Solved

# count last three characters of the record

Posted on 2014-02-27
439 Views
Hello All:

I need query help... I have following records ....

filename.a01
filename.a02
filename.a03
filename.a04
filename.a05
filename.a06
filename.a01
filename.a08
filename.a01
filename.a10
filename.a11
filename.a12
filename.a13
filename.a01
filename.a15
filename.a16
filename.a17
filename.a18
filename.a19
filename.a04
filename.a21
filename.a22
filename.a23
filename.a24
filename.a04
filename.a26
filename.a27
filename.a28
filename.a29
filename.a04
filename.a31
filename.a32
filename.a33
filename.a34
filename.a35
filename.a36

I need to count all the extents of the filename for example

a36 = 1
a35 = 1

and so on ...

How to query like this?
0
Question by:CalmSoul
• 6
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• 2

LVL 76

Expert Comment

ID: 39893739
Try the example below.  You can use regular expressions if the actual data is more complex but they are pretty expensive operations and should be avoided whenever possible,

``````drop table tab1 purge;

create table tab1(col1 varchar2(20));

insert into tab1 values('filename.a01');
insert into tab1 values('filename.a02');
insert into tab1 values('filename.a03');
insert into tab1 values('filename.a04');
insert into tab1 values('filename.a05');
insert into tab1 values('filename.a06');
insert into tab1 values('filename.a01');
insert into tab1 values('filename.a08');
commit;

select substr(col1,instr(col1,'.')+1), count(*)
from tab1
group by substr(col1,instr(col1,'.')+1)
/
``````
0

LVL 76

Expert Comment

ID: 39893741
Regular expression version:
``````select regexp_substr(col1,'[^.]+\$') , count(*)
from tab1
group by regexp_substr(col1,'[^.]+\$')
/
``````
0

LVL 5

Author Comment

ID: 39893784
How is this solution on performance ... ?
0

LVL 76

Expert Comment

ID: 39893788
>>How is this solution on performance ... ?

First one should be about as fast as I can think to make it.

Depending on how many rows and how often you need to run this query, you can create a function based index (FBI) on substr(col1,instr(col1,'.')+1) and it should fly.

The downside to the FBI is the hit on inserts and updates on the base table.

You will need to experiment to find the balance.
0

LVL 34

Accepted Solution

johnsone earned 500 total points
ID: 39894565
I would make a subtle change:

``````select substr(col1,instr(col1,'.', -1)+1), count(*)
from tab1
group by substr(col1,instr(col1,'.', -1)+1)
/
``````

This would handle a case where the file name contained a period, such as file.name.a01.  The original would process that as name.a01, which doesn't sound like the intent.  This would change to look for the last period in the name and not the first period.

If you wanted to instead look strictly at the last 3 characters of the name, as the post title suggests, then it would be:

``````select substr(col1,-3), count(*)
from tab1
group by substr(col1, -3)
/
``````
0

LVL 76

Expert Comment

ID: 39894646
>>This would handle a case where the file name contained a period

Excellent catch!
0

LVL 5

Author Comment

ID: 39895464
``````select substr(col1,instr(col1,'.', -1)+1), count(*)
from tab1
group by substr(col1,instr(col1,'.', -1)+1)
``````

Problem with this solution it should also check if "." is present in last 3 charaters. Basicially its counting where ever the period is present in the string ...

So...

Find the period in the string and verify is it the last 3 charaters from the right. If yes, then count.
0

LVL 76

Expert Comment

ID: 39895473
>>Problem with this solution it should also check if "." is present in last 3 charaters

You lost me.  Did you try the corrections from johnsone above?  his first one starts at the end of the string and takes from the LAST period.

The second one takes the last three characters no matter what they are.

Try the regular expression one I posted.  It takes anything not a '.' to the end of the string.

We can give you anything you want, we just need to know all the requirements and variations in your data.
0

LVL 5

Author Comment

ID: 39895486
with regular experssion I am getting following error

``````Lookup Error
ORA-00904: invalid column name
``````
0

LVL 5

Author Comment

ID: 39895510
I think regular expression doesn't work in oracle 8i
0

LVL 76

Expert Comment

ID: 39895608
>>I think regular expression doesn't work in oracle 8i

Probably not.

Try the ones from johnsone.  If those fail, please post additional data and expected results.
0

LVL 34

Expert Comment

ID: 39897266
I'm lost by the explanation too.  If the ones that I have already posted don't work, please provide better samples of what you are looking for.  Especially post the "hard" cases.
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