05 Feb 2008
can you try this .......select NAMEfrom TABLE_NAMEwhere NAME LIKE '%''%';
05 Feb 2008
Silva I tried '%''%' it is not working
06 Feb 2008
using two single quoteslike '%''%'seems to work. same with insert statements, i.e. insert into table (names) values ('Int''l'); would insert the string Int'l
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I was trying select NAME from TABLE_NAME where NAME '%\'%' ESCAPE '\';The above query not working.I would like to how would we handle quotation in side the data.