DRF Update Insert in ManytoMany, ForeignKey Relationship Models

Question

I have six models, they are below:

class Certificate(DateTimeLog):
    name = models.TextField(max_length=255)


class Vacancy(DateTimeLog):
    name = models.CharField(max_length=255)
    parent_position = models.ForeignKey("self", on_delete=models.CASCADE, null=True, blank=True,
                                        related_name='sub_positions')


class Region(DateTimeLog):
    name = models.CharField(max_length=255)


class MaritalStatus(DateTimeLog):
    name = models.CharField(max_length=255)


class Candidate(DateTimeLog):
    pin = models.CharField(max_length=16, unique=True)
    first_name = models.CharField(max_length=64, blank=True, null=True)
    last_name = models.CharField(max_length=64, blank=True, null=True)
    marital_status = models.ForeignKey(MaritalStatus, on_delete=models.SET_NULL, null=True, blank=True)
    certificate = models.ManyToManyField(Certificate, blank=True)


class Candidacy(DateTimeLog):
    candidate = models.ForeignKey(Candidate, on_delete=models.CASCADE)
    vacancy = models.ForeignKey(Vacancy, on_delete=models.CASCADE)
    work_region = models.ForeignKey(Region, on_delete=models.SET_NULL, null=True, blank=True)

Now I want to handle, if candidate record exists(I am checking it with pin), then check and update Candidate related data. If a candidate does not exist create it. After the candidate is created or updated assign it to the candidacy. My Serializer looks like below:

class CandidateSerializer(serializers.ModelSerializer):
    marital_status = MaritalStatusSerializer(required=False)
    certificate = CertificateSerializer(many=True, required=False)

    def create(self, validated_data):
        marital_status_data = validated_data.pop("marital_status")
        certificate_data = validated_data.pop("certificate")
        candidate = Candidate.objects.create(**validated_data)

        ms_o = MaritalStatus.objects.get(name=marital_status_data["name"])
        candidate.marital_status = ms_o

        for certificate in certificate_data:
            certificate_o = Certificate.objects.create(**certificate)
            candidate.certificate.add(certificate_o)

        candidate.save()
        return candidate

    class Meta:
        model = Candidate
        fields = '__all__'
        depth = 1


class CandidacySerializer(serializers.ModelSerializer):
    candidate = CandidateSerializer()
    vacancy = VacancySerializer()
    work_region = RegionSerializer()

    def create(self, validated_data):
        candidate_s = CandidateSerializer()
        candidate_data = validated_data.pop('candidate')

        vacancy_data = validated_data.pop('vacancy')
        work_region_data = validated_data.pop('work_region')

        vac_o = Vacancy.objects.get(name=vacancy_data['name'])
        wr_o = Region.objects.get(name=work_region_data['name'])
        candidate_o = candidate_s.create(validated_data=candidate_data)
        validated_data.update({
            'candidate': candidate_o,
            'vacancy': vac_o,
            'work_region': wr_o
        })
        candidacy = Candidacy.objects.create(**validated_data)
        return candidacy

    class Meta:
        model = Candidacy
        fields = '__all__'
        depth = 1

Currently, I am able to create it with the post request. Am I need to check it explicitly or it is possible to achieve it inside the serializer?

do you mean if the record is already in the database as in like a unique validator or..? — rediet yosef, Jan 27 '22 at 17:34
In your `CandidateSerializer.create` you can do: `candidate = Candidate.objects.update_or_create(pin=validated_data.get('pin'), defaults=validated_data)` — Brian Destura, Jan 27 '22 at 21:59
@BrianDestura when post I got 400(bad request) it says that candidate with this pin already exists — Mamed, Jan 28 '22 at 07:53
I am sorry not understand, could you please elaborate more, what do you mean by "code" exactly? — Mamed, Jan 28 '22 at 08:10
Please show the changes you have done that caused the `400(bad request)` — Brian Destura, Jan 30 '22 at 21:52

score 0 · Answer 1 · answered Jan 29 '22 at 18:02

0

so you have a more straightforward option. There is a function called get or create, which will check if the object exists and return it else make it. similar question can be found here Another use of the same question By the way, if this pin is sensitive, I recommend encrypting it.

