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<shiuhlin_at_jshape.com> wrote in message news:916el8$ci5$1_at_nnrp1.deja.com...
> Hi,
>
> I have a problem needs help. My problem is : I have a table called
"DOCS"
> which contains a field called "NEWS", and I like to perform a query to
find
> all records which contains "10%" key words in the NEWS field. The "%" in
> "10%" (10 - percent) isn't the wildcard character used in SQL. Now, can
> anyone show me how to find the "10%" in NEWS?
>
> SELECT NEWS FROM DOCS WHERE NEWS LIKE ?????
>
> Shiuh-Lin Lee
Use ESCAPE.
SELECT *
FROM docs
WHERE news LIKE '%10/%%' ESCAPE '/';
Directly from SQL Reference, Release 8.1.5 : ESCAPE identifies a single character as the escape character. The escape character can be used to cause Oracle to interpret % or _ literally, rather than as a special character. If you wish to search for strings containing an escape character, you must specify this character twice. For example, if the escape character is ’/’, to search for the string ’client/server’, you must specify, ’client//server’.
-- Michael O'Neill mjoneill_at_email.comReceived on Tue Dec 12 2000 - 20:46:14 CST