Is there any way to use HyperlinkedModelSerializer without having to define the url view_name?

[brunodantas]

This is basically a duplicate of a stackoverflow question that had no answer.

When I implement a HyperlinkedModelSerializer subclass without redefining the url, I get the following error.

django.core.exceptions.ImproperlyConfigured: Could not resolve URL for hyperlinked relationship using view name "client-detail". You may have failed to include the related model in your API, or incorrectly configured the lookup_field attribute on this field.

It only works if I manually set the url attribute as follows.

url = HyperlinkedIdentityField(view_name='myapp:myview-detail')

Is there any way to do this without having to set the url attribute for every serializer class?

[jerinpetergeorge]

Is there any way to do this without having to set the url attribute for every serializer class?

If you're not setting the url parameter, how would be the result? What is your expected behavior?

[brunodantas]

Current behavior: it is required to set the url attribute manually for every serializer class in order to specify the app name. Otherwise, I get that ImproperlyConfigured error.

Example:

class Example1Serializer(HyperlinkedModelSerializer):
    url = HyperlinkedIdentityField(view_name='myapp:example1-detail') 
    class Meta:
    ...

class Example2Serializer(HyperlinkedModelSerializer):
    url = HyperlinkedIdentityField(view_name='myapp:example2-detail') 
    class Meta:
    ...

class Example3Serializer(HyperlinkedModelSerializer):
    url = HyperlinkedIdentityField(view_name='myapp:example3-detail') 
    class Meta:
    ...

Desired behavior: some way to specify the app name only once instead of setting the url parameter manually for every class. Example:

APP_NAME = "myapp"

class Example1Serializer(HyperlinkedModelSerializer):
    class Meta:
    ...

class Example2Serializer(HyperlinkedModelSerializer):
    class Meta:
    ...

class Example3Serializer(HyperlinkedModelSerializer):
    class Meta:
    ...