ResultSet.getString(1) throws java.sql.SQLException: Invalid operation at current cursor position
java, jdbc, resultset, sqlexception
Solution
You should use the `next` statement first.
ResultSet set = statement.executeQuery();
if (set.next()) {
userName = set.getString(1);
//your logic...
}
UPDATE
As the Java 6 Documentation says
A ResultSet cursor is initially positioned before the first row; the first call to the method next makes the first row the current row; the second call makes the second row the current row, and so on.
This means when you execute the sentence
ResultSet set = statement.executeQuery();
The ResultSet `set` will be created and pointing to a row before the first result of the data. You can look it this way:
SELECT email,firstname FROM registrationinformation
email | firstname
____________________________________
0 <= set points to here
1 email1@gmail.com | Email1 Person
2 foo@bar.com | Foo Bar
So, after openning your ResulSet, you execute the method `next` to move it to the first row.
if(set.next())
Now `set` looks like this.
email | firstname
____________________________________
0
1 email1@gmail.com | Email1 Person <= set points to here
2 foo@bar.com | Foo Bar
If you need to read all the data in the ResultSet, you should use a while instead of if:
while(set.next()) {
//read data from the actual row
//automatically will try to forward 1 row
}
If the `set.next()` return false, it means that there was no row to read, so your while loop will end.
More information here.
Problem
When I run the following servlet: ``` // package projectcodes; public void doPost(HttpServletRequest request,HttpServletResponse response) throws ServletException,IOException { String UserID = request.getParameter("UserID"); String UserPassword = request.getParameter("UserPassword"); String userName = null; String Email = null; Encrypter encrypter = new Encrypter(); String hashedPassword = null; try { hashedPassword = encrypter.hashPassword(UserPassword); Context context = new InitialContext(); DataSource ds = (DataSource)context.lookup("java:comp/env/jdbc/photog"); Connection connection = ds.getConnection(); String sqlStatement = "SELECT email,firstname FROM registrationinformation WHERE password='" + hashedPassword + "'"; PreparedStatement statement = connection.prepareStatement(sqlStatement); ResultSet set = statement.executeQuery(); userName = set.getString(1); // <<---------- Line number 28 response.sendRedirect("portfolio_one.jsp"); // userName = set.getString("FirstName"); Email = set.getString(3); if(set.wasNull() || Email.compareTo(UserID) != 0) { // turn to the error page response.sendRedirect("LoginFailure.jsp"); } else { // start the session and take to his homepage HttpSession session = request.getSession(); session.setAttribute("UserName", userName); session.setMaxInactiveInterval(900); // If the request doesn't come withing 900 seconds the server will invalidate the session RequestDispatcher rd = request.getRequestDispatcher("portfolio_one.jsp"); rd.forward(request, response); // forward to the user home-page } }catch(Exception exc) { System.out.println(exc); } ``` I get the following exceptions: ``` INFO: java.sql.SQLException: Invalid operation at current cursor position. at org.apache.derby.client.am.SQLExceptionFactory40.getSQLException(Unknown Source) at org.apache.derby.client.am.SqlException.getSQLException(Unknown Source) at org.apache.derby.client.am.ResultSet.getString(Unknown Source) at com.sun.gjc.spi.base.ResultSetWrapper.getString(ResultSetWrapper.java:155) -----> at projectcodes.ValidateDataForSignIn.doPost(ValidateDataForSignIn.java:28 at javax.servlet.http.HttpServlet.service(HttpServlet.java:754) at javax.servlet.http.HttpServlet.service(HttpServlet.java:847) at org.apache.catalina.core.StandardWrapper.service(StandardWrapper.java:1539) at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:281) at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:175) at org.apache.catalina.core.StandardPipeline.doInvoke(StandardPipeline.java:655) at org.apache.catalina.core.StandardPipeline.invoke(StandardPipeline.java:595) at com.sun.enterprise.web.WebPipeline.invoke(WebPipeline.java:98) at com.sun.enterprise.web.PESessionLockingStandardPipeline.invoke(PESessionLockingStandardPipeline.java:91) at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:162) at org.apache.catalina.connector.CoyoteAdapter.doService(CoyoteAdapter.java:330) at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:231) at com.sun.enterprise.v3.services.impl.ContainerMapper.service(ContainerMapper.java:174) at com.sun.grizzly.http.ProcessorTask.invokeAdapter(ProcessorTask.java:828) at com.sun.grizzly.http.ProcessorTask.doProcess(ProcessorTask.java:725) at com.sun.grizzly.http.ProcessorTask.process(ProcessorTask.java:1019) at com.sun.grizzly.http.DefaultProtocolFilter.execute(DefaultProtocolFilter.java:225) at com.sun.grizzly.DefaultProtocolChain.executeProtocolFilter(DefaultProtocolChain.java:137) at com.sun.grizzly.DefaultProtocolChain.execute(DefaultProtocolChain.java:104) at com.sun.grizzly.DefaultProtocolChain.execute(DefaultProtocolChain.java:90) at com.sun.grizzly.http.HttpProtocolChain.execute(HttpProtocolChain.java:79) at com.sun.grizzly.ProtocolChainContextTask.doCall(ProtocolChainContextTask.java:54) at com.sun.grizzly.SelectionKeyContextTask.call(SelectionKeyContextTask.java:59) at com.sun.grizzly.ContextTask.run(ContextTask.java:71) at com.sun.grizzly.util.AbstractThreadPool$Worker.doWork(AbstractThreadPool.java:532) at com.sun.grizzly.util.AbstractThreadPool$Worker.run(AbstractThreadPool.java:513) at java.lang.Thread.run(Thread.java:722) Caused by: org.apache.derby.client.am.SqlException: Invalid operation at current cursor position. at org.apache.derby.client.am.ResultSet.checkForValidCursorPosition(Unknown Source) at org.apache.derby.client.am.ResultSet.checkGetterPreconditions(Unknown Source) ... 30 more ``` The logs above from the server show that line number 28 is the cause of the exception. But i am unable to get the reason for exception. All the columns in the table have a datatype of varchar. I have highlighted line number 28 (cause of exception according to server logs) in the servlet code.