Customer premissions based on Membership plan

2020-10-06 Thread Mayank Tripathi
Hi All,

I am working on hands-on Django web application, and for this i have to 
provide access to some videos; products based on the Customer Membership 
plan. Like if a Customer is in Basic Plan, then allow only to view 10 
videos or so.. and for Permium Customer, no limit.

For such type of requirement, i was trying to create a model class as 
Customer_membership which will be a one-to-one relationship with my custom 
Customer model class.

Please suggest, on view and templates how will i be restrict the customer 
based on their plan.

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Django django_filters DateFilter customization

2021-01-22 Thread Mayank Tripathi
Hi All,

I am look for a solution where I can restrict the user to select date from 
current month only instead of selecting any other month or year.

Currently user can select past month as well, have to restrict for current 
month only in date picker.

Please help.

Below is the code.
filters.py file screenshot is as below.


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Django - PrePopulate the Foreign Key on the web page, and use that id to save the model.

2020-04-11 Thread Mayank Tripathi
Hi All,

I am facing an issue, to pre-populate the Foreign Key on the web page.
Actually i have two models Projects and Modules. 
To create a Module, one has to select the Project and go to Module page for 
create it, but there the Project is not populated.

Below are the details... Please guide me.

*models.py*
class Project(models.Model): 
name = models.CharField(max_length = 200, null = True)
startdate = models.DateTimeField()

def __str__(self):
return self.name

class Modules(models.Model): 
project = models.ForeignKey(Project, null = True, on_delete = 
models.SET_NULL) 
modulename = models.CharField(max_length = 200, null = True)
modulestartdate = models.DateTimeField()

def __str__(self):
return self.modulename 

*forms.py*
class ProjectForm(forms.ModelForm):
class Meta:
model = Project
fields = '__all__'


class ModuleForm(forms.ModelForm): 
class Meta: 
model = Modules
fields = '__all__'  

*views.py*
def projectView (request):
if request.method == 'POST':
form = ProjectForm(request.POST) 
if form.is_valid():
form.save(commit=True)

return render(request, 'budget/projectForm.html', {'form': form})

else :
form = ProjectForm()
return render(request, 'budget/projectForm.html', {'form': form})


def modulesView (request, projectid):
project = Project.objects.get(pk=projectid)
 
if request.method == 'POST':
form = ModuleForm(request.POST) 
if form.is_valid():
form.save(commit=True)

return render(request, 'example/modulesForm.html', {'form': form})

else :
form = ModuleForm(instance=project)
return render(request, 'example/modulesForm.html', {'form': form})

*urls.py*
path('testproject/', views.projectView, name='projectView'),
path('testprojectmodule//', views.modulesView, 
name='modulesView'),

in html forms for both Project and Modules... just using {{ form.as_table 
}}.

Now after createing the Project, i used the url as 
http://127.0.0.1:8000/testprojectmodule/1/ 
then my Project should get pre-populated... as reached using the project id 
which is 1 in this case, but i am getting full drop-down. Refer attached 
image.
Either the Project should be pre-selected.. or can make it non-editable 
either will work for me.

Please help.


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django dependent drop down list without using javascript; ajax; jquery

2020-04-11 Thread Mayank Tripathi
Hi All,

Could any one please share some details on how we can do dependent drop 
down list selection.
I am seeing lot of videos or content to handle this via JavaScript; AJAX; 
jquery etc.. but just wondering if Django has anything which can be handled 
easily.

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Django - What is the best approach to handle multiple user types…and route the HTML pages based on this?

2020-04-26 Thread Mayank Tripathi


I'm making a small test project with below user types: School Admin, 
Teacher, Student, Parent. And each user type will have different 
Permissions like School Admin has full access... Parents can only view 
their Childern's Activity. Teacher can see all students but can add / edit 
marks for their respective students only. Teachers can setup Exam etc.. 
Students can take exam and submit the exam, but cannot edit / add any other 
information. Just can edit his profile detail.

