This is still not working,
please can you provide full code, this way it is complete and accurate, and not 
me guessing?

Best,
Kean

On 6 Sep 2019, at 21:13, Bhoopesh sisoudiya <[email protected]> wrote:

> Write this
> 
> Expense.objects.select_related('project').all()
> 
> 
> 
> On Sat, Sep 7, 2019, 1:37 AM Kean <[email protected]> wrote:
> Thanks,
> 
> I updated the view, and now I am getting syntax error.
> Please see below.
> 
> views.py
> 
> def projectdetail(request):
>     project = Project.objects.all()
>     return render(request, 'busprojectdetail.html', {'project': project, 
> 'expense_list': project.expenses.all()})
> 
>     projectDetails = Project.objects.select_related(Project).values(select 
> column).all()
> 
> please advise how to correct?
> 
> Best,
> 
> Kean
> 
> On 6 Sep 2019, at 21:01, Bhoopesh sisoudiya <[email protected]> wrote:
> 
>> def projectdetail(request):
>>     project = Project.objects.all()
>>     return render(request, 'busprojectdetail.html', {'project': project, 
>> 'expense_list': project.expenses.all()})
>> 
>> According to me by this archive you want expense details of each project
>> 
>> Then you write
>> 
>> projectDetails = Project.objects.select_related(your foreign key).values 
>> (select column).all()
>> 
>> Thanks
>> Bhoopesh sisoudiya
>> 
>> 
>> On Sat, Sep 7, 2019, 1:22 AM Kean <[email protected]> wrote:
>> Hi,
>> 
>> please ignore the manager it is just a naming convention ignore, it if for 
>> me to identify tables, if i have 50 table, I want to be able to know which 
>> model is holding which object information and for what I need it.
>> 
>> Please use below, I have removed the # information to reduce confusion.
>> 
>> models.py
>> 
>> class Project(models.Model):
>>     name = models.CharField(max_length=100)
>>     budget = IntegerField()
>> 
>>     def save(self, *args, **kwargs):
>>         self.name
>>         super(Project, self).save(*args, **kwargs)
>> 
>> class Category(models.Model):
>>     project = models.ForeignKey(
>>         Project, on_delete=models.CASCADE)
>>     name = models.CharField(max_length=150)
>> 
>> class Expense(models.Model):
>>     project = models.ForeignKey(
>>         Project, on_delete=models.CASCADE, related_name='expenses')
>>     title = models.CharField(max_length=100)
>>     amount = models.DecimalField(max_digits=8, decimal_places=2)
>>     category = models.ForeignKey(Category, on_delete=models.CASCADE)
>> On 6 Sep 2019, at 20:45, Bhoopesh sisoudiya <[email protected]> wrote:
>> 
>> Best wishes,
>> Kean
>> 
>>> Ok No problem but 
>>> why you want Manager?
>>> 
>>> Please send me answer
>>> 
>>> Thanks
>>> Bhoopesh sisoudiya
>>> 
>>> On Sat, Sep 7, 2019, 1:06 AM Kean <[email protected]> wrote:
>>> Hi Bhoopesh
>>> Thanks for this, please can you put it into the context of my specific 
>>> model setup, so I get it right and understand.
>>> This will be easier for me as this approach is causing me some other 
>>> errors, when i try to apply it.
>>> 
>>> Best wishes, 
>>> 
>>> Kean 
>>> On 6 Sep 2019, at 20:19, Bhoopesh sisoudiya <[email protected]> wrote:
>>> 
>>>> Hi Kean,
>>>> 
>>>>  Write manager like this ,I give you reference
>>>> 
>>>> from django.db import models
>>>>  # First, define the Manager subclass. 
>>>> class BhoopeshBookManager(models.Manager):
>>>>    def get_queryset(self): 
>>>>         return super(BhoopeshBookManager, 
>>>> self).get_queryset().filter(author='Bhoopesh') 
>>>> 
>>>> # Then hook it into the Book model explicitly. class Book(models.Model): 
>>>> title = models.CharField(max_length=100) 
>>>> author = models.CharField(max_length=50)
>>>>  # ... 
>>>> objects = models.Manager()
>>>>  # The default manager. 
>>>> 
>>>> bhoopesh_objects = BhoopeshBookManager()
>>>>  # The Bhoopesh-specific manager.
>>>> 
>>>> 
>>>> Book.bhoopesh_objects.all() # use manager
>>>> 
>>>> 
>>>> Thanks 
>>>> Bhoopesh Kumar
>>>> 
>>>> 
>>>> On Sat, Sep 7, 2019, 12:34 AM Bhoopesh sisoudiya <[email protected]> 
>>>> wrote:
>>>> Hi kean,
>>>>     Please check Expenses Class is not Manager
>>>>     Please make Manager Class.
>>>> 
>>>> 
>>>> Thanks
>>>> Bhoopesh sisoudiya
>>>> 
>>>> On Sat, Sep 7, 2019, 12:20 AM Kean <[email protected]> wrote:
>>>> Hi, 
>>>> 
>>>> New to Django. when navigating to the reqired view, I get the error above.
>>>> 
>>>> urls.py
>>>> 
>>>> path('businesslogin/businessadmin/busproject', views.projectdetail, 
>>>> name='projectdetail'),
>>>> 
>>>> views.py
>>>> 
>>>> def projectdetail(request):
>>>>     project = Project.objects.all()
>>>>     return render(request, 'busprojectdetail.html', {'project': project, 
>>>> 'expense_list': project.expenses.all()})
>>>> 
>>>> models.py
>>>> 
>>>> # Project model
>>>> 
>>>> class Project(models.Model):
>>>>     name = models.CharField(max_length=100)
>>>>     budget = IntegerField()
>>>> 
>>>>     def save(self, *args, **kwargs):
>>>>         self.name
>>>>         super(Project, self).save(*args, **kwargs)
>>>> 
>>>> # Category manager model
>>>> 
>>>> 
>>>> class Category(models.Model):
>>>>     project = models.ForeignKey(
>>>>         Project, on_delete=models.CASCADE)
>>>>     name = models.CharField(max_length=150)
>>>> 
>>>> 
>>>> # Expenses manager model
>>>> 
>>>> class Expense(models.Model):
>>>>     project = models.ForeignKey(
>>>>         Project, on_delete=models.CASCADE, related_name='expenses')
>>>>     title = models.CharField(max_length=100)
>>>>     amount = models.DecimalField(max_digits=8, decimal_places=2)
>>>>     category = models.ForeignKey(Category, on_delete=models.CASCADE)
>>>> 
>>>> the error is as follows:
>>>> 
>>>> 
>>>> Exception Type:    AttributeError
>>>> Exception Value:   
>>>> 'QuerySet' object has no attribute 'expenses'
>>>> Exception Location:        
>>>> /Users/ProductionEnv/Desktop/test/test1/dev/core/views.py in 
>>>> projectdetail, line 238
>>>> 
>>>> Please can anyone help?
>>>> 
>>>> Best 
>>>> 
>>>> K
>>>> 
>>>> 
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>>> 
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