces; use 
bools.) In other words, the list looks like:

[
  [skill1, [True, False, False...],
  [skill2, [True, False, True...].
  ...
]

So, in the view:

for s in skills:
v_skill = []
valeur = False
for a in achievements:
...
if...
valeur = True
v_skill.append(valeur)
v_table.append(skill, v_skill)

Now in the template you can simply do:

{% for skill, achievements in skills %}
  
  {{ skill.code }}
  {{ skill.name }}

  {% for achievement in achievements %}

  {{ achievement|yesno:"X, " }}

  {% endfor %}
  
{% endfor %}

-- 
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Re: how to see the list of created tables in sqlite

[Joe Reitman]

A good way to view your tables is to use https://sqlitebrowser.org/

On Thursday, July 18, 2019 at 8:28:31 AM UTC-5, yasar arafath Kajamydeen 
wrote:
>
> Hi all,
>
> After executing the  command  -  python manage.py migrate
> the command is successful and then i would like to see the list of created 
> tables so  i enter command like  .schema  but it showing error.
>
> Please try to help me .
>
> Note - Am beginer to python 
>
>

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Re: Email not sent via model form

[Kasper Laudrup]

On 18/07/2019 16.40, ahadujjaman nur wrote:

Error: socket.gaierror: [Errno -2] Name or service not known



https://en.wikipedia.org/wiki/Domain_Name_System

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Email not sent via model form

[ahadujjaman nur]

Error: socket.gaierror: [Errno -2] Name or service not known 


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Re: sqlite update on CentOS 7 for latest django version

[Tal]

Got it.
For anyone also trying this, on a fresh CentOS 7 minimal install, do this 
as root:

   - yum groupinstall "Development Tools"
   - yum install tcl
   - curl -O https://www.sqlite.org/src/tarball/sqlite.tar.gz
   - tar -xvf sqlite.tar.gz
   - cd sqlite/
   - ./configure
   - make
   - cp -v .libs/libsqlite3.so.0.8.6 /usr/local/lib64/
   - Dont miss the dot in .libs
   - echo "/usr/local/lib64" > /etc/ld.so.conf.d/sqlite-x68_64.conf
   - ldconfig

We basically just compiled the sqlite library from source, moved it to 
/usr/local/lib64/, and told the system to use it instead of the original 
/usr/lib64/libsqlite3.so.0.8.6 library it was using.

At this point, if you want, you can delete the sqlite.tar.gz file, and 
sqlite directory with all the source code.
If you ever need to undo this, delete /etc/ld.so.conf.d/sqlite-x68_64.conf, 
delete /usr/local/lib64/libsqlite3.so.0.8.6, and run "ldconfig" as root.

I'll try to add this to the djangoproject ticket as well.

Thanks

On Thursday, July 18, 2019 at 9:07:50 AM UTC-6, Tal wrote:
>
> Yoo - this is a known problem. 
>  It has nothing to do with 
> running "migrate", being in a virtual environment, or settings.py. 
>
> Jani - I tried compiling the latest Python manually - it was actually 
> super easy - but even with the latest Python, importing _sqlite3 and 
> checking the version, I still see the old version. I don't think sqlite is 
> built into Python anymore - I think sqlite is some sort of library that 
> Python uses. It seems not to be /usr/bin/sqlite3. I think that's just the 
> binary that lets you interact with an sqlite database in the terminal - not 
> the library.
>
> There's an article here 
> 
>  
> that tries to cover updating sqlite so your Python picks up on it, but so 
> far, I haven't had any luck getting it to work.
>
>
> On Thursday, July 18, 2019 at 1:14:42 AM UTC-6, Jani Tiainen wrote:
>>
>> Hi.
>>
>> I think Python has built in version so you might need to build custom 
>> version of python.
>>
>> Pyenv is pretty neat tool to handle custom python builds.
>>
>> ke 17. heinäk. 2019 klo 21.36 Tal  kirjoitti:
>>
>>> Or is Python's sqlite built into Python, not depending on 
>>> /usr/bin/sqlite3 at all?
>>> Would I have to recompile Python for a newer version of sqlite that 
>>> django can use?
>>>
>>> PS. I know I can downgrade django. I'm wondering how complex it is to 
>>> compile a new version of sqlite for it to use.
>>>
>>> On Wednesday, July 17, 2019 at 12:14:54 PM UTC-6, Tal wrote:

 When using the latest django from PyPI in CentOS 7, running 
 "./manage.py runserver" gives an error about sqlite being too old.
 Since there's no newer sqlite version in the CentOS repos, I tried 
 building sqlite from scratch:

 curl -L https://www.sqlite.org/2019/sqlite-amalgamation-329.tar.gz 
 > sqlite-amalgamation-329.tar.gz
 tar -xvf sqlite-amalgamation-329.tar.gzcd sqlite-autoconf-329
 ./configure
 make
 make install


 This sets up the latest sqlite3 to /usr/local/bin/.
 Since /usr/local/bin is ahead of /usr/bin in my PATH, just running 
 "sqlite3" in the terminal runs the latest sqlite.
 It runs without issues, and shows that it's the latest version:

 my_hostname# sqlite3

 SQLite version 3.29.0 2019-07-10 17:32:03 
 Enter ".help" for usage hints. 
 Connected to a transient in-memory database. 
 Use ".open FILENAME" to reopen on a persistent database. 
 sqlite>


 Running "./manage.py runserver" again, it still tries to use the old 
 version in /usr/bin, and fails.
 My django is running in a pipenv virtual environment, where PATH still 
 has /usr/local/bin/ ahead of /usr/bin, and running "sqlite3" in terminal 
 still shows the latest version.

