Hi Juan,

this will do it in version 10.5.8 (and probably earlier:

nested_category = Table(
    'nested_category',
    MetaData(),
    Column('category_id', Integer, primary_key=True),
    Column('name', Text, nullable=False),
    Column('lft', Integer, nullable=False),
    Column('rgt', Integer, nullable=False)
    )
node = nested_category.alias('node')
parent = nested_category.alias('parent')
query = select([node.c.name, (func.count(node.c.name) - text
('1')).label('level')],
    from_obj=join(node, parent,
                  node.c.lft.between(parent.c.lft, parent.c.rgt)
                  )
    ).group_by(node.c.name)

str(query) will show that it is correct (it uses a JOIN expression
instead of the WHERE condition, but that's equivalent and more
explicit)

The text('1') instead of just 1 is so that the literal constant 1 is
not needlessly replaced by a bind param. It works either way though.

Regards,

    - Gulli


On Jan 29, 8:53 am, Juan Dela Cruz <juandelacru...@gmail.com> wrote:
> Can someone please help me to figure out the equivalent of this sql query to
> sqlalchemy
>
> This my nested_category table:
>
> +-------------+----------------------+-----+-----+
> | category_id | name                 | lft | rgt |
> +-------------+----------------------+-----+-----+
> |           1 | ELECTRONICS          |   1 |  20 |
> |           2 | TELEVISIONS          |   2 |   9 |
> |           3 | TUBE                 |   3 |   4 |
> |           4 | LCD                  |   5 |   6 |
> |           5 | PLASMA               |   7 |   8 |
> |           6 | PORTABLE ELECTRONICS |  10 |  19 |
> |           7 | MP3 PLAYERS          |  11 |  14 |
> |           8 | FLASH                |  12 |  13 |
> |           9 | CD PLAYERS           |  15 |  16 |
> |          10 | 2 WAY RADIOS         |  17 |  18 |
> +-------------+----------------------+-----+-----+
>
> SELECT node.name, (COUNT(node.name)-1) AS level
>     FROM nested_category AS node, nested_category AS parent
>         WHERE node.lft BETWEEN parent.lft AND parent.rgt
>             GROUP BY node.name;
>
> The result will be:
> +----------------------+-------+
>
> | name                 | depth |
> +----------------------+-------+
> | ELECTRONICS          |     0 |
> | TELEVISIONS          |     1 |
> | TUBE                 |     2 |
> | LCD                  |     2 |
> | PLASMA               |     2 |
> | PORTABLE ELECTRONICS |     1 |
> | MP3 PLAYERS          |     2 |
> | FLASH                |     3 |
> | CD PLAYERS           |     2 |
> | 2 WAY RADIOS         |     2 |
> +----------------------+-------+

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