Django: Saving a URL request parameter Into Form

Question

I'm Using a Form to save data into it.

models.py

class presciptiontemplates(models.Model):
    templateid = models.AutoField(primary_key=True)
    template = tinymce_models.HTMLField()
    draft = models.BooleanField(default=False)
    savedate = models.DateTimeField(default=datetime.now())
    patientid = models.ForeignKey('Patient')

I'm Passing the parameter Patient Id From the URL.

url(r'^addprescription/(?P<patid>\d+)/$', views.viewtemplate

How do I save this parameter into form

views.py

def viewtemplate(request, patid):
    userid = patid
    form = templateform(request.POST)
    if request.method == "POST":
        if form.is_valid():
            presciptiontemplates.patientid = userid
            form.save()
        return redirect('index')
    else:
        form = templateform()
    return render(request, 'presapp/prescription.html', {'form': form})

forms.py

class templateform(forms.ModelForm):
class Meta:
    model = presciptiontemplates
    fields = ['template', 'draft']
    widgets = {
    'template': TinyMCE(),
    'draft': forms.CheckboxInput()
    }
    labels = {
        "patientid": _("Patient ID"),
        }
    def __init__(self, *args, **kwargs):
        super(ThatForm, self).__init__(*args, **kwargs)
        self.fields['template'].required = True

This gives me an error

[Microsoft][SQL Server Native Client 11.0][SQL Server]Cannot insert the value NULL into column 'patientid'

Django  1.10.4
Python  3.5.2
Windows 7

lots of things wrong with this question. Where is your modelform. What about the indentation. What on earth is presciptiontemplates which appears to be undefined? — e4c5
– e4c5, Commented Jan 5, 2017 at 6:39
Show the prescription.html and also tell us what is presciptiontemplates ? — Prakhar Trivedi
– Prakhar Trivedi, Commented Jan 5, 2017 at 6:46
Added modelform , presciptiontemplates is the model in which I Intend to save the Data from Rich text Editor — Romil Bahukhandi
– Romil Bahukhandi, Commented Jan 5, 2017 at 7:03
Add full traceback of the error. The error is coming from forms.py or from this line presciptiontemplates.patientid = userid ? — Prakhar Trivedi
– Prakhar Trivedi, Commented Jan 5, 2017 at 7:11

Prakhar Trivedi · Accepted Answer · 2017-01-05 07:42:37Z

2

The problem is in this line :

presciptiontemplates.patientid = userid

In your model, patientid is a foreign key to Patient Model, So it is expecting a Patient model object, and not the id of the patient as patid. So firstly get the Patient object and then assign it to patientid,Like this :

def viewtemplate(request, patid):

    patient = Patient.object.get(id=patid)

    form = templateform(request.POST)
    if request.method == "POST":
        if form.is_valid():
            form = form.save(commit=False)
            form.patientid = patient
            form.save()
        return redirect('index')
    else:
        form = templateform()
    return render(request, 'presapp/prescription.html', {'form': form})

edited Jan 5, 2017 at 7:42

answered Jan 5, 2017 at 6:58

Prakhar Trivedi

8,5563 gold badges30 silver badges36 bronze badges

Sign up to request clarification or add additional context in comments.

8 Comments

Romil Bahukhandi Over a year ago

It gives me an error int() argument must be a string, a bytes-like object or a number, not 'Patient' Any suggestions

Prakhar Trivedi Over a year ago

@RomilBahukhandi Error is in which line of this code ?

Prakhar Trivedi Over a year ago

@RomilBahukhandi updating my answer. Try form.patientid = patid in the place of patient object.

Romil Bahukhandi Over a year ago

The error trace-back is to form.save() Im Using MSSQL As the backend using Pyodbc-azure , Can it be the Problem

Prakhar Trivedi Over a year ago

Use this line also form = form.save(commit=False) Then the error should be gone.

|

Rohan · Accepted Answer · 2017-01-05 06:50:33Z

2

Assuming templateform is a model form.

You can use commit=False parameter in the save method and then save the object again with required parameters.

Short snippet to change as per your code

...
if request.method == "POST":
    if form.is_valid():
        obj = form.save(commit=False)
        obj.patientid = userid
        obj.save()
    return redirect('index')
else:
...

answered Jan 5, 2017 at 6:50

Rohan

53.5k12 gold badges96 silver badges89 bronze badges

Comments

neverwalkaloner · Accepted Answer · 2017-01-05 06:50:01Z

0

To add userid value into new object you can use ModelForm.

Just make templateform inherited from ModelForm not Form class.

class templateform(forms.ModelForm):
    class Meta:
        model = presciptiontemplates

now in view you can do following:

form = templateform(request.POST or None, request.FILES or None)
if form.is_valid():
    instance = form.save(commit=False)
    instance.patientid = userid
    instance.save()

answered Jan 5, 2017 at 6:50

neverwalkaloner

47.6k7 gold badges101 silver badges103 bronze badges

Collectives™ on Stack Overflow

Django: Saving a URL request parameter Into Form

3 Answers 3

8 Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

8 Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related