CODEWITHSUNDEEP
javamysqlsql-insert
Klayd Pro
Klayd Pro

Reputation: 11

Insert values to mysql Java

It gives this error:

"java.sql.SQLException: Parameter index out of range (1 > number of parameters, which is 0). "

This is the code:

public void insertData(int id,String name, String status, String age){



     String sql = "INSERT INTO student_table (id,name,satus,age) VALUES  (id,name,status,age)";

     try (Connection conn1 = this.connects();PreparedStatement Prepst= conn1.prepareStatement(sql)){

     Prepst.setInt(1, id);
     Prepst.setString(2, name);
     Prepst.setString(3, status);
     Prepst.setString(4, age);
     Prepst.executeUpdate();

     }catch(Exception e){
            System.err.println(e);

      }

If I change the parameter index to another value i.e.:

Prepst.setInt(0, id);

then it gives this error:

" java.sql.SQLException: Parameter index out of range (0 < 1 ). "

----UPDATE -----

I inserted this code: 

String sql = "INSERT INTO student_table (id,name,satus,age) VALUES  (?,?,?,?)";

It produces the same:

RESULT

Table Structure:

HERE

Upvotes: 0

Views: 92

Answers (2)

Gayathri Rajan
Gayathri Rajan

Reputation: 469

Try this...

public void insertData(int id,String name, String status, String age){



     String sql = "INSERT INTO student_table (id,name,satus,age) VALUES  (?,?,?,?)";

     try (Connection conn1 = this.connection;PreparedStatement Prepst= conn1.prepareStatement(sql)){

     Prepst.setInt(1, id);
     Prepst.setString(2, name);
     Prepst.setString(3, status);
     Prepst.setString(4, age);
     Prepst.executeUpdate();

     }catch(Exception e){
            System.err.println(e);

      }
}

Upvotes: 0

lexicore
lexicore

Reputation: 43709

Your SQL expression

INSERT INTO student_table (id,name,satus,age) VALUES  (id,name,status,age)

actually does not have any parameters. Try with:

INSERT INTO student_table (id,name,satus,age) VALUES  (?,?,?,?)

Upvotes: 3

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