answered Jan 29 '22 at 18:02

shaked

561
7
19

I tried it, but not worked! same exception. – Mamed Jan 29 '22 at 20:24
What version of the Django rest framework is the error you are getting? – shaked Jan 29 '22 at 22:40

score 0 · Answer 2 · answered Jan 29 '22 at 18:59

0

Since you have already made pin unique with

pin = models.CharField(max_length=16, unique=True)

there is no chance of having two similar pins. and now to check a candidate based on pin you will have to manually do something like this in your create method.

if Candidate.objects.filter(pin = revievedPinData):
    here candidate is already present so update the information information
    and again check if Canditate is in Candidacy or not
    first get the candidate object
    c = Candidate.objects.get(pin = revievedPinData)
    then
    if Candidacy.objects.filter(candidate = c):
       if true candidate is already assigned and the updated value of candidate will reflect here
    else: create the Candidacy with the current candidate and other values
else:
    if candidate is not there (from the given pin)
    create a new candidate
    aa = candidate.objects.create()
    create a new Candidacy with candidate aa
    Candidacy.objects.create(candidate=aa, **otherdata)

answered Jan 29 '22 at 18:59

Ranu Vijay

1,137
9
18

it is not getting errors in create method. Not worked! I got an exception from validations. – Mamed Jan 29 '22 at 20:24
What exceptions from validators? – Ranu Vijay Jan 29 '22 at 20:30
Are you getting error in candidate create, candidate update or candidancy create ? – Ranu Vijay Jan 29 '22 at 20:37
candidate create – Mamed Jan 29 '22 at 21:02
and what is the error ? – Ranu Vijay Jan 29 '22 at 21:02
{ "candidate": { "pin": [ "candidate with this pin already exists." ] } } – Mamed Jan 29 '22 at 21:10
so if a candidate with that pin is already there, why are you creating a new candidate with the same pin ? just see whether that candidate is in Candidacy or not. If not create a new candidacy with the same candidate having this pin. – Ranu Vijay Jan 29 '22 at 21:12
Check wheteher this -> **Candidate.objects.filter(pin = revievedPinData):** is returning true or false. – Ranu Vijay Jan 29 '22 at 21:14

score 0 · Accepted Answer · answered Feb 01 '22 at 19:15

First of all, you can use update_or_create in your CandidateSerializer like this:

class CandidateSerializer(serializers.ModelSerializer):
    marital_status = MaritalStatusSerializer(required=False)
    certificate = CertificateSerializer(many=True, required=False)

    def create(self, validated_data):
        marital_status_data = validated_data.pop("marital_status")
        certificate_data = validated_data.pop("certificate")
        candidate, created = Candidate.objects.update_or_create(
            pin=validated_data['pin'],
            **validated_data
        )

        ms_o = MaritalStatus.objects.get(name=marital_status_data["name"])
        candidate.marital_status = ms_o

        for certificate in certificate_data:
            certificate_o = Certificate.objects.create(**certificate)
            candidate.certificate.add(certificate_o)

        candidate.save()
        return candidate

    class Meta:
        model = Candidate
        fields = '__all__'
        depth = 1

Secondly, you should not call the method create() directly. You have to use save() instead. You have to replace:

candidate_s = CandidateSerializer()
candidate_o = candidate_s.create(validated_data=candidate_data)

by

candidate_s = CandidateSerializer(data=candidate_data)
candidate_o = candidate_s.save()

You can read more about saving instances here.

thank you very much for the answer. I will test it. – Mamed Feb 01 '22 at 20:40 — Mamed, Feb 01 '22 at 20:40

DRF Update Insert in ManytoMany, ForeignKey Relationship Models

3 Answers3