Selecting all data using LIKE in java by getting the string in jtextfield

Question

try{

        String Sql ="Select * from learners_info where lrn LIKE %?% ";
        pst=conn.prepareStatement(Sql);
        pst.setString(1, txtSearch.getText());
        System.out.print(pst);
        rs = pst.executeQuery();



            DefaultTableModel ts = (DefaultTableModel)tableSearch.getModel();
            ts.setRowCount(0);


            while (rs.next()){
            Object searchTable[] = {
            rs.getInt("lrn"),
            rs.getString("learner_firstname"),
            rs.getString("learner_middlename"),
            rs.getString("learner_lastname"),
            };
            ts.addRow(searchTable);
            }
        }

        catch(Exception e){
        JOptionPane.showMessageDialog(null, "Invalid info idnumber");
        e.printStackTrace();
        }
        //
        finally {
        try {
            rs.close(); 
        } 
        catch (Exception e) {
        e.printStackTrace();
        }

        try {
            pst.close(); 
        } catch (Exception e) {
            e.printStackTrace(); 
        }

        try {

        } catch (Exception e) {
            e.printStackTrace(); 
        }
    }

The problem is I cant find the proper String to use for me to get the LIKE terms in the SQL from text field

**Using this String **

String Sql ="Select * from learners_info where lrn LIKE %?% ";

I seem to get this error :

---

java.sql.SQLException: Parameter index out of range (1 > number of parameters, which is 0).
at com.mysql.jdbc.SQLError.createSQLException(SQLError.java:1074)
at com.mysql.jdbc.SQLError.createSQLException(SQLError.java:988)
at com.mysql.jdbc.SQLError.createSQLException(SQLError.java:974)
at com.mysql.jdbc.SQLError.createSQLException(SQLError.java:919)
at com.mysql.jdbc.PreparedStatement.checkBounds(PreparedStatement.java:3813)
at com.mysql.jdbc.PreparedStatement.setInternal(PreparedStatement.java:3795)
at com.mysql.jdbc.PreparedStatement.setString(PreparedStatement.java:4616)

---

And if I try this String

String Sql ="Select * from learners_info where lrn LIKE '%'?'%' ";

I printed what kind of string is being used using system.out and it shows this.

com.mysql.jdbc.JDBC4PreparedStatement@414a0b8f: Select * from learners_info where lrn LIKE '%''1411237''%'

---

this query works but it doesn't seem to return any values

I'm sorry cause I'm still new to database in java.

Answer 1

Remove the '%' parts from your SQL String. Try with this :

String Sql ="Select * from learners_info where lrn LIKE ?";

Then try passing the %s along with your String value in your PreparedStatement setString() method