问题
If I used the following expression, the result should be 1.
regexp_instr('500 Oracle Parkway, Redwood Shores, CA','[[:digit:]]')
Is there a way to make this look for the last number in the string? If I were to look for the last number in the above example, it should return 3.
回答1:
If you were using 11g, you could use regexp_count to determine the number of times that a pattern exists in the string and feed that into the regexp_instr
regexp_instr( str,
'[[:digit:]]',
1,
regexp_count( str, '[[:digit:]]')
)
Since you're on 10g, however, the simplest option is probably to reverse the string and subtract the position that is found from the length of the string
length(str) - regexp_instr(reverse(str),'[[:digit:]]') + 1
Both approaches should work in 11g
SQL> ed
Wrote file afiedt.buf
1 with x as (
2 select '500 Oracle Parkway, Redwood Shores, CA' str
3 from dual
4 )
5 select length(str) - regexp_instr(reverse(str),'[[:digit:]]') + 1,
6 regexp_instr( str,
7 '[[:digit:]]',
8 1,
9 regexp_count( str, '[[:digit:]]')
10 )
11* from x
SQL> /
LENGTH(STR)-REGEXP_INSTR(REVERSE(STR),'[[:DIGIT:]]')+1
------------------------------------------------------
REGEXP_INSTR(STR,'[[:DIGIT:]]',1,REGEXP_COUNT(STR,'[[:DIGIT:]]'))
-----------------------------------------------------------------
3
3
回答2:
Another solution with less effort is
SELECT regexp_instr('500 Oracle Parkway, Redwood Shores, CA','[^[:digit:]]*$')-1
FROM dual;
this can be read as.. find the non-digits at the end of the string. and subtract 1. which will give the position of the last digit in the string..
REGEXP_INSTR('500ORACLEPARKWAY,REDWOODSHORES,CA','[^[:DIGIT:]]*$')-1
--------------------------------------------------------------------
3
which i think is what you want.
(tested on 11g)
来源:https://stackoverflow.com/questions/9755681/use-regexp-instr-to-get-the-last-number-in-a-string