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find word greater than 17
comp.databases.oracle.misc comp.databases.oracle.misc
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#1
 
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find word greater than 17 -
04-23-2008
, 09:40 AM
#2
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i am looking to find the first word of a field greater than 17 |
Quote:
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select businessname from tablename where CHAR(' ', businessname + ' ') > 17 |
where length(trim(businessname)) > 17
or...
-- has any (space delimited) word of length > 17
select regexp_substr(businessname, '[^[:space:]]{18,}')
from tablename
where regexp_like(businessname, '[^[:space:]]{18,}')
or...
-- first (space delimited) word has length > 17
select regexp_substr(businessname, '[^[:space:]]{18,}')
from tablename
where regexp_like(businessname, '^[^[:space:]]{18,}')
--
Peter
#3
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|
|
i am looking to find the first word of a field greater than 17 |
Quote:
|
select businessname from tablename where CHAR(' ', businessname + ' ') > 17 |
where length(trim(businessname)) > 17
or...
-- has any (space delimited) word of length > 17
select regexp_substr(businessname, '[^[:space:]]{18,}')
from tablename
where regexp_like(businessname, '[^[:space:]]{18,}')
or...
-- first (space delimited) word has length > 17
select regexp_substr(businessname, '[^[:space:]]{18,}')
from tablename
where regexp_like(businessname, '^[^[:space:]]{18,}')
--
Peter
#4
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|
|
i am looking to find the first word of a field greater than 17 |
Quote:
|
select businessname from tablename where CHAR(' ', businessname + ' ') > 17 |
where length(trim(businessname)) > 17
or...
-- has any (space delimited) word of length > 17
select regexp_substr(businessname, '[^[:space:]]{18,}')
from tablename
where regexp_like(businessname, '[^[:space:]]{18,}')
or...
-- first (space delimited) word has length > 17
select regexp_substr(businessname, '[^[:space:]]{18,}')
from tablename
where regexp_like(businessname, '^[^[:space:]]{18,}')
--
Peter
#5
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|
|
i am looking to find the first word of a field greater than 17 |
Quote:
|
select businessname from tablename where CHAR(' ', businessname + ' ') > 17 |
where length(trim(businessname)) > 17
or...
-- has any (space delimited) word of length > 17
select regexp_substr(businessname, '[^[:space:]]{18,}')
from tablename
where regexp_like(businessname, '[^[:space:]]{18,}')
or...
-- first (space delimited) word has length > 17
select regexp_substr(businessname, '[^[:space:]]{18,}')
from tablename
where regexp_like(businessname, '^[^[:space:]]{18,}')
--
Peter
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