How get the value by passing first letter

Hi All,
I have to display values based on first digit .
Here is the query. " SELECT assetid,version_type FROM EDECO_ASSETS WHERE version_type like '1%'; "
Sample data is.
AssetId,Version_type.
1201 1
1202 1,2
1203 1,2,3
1204 10,11
1205 10,12
if I pass version_type is '1' then it should display 1201,1202,1203,
if I pass vertsion_type '10' then it shoud display 1204,1205 records. DB version is 11g.
Please let me know how to achieve this. Thanks in Advance.
Regards,
Rajasekhar
Edited by: rajasekhar_n on Nov 17, 2008 4:53 AM

in oracle 7i onwards you can do
SELECT assetid,version_type FROM EDECO_ASSETS WHERE substr(version_type,1,instr(version_type,',')-1) = :p_parameter; In oracle 10 you could do
SELECT assetid,version_type FROM EDECO_ASSETS where regexp_like (version_type,'^'||:p_parameter||'(,|$)+')
No absolutly sure if I got the restriction correctly. Maybe some of the regexp_gurus can recheck.
:p_parameter can be replaced with '1' or '10' as you like.
Edited by: Sven W. on Nov 17, 2008 2:12 PM - changed repexp_like comparison

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i am experimenting with oops concept. first have a look at these coding and then tell me this.
what do i have to do to get the value of an object created. i want to use the get method to get the name, food and quantity, without the use of the created object. no. wrong.
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