 I followed the traceback django gives me to the dbapi2.py module, where 
 to figure out the sqlite version it does this:

 import _sqlite3
 _sqlite3.sqlite_version


 If I run "python" in my virtualenv, and type those 2 lines, it shows 
 the old version of sqlite too.
 _sqlite3 is not written in python - it's a compiled binary, so I can't 
 examine it to see where it looks.

 Am I missing something?
 How can I tell _sqlite3 that there's a newer version of sqlite 
 available on the system?
 Does _sqlite3 even care about /usr/local/bin/sqlite3? Or is there some 
 sqlite library it's looking for?

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Re: sqlite update on CentOS 7 for latest django version

[Tal]

Yoo - this is a known problem.  It 
has nothing to do with running "migrate", being in a virtual environment, 
or settings.py. 

Jani - I tried compiling the latest Python manually - it was actually super 
easy - but even with the latest Python, importing _sqlite3 and checking the 
version, I still see the old version. I don't think sqlite is built into 
Python anymore - I think sqlite is some sort of library that Python uses. 
It seems not to be /usr/bin/sqlite3. I think that's just the binary that 
lets you interact with an sqlite database in the terminal - not the library.

There's an article here 

 
that tries to cover updating sqlite so your Python picks up on it, but so 
far, I haven't had any luck getting it to work.


On Thursday, July 18, 2019 at 1:14:42 AM UTC-6, Jani Tiainen wrote:
>
> Hi.
>
> I think Python has built in version so you might need to build custom 
> version of python.
>
> Pyenv is pretty neat tool to handle custom python builds.
>
> ke 17. heinäk. 2019 klo 21.36 Tal > 
> kirjoitti:
>
>> Or is Python's sqlite built into Python, not depending on 
>> /usr/bin/sqlite3 at all?
>> Would I have to recompile Python for a newer version of sqlite that 
>> django can use?
>>
>> PS. I know I can downgrade django. I'm wondering how complex it is to 
>> compile a new version of sqlite for it to use.
>>
>> On Wednesday, July 17, 2019 at 12:14:54 PM UTC-6, Tal wrote:
>>>
>>> When using the latest django from PyPI in CentOS 7, running "./manage.py 
>>> runserver" gives an error about sqlite being too old.
>>> Since there's no newer sqlite version in the CentOS repos, I tried 
>>> building sqlite from scratch:
>>>
>>> curl -L https://www.sqlite.org/2019/sqlite-amalgamation-329.tar.gz 
>>> > sqlite-amalgamation-329.tar.gz
>>> tar -xvf sqlite-amalgamation-329.tar.gzcd sqlite-autoconf-329
>>> ./configure
>>> make
>>> make install
>>>
>>>
>>> This sets up the latest sqlite3 to /usr/local/bin/.
>>> Since /usr/local/bin is ahead of /usr/bin in my PATH, just running 
>>> "sqlite3" in the terminal runs the latest sqlite.
>>> It runs without issues, and shows that it's the latest version:
>>>
>>> my_hostname# sqlite3
>>>
>>> SQLite version 3.29.0 2019-07-10 17:32:03 
>>> Enter ".help" for usage hints. 
>>> Connected to a transient in-memory database. 
>>> Use ".open FILENAME" to reopen on a persistent database. 
>>> sqlite>
>>>
>>>
>>> Running "./manage.py runserver" again, it still tries to use the old 
>>> version in /usr/bin, and fails.
>>> My django is running in a pipenv virtual environment, where PATH still 
>>> has /usr/local/bin/ ahead of /usr/bin, and running "sqlite3" in terminal 
>>> still shows the latest version.
>>>
>>> I followed the traceback django gives me to the dbapi2.py module, where 
>>> to figure out the sqlite version it does this:
>>>
>>> import _sqlite3
>>> _sqlite3.sqlite_version
>>>
>>>
>>> If I run "python" in my virtualenv, and type those 2 lines, it shows the 
>>> old version of sqlite too.
>>> _sqlite3 is not written in python - it's a compiled binary, so I can't 
>>> examine it to see where it looks.
>>>
>>> Am I missing something?
>>> How can I tell _sqlite3 that there's a newer version of sqlite available 
>>> on the system?
>>> Does _sqlite3 even care about /usr/local/bin/sqlite3? Or is there some 
>>> sqlite library it's looking for?
>>>
>> -- 
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>> "Django users" group.
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>> email to django...@googlegroups.com .
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>> .
>> Visit this group at https://groups.google.com/group/django-users.
>> To view this discussion on the web visit 
>> https://groups.google.com/d/msgid/django-users/3d011ce9-c849-48a7-82be-0ea08edb4566%40googlegroups.com
>>  
>> 
>> .
>> For more options, visit https://groups.google.com/d/optout.
>>
>

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Using a 2-dimensional array in a template

[Daniel Roseman]

You haven't really given enough context to answer the question. What are 
`skills` and `achievements`? How are they related, and how are either of them 
related to `v_table`? Normally in Django you would iterate directly over a 
relationship by following foreign keys - why can't you do that? Post your 
models and view.
-- 
DR.

