Reply-To: "Michael O'Neill" From: "Michael O'Neill" Newsgroups: comp.databases.oracle.server References: <916el8$ci5$1@nnrp1.deja.com> Subject: Re: SQL query "10%" in a field Lines: 35 Organization: Bottom Line Party X-Priority: 3 X-MSMail-Priority: Normal X-Newsreader: Microsoft Outlook Express 5.50.4133.2400 X-MimeOLE: Produced By Microsoft MimeOLE V5.50.4133.2400 Message-ID: Date: Wed, 13 Dec 2000 02:46:14 GMT NNTP-Posting-Host: 24.11.25.230 X-Complaints-To: abuse@home.net X-Trace: news2.rdc2.tx.home.com 976675574 24.11.25.230 (Tue, 12 Dec 2000 18:46:14 PST) NNTP-Posting-Date: Tue, 12 Dec 2000 18:46:14 PST wrote in message news:916el8$ci5$1@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@email.com