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I'm trying to generate my JasperReports's report using a parameter enter in my textfield. It seems that its not working using query with parameter.

Is there any other way to generate the report using parameter from Java via JasperReports API? I set my report that accept a parameter values as string. 

public void reportDefault() {
    try {
        String searchreport = searchtextField.getText();
        JasperDesign jd = JRXmlLoader.load("D:\\reports\\supplierlist.jrxml");
        String sql = "SELECT * FROM businfo WHERE busname='%"+searchreport+"%'";
        JRDesignQuery newQuery = new JRDesignQuery();
        newQuery.setText(sql);      
        jd.setQuery(newQuery);
        JasperReport jr = JasperCompileManager.compileReport(jd);
        JasperPrint jp = JasperFillManager.fillReport(jr,null,conn);
        JasperViewer jv = new JasperViewer(jp,false);
        jv.viewReport(jp, false);
    } catch(Exception e) {
        e.printStackTrace();
    }
}
Alex K
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agentOrange002
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1 Answers1

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You seem to have an SQL syntax error - you have used the wildcard operator % with = instead of LIKE. See http://www.w3schools.com/sql/sql_wildcards.asp

Robert Mugattarov
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