Number of the aid range

Oracle 11.2.0.1

create table test (col1 varchar2 (3), col2 varchar2 (4));

insert into test ('201 ', ' 1234') values;

insert into test ('201 ', ' 1235') values;

insert into test ('201 ', ' 1236') values;

insert into test ('201 ', ' 1240') values;

insert into test ('201 ', ' 1241') values;

insert into test ('201 ', ' 1250') values;

insert into test ('202 ', ' 1251') values;

insert into test ('202 ', ' 1252') values;

insert into test ('202 ', ' 1253') values;

insert into test ('202 ', ' 1254') values;

insert into test ('203 ', ' 1255') values;

insert into test ('204 ', ' 1256') values;

insert into test values ("204", "1257 ');

insert into test ("204", "1258") values;

Power required:

col1 start end total

201 1234 1236 3 <-because the numbers are of 1234-1236

201, 1240, 1241, 2 <- then from 1240 to 1241

201 1250 1250 1 <-one number for the same value of col1

202 1251 1254 4 <-same as above.

203 1255 1255 1

204 1256 and 1258 3

As indicated above required output, I see numbers continue as starting point at the end of column for the value of col1.  If there is no gap or a single value, it should come in another line.  Kindly help me.

Hello

This is a classic method of Tabibitosan: Tabibitosan tutorial Aketi Jyuuzou method

SQL> with got_grp as
  2  (
  3  select col1
  4        , col2
  5        , row_number() over (partition by col1 order by col2)
  6          - col2 as grp
  7    from test
  8  )
  9  select col1, min(col2) col2_start, max(col2) col2_end, count(*) total
 10    from got_grp
 11   group by col1, grp
 12   order by col1, col2_start;

COL COL2 COL2      TOTAL
--- ---- ---- ----------
201 1234 1236          3
201 1240 1241          2
201 1250 1250          1
202 1251 1254          4
203 1255 1255          1
204 1256 1258          3                                                        

6 rows selected.

SQL>

Kind regards.

Alberto

Tags: Database

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