SQL Server Equivalent of Oracle 'CONNECT BY PRIOR', and 'ORDER SIBLINGS BY'

Question

I\'ve got this Oracle code structure I\'m trying to convert to SQL Server 2008 (Note: I have used generic names, enclosed column names and tabl

Answer 1

The option given by the user "a_horse_with_no_name" worked for me. I changed the code and applied it to a menu generator query and it worked the first time. Here is the code:

WITH tree(option_id,
       option_description,
      option_url,
      option_icon,
      option_level,
      sort_path)
     AS (
     SELECT ppo.option_id,
            ppo.option_description,
          ppo.option_url,
          ppo.option_icon,
          1 AS option_level,
          CAST('/' + RIGHT('00' + CONVERT(VARCHAR, ppo.option_index), 6) AS VARCHAR(MAX))
     FROM security.options_table_name ppo
     WHERE ppo.option_parent_id IS NULL
     UNION ALL
     SELECT co.option_id,
            co.option_description,
          co.option_url,
          co.option_icon,
          po.option_level + 1,
          po.sort_path + '/' + RIGHT('00' + CONVERT(VARCHAR, co.option_index), 6)
     FROM security.options_table_name co,
          tree AS po
     WHERE po.option_id = co.option_parent_id)
     SELECT *
     FROM tree
    ORDER BY sort_path;

Answer 2

to get dates for last 10 days:

SELECT DISTINCT RecordDate = DATEADD(DAY,-number,CAST(GETDATE() AS DATE)) 
FROM master..[spt_values] 
WHERE number BETWEEN 1 AND 10

Answer 3

Simulating the LEVEL column

The level column can easily be simulated by incrementing a counter in the recursive part:

WITH tree (empid, name, level) AS  (
  SELECT empid, name, 1 as level
  FROM emp
  WHERE name = 'Joan'

  UNION ALL

  SELECT child.empid, child.name, parent.level + 1
  FROM emp as child
    JOIN tree parent on parent.empid = child.mgrid
)
SELECT name 
FROM tree;

---

Simulating order siblings by

Simulating the order siblings by is a bit more complicated. Assuming we have a column sort_order that defines the order of elements per parent (not the overall sort order - because then order siblings wouldn't be necessary) then we can create a column which gives us an overall sort order:

WITH tree (empid, name, level, sort_path) AS  (
  SELECT empid, name, 1 as level, 
         cast('/' + right('000000' + CONVERT(varchar, sort_order), 6) as varchar(max))
  FROM emp
  WHERE name = 'Joan'

  UNION ALL

  SELECT child.empid, child.name, parent.level + 1, 
         parent.sort_path + '/' + right('000000' + CONVERT(varchar, child.sort_order), 6) 
  FROM emp as child
    JOIN tree parent on parent.empid = child.mgrid
)
SELECT * 
FROM tree
order by sort_path;

The expression for the sort_path looks so complicated because SQL Server (at least the version you are using) does not have a simple function to format a number with leading zeros. In Postgres I would use an integer array so that the conversion to varchar isn't necessary - but that doesn't work in SQL Server either.