How can I grab this in oracle in SQL

Hi,

here is sqlfidlle

http://sqlfiddle.com/#!4/8089b/1

I would like to get the result below out of that...

fist  |  second
100  | 200
101  | 201
102  | 202
103  | 203
104  | 204
105  | 205

regards
LVL 1
hi4pplAsked:
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Gerwin Jansen, EE MVEConnect With a Mentor Topic Advisor Commented:
Already some replies above but  I'd first look for that part that contains the numbers:
select regexp_substr (ThatField,'{name,\d{3},\d{3}}') from Table1;

Open in new window


which would result in:
{name,100,200}
{name,101,201}
{name,102,202}
{name,103,203}
{name,104,204}
{name,105,205}

and then 'translate' using regexp_replace it the way you want it:
select
regexp_replace((regexp_substr (ThatField,'{name,\d{3},\d{3}}')),
'({name,)(\d{3}),(\d{3})(})',
'\2|\3')
as "first|second"
from Table1;

Open in new window


(the regexp_replace is finding 4 patterns, I'm only printing 2 and 3 with a | in between)

This results in:

FIRST|SECOND
100|200
101|201
102|202
103|203
104|204
105|205
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Gerwin Jansen, EE MVETopic Advisor Commented:
What have you tried already, besides 'select * from ....' ?
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hi4pplAuthor Commented:
Hi,

I did try select regexp_substr(thatfield,'[^,]+]',2,2) as a table1 but it only pick if it will be the start if it's at end it won't pick it ...
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awking00Commented:
select
substr(thatfield,instr(thatfield,'{name') + 6,3) as first,
substr(thatfield,instr(thatfield,'{name') + 10,3) as second
from table1;
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awking00Connect With a Mentor Commented:
With regexp -
select ltrim(regexp_substr(thatfield,',[^,]+',instr(thatfield,'{name')),',') as first,
ltrim(regexp_substr(thatfield,',[^}]+',instr(thatfield,'{name') + 6),',') as second
from table1;
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PaulCommented:
will the "field" widths alter in the data? e.g.

THATFIELD
{blahblah,77100,6200}{secondname,200001}
{muchlongerstringinthislocationwithembeddednumbers123name,17801,201234}
{can spaces exist in this location - question to you,1941,51244}


when preparing sample data make sure it covers all situations (well, at least the most likely)
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