oracle - Sql query to fetch minimun date start against each employee -


i have 2 table asg , work rship

in work rship table there date_start each employee. few employee date_start duplicate have choose min(date_start) work rship table

for wrote query :

select assignment_name,   regexp_substr(assignment_name, '[0-9]+') person_number,   nvl( wrk.date_start,t.effective_start_date) ,   nvl( wrk.worker_type,'e'),   nvl( wrk.legal_employer_name, 'n/a')   (select apps.assignment_table.*,     count(*) on (partition assignment_name, effective_start_date, effective_end_date, effective_latest_change) c   apps.assignment_table   ) t left outer join   (select *   apps.work_table   date_start =     (select min(date_start) apps.work_table     )   ) wrk on regexp_substr(t.assignment_name, '[0-9]+')=wrk.person_number c   =1;

but in above query did mistake of selecting select min(date_start) apps.work_table means min(date_start) entire table. instead of should have included regexp_substr(t.assignment_name, '[0-9]+')=wrk.person_number inside inline query itself.

but when including it.. not working. can point out erroe.

left outer join   (select *   apps.work_table   date_start =     (select min(date_start) apps.work_table     )   ) wrk on

unless start_date person minimum value across people in entire table (and not minimum date person) code above not find rows match person want.

what meant like:

left outer join ( select *     (     select t.*,            row_number() on ( partition person_number order date_start asc ) rn       apps.work_table t   )    rn = 1 ) wrk on

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