Approach 1: Do i need to create a Custom User calss and apply to each user 
type (ie one Custome Class.. and 4 user type calss).. and similarly have 4 
different view and html pages?

Approach 2: Just have one custome class and have an field with UserType 
which will have the have as "SchoolAdmin", "Teacher", "Student","Parent".. 
and one view and html (as the data page would remain same and only data 
record will restrict), and somehow identify the User Type in View, and 
filter the record?

Definately some view or html pages will be specific to one user type only 
which is i am able to handle, the issue is to use same view / html page to 
handle all user type.

Please suggest... and any code snippet will will more helpful.


Below is the thing from approach 1, i was trying.

# models.py-class CustomUser(AbstractUser):
USER_TYPE_CHOICES = (
('SchoolAdmin'),
('Teacher'),
('Student'),
('Parents'),
)

user_type = models.CharField(blank=False, choices=USER_TYPE_CHOICES)
name = models.CharField(blank=False, max_length=255)
country = models.CharField(blank=False, max_length=255)
city = models.CharField(blank=False, max_length=255)
phone = models.CharField(blank=True, max_length=255)
created_at = models.DateField(auto_now_add=True)

def __str__(self):
return self.name

class SchoolAdmin(models.Model):
user = models.OneToOneField(
CustomUser, on_delete=models.CASCADE, primary_key=True)

class Teacher(models.Model):
user = models.OneToOneField(
CustomUser, on_delete=models.CASCADE, primary_key=True)
photo = models.ImageField(upload_to='photos/%Y/%m/%d/', blank=True)

class Student(models.Model):
user = models.OneToOneField(
CustomUser, on_delete=models.CASCADE, primary_key=True)
teacher = models.ForeignKey(Teacher)
photo = models.ImageField(upload_to='photos/%Y/%m/%d/', blank=True)
class Parent(models.Model):
user = models.OneToOneField(
CustomUser, on_delete=models.CASCADE, primary_key=True)
student= models.ForeignKey(Student)

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Django Type Error with POST method

2020-05-01 Thread Mayank Tripathi
Hi All,

I am working on small hands-on project.. and having an issue when calling 
the POST method with form.py

Below is the code.
Here i have two models.. one is Category and another is SubCategory.
For SubCategory i have created form "SubCategoryFormv2" in forms.py... 
where i am checking if categoryid is passed then pre-select that Category, 
else display all Category for selection.
On submit the form from html... i am getting TypeError. (full error is 
displayed below)

Please help... What i could identify is with the POST method, when the 
statement *form = SubCategoryFormv2(request.POST)* is executed, it is 
expecting some id.. not sure what is this id.

# models.py

class SubCategory(models.Model) : 
category = models.ForeignKey(Category, null = True, on_delete = 
models.SET_NULL)
name = models.CharField(max_length = 200, null = True)
dateCreated = models.DateTimeField(default = timezone.now)
createdBy = models.ForeignKey(User, on_delete = models.PROTECT )

# forms.py
--
class SubCategoryFormv2(forms.ModelForm): 
class Meta:
model = SubCategory
fields = '__all__' 
widgets = {'dateCreated' : DateInput()}

def __init__(self, categoryid=None, *args, **kwargs):
super(SubCategoryFormv2, self).__init__(*args, **kwargs)
if categoryid is not None:
self.fields['category'] = forms.ModelChoiceField(initial=categoryid, 
queryset=Category.objects.filter(id=categoryid))

# views.py

def createSubCategoryV2(request, categoryid=None):
context = {
'title':'Add SubCategory'
}

if request.method =='POST':
print('POST METHOD')
form = SubCategoryFormv2(request.POST)
print('POST METHOD - SubCategoryFormv2 called') 
if form.is_valid():
ins = form.save(commit=True)
print('Inside for form validation')
messages.success(request, f'SubCategory added Successfully')
return redirect('listsubcategory')
else: # for GET method.
print('GET METHOD')
form = SubCategoryFormv2(categoryid)
context["form"]= form
return render(request, 'DailyBudget/subcategory_form.html', context) 

Error : 
Exception Value: 

Field 'id' expected a number but got .

