RE: question about regexp_substr
- From: Jackie Brock <J.Brock@xxxxxxxxxxxxx>
- To: amonte <ax.mount@xxxxxxxxx>
- Date: Tue, 16 Dec 2014 21:42:56 +0000
☺ Run it with various values where the three is – get the first occurrence,
then the second, then the third, then the fourth – you’ll easily see what it’s
doing.
From: amonte [mailto:ax.mount@xxxxxxxxx]
Sent: Tuesday, December 16, 2014 2:30 PM
To: Jackie Brock
Cc: Oracle-L Group
Subject: Re: question about regexp_substr
Hello Jackie
I know + means > 1 and * > 0 occurence. But I dont see why they give different
results in my example.
I understand that what query is asking with * is
"find in the string any non-numeric character pattern, no matter if the there
is character or not in the third occurence". I dont see why B satisfies such
condition?
Thanks in advance
Alex
2014-12-16 22:23 GMT+01:00 Jackie Brock
<J.Brock@xxxxxxxxxxxxx<mailto:J.Brock@xxxxxxxxxxxxx>>:
The plus sign indicates that it expects at least 1 digit (1 or more). The *
means 0 or more.
From: oracle-l-bounce@xxxxxxxxxxxxx<mailto:oracle-l-bounce@xxxxxxxxxxxxx>
[mailto:oracle-l-bounce@xxxxxxxxxxxxx<mailto:oracle-l-bounce@xxxxxxxxxxxxx>] On
Behalf Of amonte
Sent: Tuesday, December 16, 2014 2:07 PM
To: Oracle-L Group
Subject: question about regexp_substr
Hi people
I have some difficulty understanding applying an operator to the pattren in
regexp_substr. Not sure if anyone can help ?
The question is, what is the difference between these two queries:
select regexp_substr('A1B2C3D4E', '[^0-9]+', 1, 3) from dual;
R
-
C
select regexp_substr('A1B2C3D4E', '[^0-9]*', 1, 3) from dual;
R
-
B
Why * and + gives different answers?
Thanks in advance
Alex
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