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how to see the list of created tables in sqlite

[yasar arafath Kajamydeen]

Hi all,

After executing the  command  -  python manage.py migrate
the command is successful and then i would like to see the list of created 
tables so  i enter command like  .schema  but it showing error.

Please try to help me .

Note - Am beginer to python 

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UserCreationForm decode str object has not decode attribute

[Ahmet İnal]

What's wrong taht code if i try register i seeing.

AttributeError at /register/

'str' object has no attribute 'decode'

'uid':urlsafe_base64_encode(force_bytes(user.pk)).decode('utf-8'), whats 
problem on this line.

class MyUserCreationForm(UserCreationForm):
class Meta:
# Yeni Model
model = User
# Yeni alanlar
fields = { 'username', 'password1', 'password2', 'email',}
exclude = ['Avatar','gender','birth_date',]





def clean_username(self):
username = self.cleaned_data['username'].lower()
if username in blacklist:
raise ValidationError("Lütfen başka bir kullanıcı adı seçin.")
return username

def save(self, commit=True):
user = super(UserCreationForm, self).save(commit=True) 
current_site = Site.objects.get_current()
mail_subject = 'Activate your blog account.'
message = render_to_string('includes/emails/activation-mail.html', {
'user': user,
'domain': current_site.domain,
'uid':urlsafe_base64_encode(force_bytes(user.pk)).decode('utf-8'),
'token':account_activation_token.make_token(user),
})
to_email = form.cleaned_data.get('email')
email = EmailMessage(
mail_subject, message, to=[to_email]
)
email.send() 

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Re: sqlite update on CentOS 7 for latest django version

[Jani Tiainen]

Hi.

I think Python has built in version so you might need to build custom
version of python.

Pyenv is pretty neat tool to handle custom python builds.

ke 17. heinäk. 2019 klo 21.36 Tal  kirjoitti:

> Or is Python's sqlite built into Python, not depending on /usr/bin/sqlite3
> at all?
> Would I have to recompile Python for a newer version of sqlite that django
> can use?
>
> PS. I know I can downgrade django. I'm wondering how complex it is to
> compile a new version of sqlite for it to use.
>
> On Wednesday, July 17, 2019 at 12:14:54 PM UTC-6, Tal wrote:
>>
>> When using the latest django from PyPI in CentOS 7, running "./manage.py
>> runserver" gives an error about sqlite being too old.
>> Since there's no newer sqlite version in the CentOS repos, I tried
>> building sqlite from scratch:
>>
>> curl -L https://www.sqlite.org/2019/sqlite-amalgamation-329.tar.gz >
>> sqlite-amalgamation-329.tar.gz
>> tar -xvf sqlite-amalgamation-329.tar.gzcd sqlite-autoconf-329
>> ./configure
>> make
>> make install
>>
>>
>> This sets up the latest sqlite3 to /usr/local/bin/.
>> Since /usr/local/bin is ahead of /usr/bin in my PATH, just running
>> "sqlite3" in the terminal runs the latest sqlite.
>> It runs without issues, and shows that it's the latest version:
>>
>> my_hostname# sqlite3
>>
>> SQLite version 3.29.0 2019-07-10 17:32:03
>> Enter ".help" for usage hints.
>> Connected to a transient in-memory database.
>> Use ".open FILENAME" to reopen on a persistent database.
>> sqlite>
>>
>>
>> Running "./manage.py runserver" again, it still tries to use the old
>> version in /usr/bin, and fails.
>> My django is running in a pipenv virtual environment, where PATH still
>> has /usr/local/bin/ ahead of /usr/bin, and running "sqlite3" in terminal
>> still shows the latest version.
>>
>> I followed the traceback django gives me to the dbapi2.py module, where
>> to figure out the sqlite version it does this:
>>
>> import _sqlite3
>> _sqlite3.sqlite_version
>>
>>
>> If I run "python" in my virtualenv, and type those 2 lines, it shows the
>> old version of sqlite too.
>> _sqlite3 is not written in python - it's a compiled binary, so I can't
>> examine it to see where it looks.
>>
>> Am I missing something?
>> How can I tell _sqlite3 that there's a newer version of sqlite available
>> on the system?
>> Does _sqlite3 even care about /usr/local/bin/sqlite3? Or is there some
>> sqlite library it's looking for?
>>
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