Python Django uses wrong query

Question

I have a problem, to solve it I tried going into the shell but I couldn't find out the solution so I hope that you can help me. I'm importing my Model Images which contains an id and the image URL but I can't get it because of this error. The code:

from recipes_database.models import Images 
Images.objects.all()

And here the error:

Traceback (most recent call last):   File "F:\Developement\Projects\WhatToCookWeb\venv\lib\site-packages\django\db\backends\utils.py", line 84, in _execute
    return self.cursor.execute(sql, params)   File "F:\Developement\Projects\WhatToCookWeb\venv\lib\site-packages\django\db\backends\sqlite3\base.py", line 413, in execute
    return Database.Cursor.execute(self, query, params) sqlite3.OperationalError: no such column: Images.id

The above exception was the direct cause of the following exception:

Traceback (most recent call last):   File "<console>", line 1, in <module>   File "F:\Developement\Projects\WhatToCookWeb\venv\lib\site-packages\django\db\models\query.py", line 263, in __repr__
    data = list(self[:REPR_OUTPUT_SIZE + 1])   File "F:\Developement\Projects\WhatToCookWeb\venv\lib\site-packages\django\db\models\query.py", line 269, in __len__
    self._fetch_all()   File "F:\Developement\Projects\WhatToCookWeb\venv\lib\site-packages\django\db\models\query.py", line 1303, in _fetch_all
    self._result_cache = list(self._iterable_class(self))   File "F:\Developement\Projects\WhatToCookWeb\venv\lib\site-packages\django\db\models\query.py", line 53, in __iter__
    results = compiler.execute_sql(chunked_fetch=self.chunked_fetch, chunk_size=self.chunk_size)   File "F:\Developement\Projects\WhatToCookWeb\venv\lib\site-packages\django\db\models\sql\compiler.py", line 1154, in execute_sql
    cursor.execute(sql, params)   File "F:\Developement\Projects\WhatToCookWeb\venv\lib\site-packages\django\db\backends\utils.py", line 98, in execute
    return super().execute(sql, params)   File "F:\Developement\Projects\WhatToCookWeb\venv\lib\site-packages\django\db\backends\utils.py", line 66, in execute
    return self._execute_with_wrappers(sql, params, many=False, executor=self._execute)   File "F:\Developement\Projects\WhatToCookWeb\venv\lib\site-packages\django\db\backends\utils.py", line 75, in _execute_with_wrappers
    return executor(sql, params, many, context)   File "F:\Developement\Projects\WhatToCookWeb\venv\lib\site-packages\django\db\backends\utils.py", line 84, in _execute
    return self.cursor.execute(sql, params)   File "F:\Developement\Projects\WhatToCookWeb\venv\lib\site-packages\django\db\utils.py", line 90, in __exit__
    raise dj_exc_value.with_traceback(traceback) from exc_value   File "F:\Developement\Projects\WhatToCookWeb\venv\lib\site-packages\django\db\backends\utils.py", line 84, in _execute
    return self.cursor.execute(sql, params)   File "F:\Developement\Projects\WhatToCookWeb\venv\lib\site-packages\django\db\backends\sqlite3\base.py", line 413, in execute
    return Database.Cursor.execute(self, query, params) django.db.utils.OperationalError: no such column: Images.id

But when I do the same thing with my other models it works and I am wondering whats different this time. Here is my Images Model:

class Images(models.Model):
    # Field name made lowercase.
    rezept_id = models.ForeignKey(
        'Rezepte', models.DO_NOTHING, db_column='Rezept_ID', blank=True, null=True)
    # Field name made lowercase.
    image_url = models.CharField(
        db_column='Image_URL', blank=True, null=True, max_length=1000)

    class Meta:
        managed = False
        db_table = 'Images'

I think that Django somehow uses a wrong query because there is no Images.id just an Images.recipe_id. Hope you can help me Thank you!

Is recipe_id the PRIMARY KEY? What is the primary key of this table? — willeM_ Van Onsem
– willeM_ Van Onsem, Commented Sep 2, 2020 at 18:16

willeM_ Van Onsem · Accepted Answer · 2020-09-03 15:27:14Z

Django automatically adds a primary key if you did not specify one yourself. Indeed, as the documentation says:

By default, Django gives each model the following field:
id = models.AutoField(primary_key=True)
This is an auto-incrementing primary key.

If you’d like to specify a custom primary key, specify primary_key=True on one of your fields. If Django sees you’ve explicitly set Field.primary_key, it won’t add the automatic id column.

In your model, you did not set a primary key, so Django will automatically add an extra field id, and this will thus be used in the query as well.

For example if the image_url is the primary key, you can specify this as:

# image_url is the primary key

class Images(models.Model):
    # Field name made lowercase.
    rezept = models.ForeignKey(
        'Rezepte',
        models.DO_NOTHING,
        db_column='Rezept_ID',
        blank=True,
        null=True
    )
    # Field name made lowercase.
    image_url = models.CharField(
        db_column='Image_URL',
        blank=True,
        null=True,
        max_length=1000,
        primary_key=True
    )

    class Meta:
        managed = False
        db_table = 'Images'

In case the rezept is the primary key, it thus also means that it is unique, in that case, it makes more sense to make it a OneToOneField [Django-doc], since that is a ForeignKey with a uniqness constraint, and furthermore it has some implications on the reverse relation:

# rezept is the primary key

class Images(models.Model):
    # Field name made lowercase.
    rezept = models.OneToOneField(
        'Rezepte',
        db_column='Rezept_ID',
        primary_key=True
    )
    # Field name made lowercase.
    image_url = models.CharField(
        db_column='Image_URL',
        blank=True,
        null=True,
        max_length=1000
    )

    class Meta:
        managed = False
        db_table = 'Images'

Note: Normally one does not add a suffix _id to a ForeignKey field, since Django will automatically add a "twin" field with an _id suffix. Therefore it should be rezept, instead of ~~rezept_id~~.

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