2004 IR Help Posted September 2, 2006 Share Posted September 2, 2006 By: allan - allanpanaguiton parameter in Report sql query with "like" 2004-08-27 02:31 Anybody who can help me with this type of sql? select podm_id, podm_date, supp_name, podm_incharge from tps_pomas left outer join tco_supplier ON podm_suppid = supp_id where podm_suppid LIKE '% $P{SupId} %' this entry will cause error because the LIKE with "%" (for wildcard) portion requires me to place a single quote " ' " on the syntax which causes the error. Thanx for you help!! :-) By: C-Box - c-box RE: parameter in Report sql query with "like" 2004-08-27 03:46 Hi, Have you already tried to put the whole query into a parameter and use that parameter in the query instead? C-Box By: Rick Millar - rmillar RE: parameter in Report sql query with "like" 2004-08-27 09:25 Try ...LIKE '%$P!{SupId}%' --don't forget to include the !. Make sure SupId == "yourSupIdVal" By: Mark Rhodes - mrhodes2 RE: parameter in Report sql query with "like" 2004-08-27 09:38 can you pass the % in as part of the parameter. Eg: $P{SupId} = new String("%123%"). Then your query is ... where podm_suppid LIKE $P{SupId} This works for me when users are looking for descriptions. Also, I'm using MS SQL 2000. May not work in all flavors of SQL. Link to comment Share on other sites More sharing options...
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