Question

I am selecting from an oracle table (using sql) that contains the following values:

ae0.32767  
bcm0.0  
em0.0  
ge-0/1/0.32767  
ge-7/1/0.100  
ge-7/1/0.32767  
lo0.0  
lsi.0  
pc-0/0/0.16383  
pc-1/0/0.16383  
pc-5/0/0.16383  
xe-0/0/0.838 

How do I extract values (using regexp) after and not including the decimal point.

For example:

32767  
0  
0  
32767  
100  
32767  
0  
0  
16383  
16383  
16383  
838  

Answer 1

The regexp_substr function can be used to extract digits after a period at the end of a string. Here is an example:

 -- sample of data from your question
 SQL> with t1  as (
  2    select 'ae0.32767' as col from dual union all
  3    select 'bcm0.0'          from dual union all
  4    select 'em0.0'           from dual union all
  5    select 'ge-0/1/0.32767'  from dual union all
  6    select 'ge-7/1/0.100'    from dual union all
  7    select 'ge-7/1/0.32767'  from dual union all
  8    select 'lo0.0'           from dual union all
  9    select 'lsi.0'           from dual union all
 10    select 'pc-0/0/0.16383'  from dual union all
 11    select 'pc-1/0/0.16383'  from dual union all
 12    select 'pc-5/0/0.16383'  from dual union all
 13    select 'xe-0/0/0.838'    from dual
 14  )
 15  select regexp_substr(col, '(\.)([[:digit:]]+)$', 1,1,'i',2) res
 16    from t1
 17  ;

RES
--------------------------------------------------------
32767
0
0
32767
100
32767
0
0
16383
16383
16383
838

12 rows selected