Code review fixes

Author: aclark4life

django_mongodb_backend/base.py

@@ -336,10 +336,6 @@ def cursor(self):
 
     def get_database_version(self):
         """Return a tuple of the database's version."""
-        # TODO: Remove this workaround and replace with
-        # `tuple(self.connection.server_info()["versionArray"])` when the minimum
-        # supported version of pymongocrypt is >= 1.14.2 and PYTHON-5429 is resolved.
-        # See: https://jira.mongodb.org/browse/PYTHON-5429
         return tuple(self.connection.admin.command("buildInfo")["versionArray"])
 
     ## Transaction API for django_mongodb_backend.transaction.atomic()

django_mongodb_backend/features.py

@@ -596,10 +596,6 @@ def mongodb_version(self):
     def is_mongodb_6_3(self):
         return self.mongodb_version >= (6, 3)
 
-    @cached_property
-    def is_mongodb_7_0(self):
-        return self.mongodb_version >= (7, 0)
-
     @cached_property
     def is_mongodb_8_0(self):
         return self.mongodb_version >= (8, 0)

django_mongodb_backend/management/commands/showencryptedfieldsmap.py

@@ -32,4 +32,4 @@ def handle(self, *args, **options):
                     if model_has_encrypted_fields(model):
                         fields = editor._get_encrypted_fields(model, create_data_keys=False)
                         encrypted_fields_map[model._meta.db_table] = fields
-        self.stdout.write(json_util.dumps(encrypted_fields_map, indent=2))
+        self.stdout.write(json_util.dumps(encrypted_fields_map, indent=4))

docs/ref/contrib/admin.rst

@@ -20,27 +20,8 @@ that you have :ref:`specified the default pk field
     class to ensure that encrypted fields are handled correctly in the admin
     interface.
 
-    Define a model with encrypted fields:
-
-    .. code-block:: python
-
-        # myapp/models.py
-        from django.db import models
-        from django_mongodb_backend.fields import EmbeddedModelField
-
-
-        class Patient(models.Model):
-            patient_name = models.CharField(max_length=255)
-            patient_id = models.BigIntegerField()
-            patient_record = EmbeddedModelField("PatientRecord")
-
-            def __str__(self):
-                return f"{self.patient_name} ({self.patient_id})"
-
-    Register it with the Django admin using the ``EncryptedModelAdmin`` as shown
-    below:
-
-    .. code-block:: python
+    Register encrypted models with the Django admin using the
+    ``EncryptedModelAdmin`` as shown below::
 
         # myapp/admin.py
         from django.contrib import admin