SQL Query: Need order by count, most must be on top, the rest follows

Question

TABLEA

JobCode Job1 Job2 Job3 zip
------- ---- ---- ---- ----------
F       F    S    NULL 90030
F       F    S    NULL 90031
F       F    S    NULL 90031
F              


        

Answer 1

SQL Server 2008 using COUNT() OVER

select *, c = count(1) over (partition by zip)
from tbl
order by c desc;

If you don't need to see the additional column, then you can move the COUNT() OVER clause into the ORDER BY clause.

select JobCode, Job1, Job2, Job3, zip
from tbl
order by count(1) over (partition by zip) desc;

Answer 2

SELECT 
  JobCode, Job1, Job2, Job3, order_jobs.zip
FROM
  jobs
  JOIN (SELECT zip, COUNT(*) AS zipcount FROM jobs GROUP BY zip) ordering 
ON jobs.zip = ordering.zip
ORDER BY zipcount DESC

Answer 3

To accomplish this, you should join against a subquery which returns the count per zipcode. The joined subquery is only needed to provide the counts (even if not displayed), while the main table yourtable provides all the rest of the columns.

SELECT 
  JobCode, 
  Job1,
  Job2,
  Job3,
  subq.zip
FROM
  yourtable
  JOIN (
     /* Subquery returns count per zip group */
     SELECT zip, COUNT(*) AS numzip 
     FROM yourtable 
     GROUP BY zip
  ) subq ON yourtable.zip = subq.zip
ORDER BY numzip DESC