java.sql.SQLException Parameter index out of range

Question

While am using this code it shows java.sql.SQLException Parameter index out of range (1 > number of parameters, which is 0):

private void cmd_searchActionPerformed(java.awt.event.ActionEvent evt) {                                           
try{

 String sql = "select * from STD where Name like '%?%' ";
      pst=conn.prepareStatement(sql);
      pst.setString(1,TXT_STUDENTNAME.getText());
      String value=TXT_STUDENTNAME.getText();
      rs=pst.executeQuery();
       jTable1.setModel(DbUtils.resultSetToTableModel(rs));

             }catch(Exception e){
       JOptionPane.showMessageDialog(null,e);
   }        
}

How can i recover from this exception?

score 2 · Answer 1 · edited May 23 '17 at 12:15

2

Try to remove the wildcard from the sql and add it to the value:

String sql = "select * from STD where Name like ? ";
pst=conn.prepareStatement(sql);
pst.setString(1,"%"+TXT_STUDENTNAME.getText()+"%");

Similar question here

edited May 23 '17 at 12:15

Community

1
1

answered Nov 10 '12 at 05:41

iTech

18,192
4
57
80

score 0 · Answer 2 · answered Nov 10 '12 at 06:27

private void cmd_searchActionPerformed(java.awt.event.ActionEvent evt) {                                           
try{

 String sql = "select * from STD where Name like '%?%' ";
      String strStudentName = TXT_STUDENTNAME.getText();
      pst=conn.prepareStatement(sql);
      pst.setString(1,strStudentName );
      rs=pst.executeQuery();
       jTable1.setModel(DbUtils.resultSetToTableModel(rs));

             }catch(Exception e){
       JOptionPane.showMessageDialog(null,e);
   }        
}

I think this will help you..

String strStudentName = TXT_STUDENTNAME.getText(); pst.setString(1,strStudentName );

java.sql.SQLException Parameter index out of range

2 Answers2