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overriding __init__ in forms.ModelForm is erroring with POST method

2020-05-01 Thread Mayank Tripathi
Hi All,

I am working on small hands-on project.. and having an issue when calling 
the POST method with form.py overriding __init__.

Below is the code.
Here i have two models.. one is Category and another is SubCategory.
For SubCategory i have created form "SubCategoryFormv2" in forms.py... 
where i am checking if categoryid is passed then pre-select that Category, 
else display all Category for selection.
On submit the form from html... i am getting TypeError. (full error is 
displayed below)

Please help... What i could identify is with the POST method, when the 
statement *form = SubCategoryFormv2(request.POST)* is executed, it is 
expecting some id.. not sure what is this id.

# models.py

class SubCategory(models.Model) : 
category  = models.ForeignKey(Category, null = True, on_delete = 
models.SET_NULL)
name  = models.CharField(max_length = 200, null = True)
dateCreated  = models.DateTimeField(default = timezone.now)
createdBy  = models.ForeignKey(User, on_delete = models.PROTECT )

# forms.py
--
class SubCategoryFormv2(forms.ModelForm): 
class Meta:
model = SubCategory
fields = '__all__' 
widgets = {'dateCreated' : DateInput()}

def __init__(self, categoryid=None, *args, **kwargs):
super(SubCategoryFormv2, self).__init__(*args, **kwargs)
if categoryid is not None:
self.fields['category'] = forms.ModelChoiceField(initial=categoryid, 
queryset=Category.objects.filter(id=categoryid))

# views.py

def createSubCategoryV2(request, categoryid=None):
context = {
'title':'Add SubCategory'
}

if request.method =='POST':
print('POST METHOD')
form = SubCategoryFormv2(request.POST)
print('POST METHOD - SubCategoryFormv2 called') 
if form.is_valid():
ins = form.save(commit=True)
print('Inside for form validation')
messages.success(request, f'SubCategory added Successfully')
return redirect('listsubcategory')
else: # for GET method.
print('GET METHOD')
form = SubCategoryFormv2(categoryid)
context["form"]= form
return render(request, 'DailyBudget/subcategory_form.html', context) 

Error : 
Exception Value: 

Field 'id' expected a number but got .

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Can we use Database with Django without creating models

2020-05-25 Thread Mayank Tripathi
Hi Team,

I am trying to get options to connect to Database from Python-Django App.
But here the requirement is I already have a Database setup (MySQL) with 
all required tables and fields.
So am not willing to use models.py and create tables.
Is there a way still i can use the tables... and still be able to do 
similar query which is done using ORM (i mean Django default querysets).

Please suggest.

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CORS policy issue when using with JSON and DJANGO

2019-07-08 Thread Mayank Tripathi
Hi Team,

I am making quiz app, for which i wrote a function in view which will 
generate the json file and place it under the same location as of my all 
teamplates (C:/Desktop/Environment/envQuiz/quizsetup/templates/quizsetup)
Now I am trying to read this json file in my HTML page via javascript file, 
and getting error as attached "Issue1.PNG".

Please help... I also tried using header.. but no luck.

Below is the js code i am using.

var requestURL = 
'C:/Desktop/Environment/envQuiz/quizsetup/templates/quizsetup/json_data.json';
let questions;
let questionsCount;
let currentQuestion;
let score = 0;

let question_title_elem = document.getElementById("title");
let answers_elem = document.getElementById("answers");
let action_btn = document.getElementById("action_btn");

function getQuestions () {
let request = new XMLHttpRequest();
request.onreadystatechange = function () {
if (this.readyState == 4 && this.status == 200) { 
questions = JSON.parse(this.responseText).questions;
questionsCount = questions.length;
currentQuestion = 0;
}
}

/*json_data -- This is from view takeQuiz */
request.open("GET", requestURL, false);
request.send();

.
..
.
.
.
...




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