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i m saving name and email id of student in database

i dont want to enter already registered students

String ans1=(request.getParameter("R1"));

//String name=(request.getParameter("T2"));

Class.forName("sun.jdbc.odbc.JdbcOdbcDriver");

Connection con=DriverManager.getConnection("jdbc:odbc:exam","","");

Statement s=con.createStatement();

Result rs;

String value = (session.getValue("s1")).toString();

String name=(request.getParameter("T1"));

rs=s.executeQuery("select emailid from check");

while(rs.next())

{

if(ans1==rs.getString(1))

{

out.println("ALREADY REGISTERED");

}

else

{

PreparedStatement st=con.prepareStatement("update check set ans1=? where name Like ?");

st.setString(1,ans1);

st.setString(2,value);

st.executeUpdate();

out.println(value);

}

this is code but its not working

plz answer if anyone know the probblem

[998 byte] By [rachna_arora82a] at [2007-11-26 13:34:06]

«« Problem running Java Problem
»» Can not access resource after tomcat restart

# 1

> if(ans1==rs.getString(1))You cannot use == to compare objects (unless you want to compare their physical address in memory). You need to use the String equals() method.

gimbal2a at 2007-7-7 22:15:11 >

top of Java-index,Java Essentials,New To Java...

# 2

hey rachna,

The logic of would bomb at runtime at a later stage ......... and as suggested if ur trying to compare the contents of the String object then use the equals method

Gimme more details of the design of the table that would be helpful.

thanks n regards

naveen m

Message was edited by:

Novice_inJAVA

Novice_inJAVAa at 2007-7-7 22:15:11 >

# 3

String ans1=(request.getParameter("R1"));

//String name=(request.getParameter("T2"));

Class.forName("sun.jdbc.odbc.JdbcOdbcDriver");

Connection con=DriverManager.getConnection("jdbc:odbc:exam","","");

PreparedStatement s;

ResultSet rs;

String value = (session.getValue("s1")).toString();

String name=(request.getParameter("T1"));

s=con.prepareStatement("select emailid from check where emailid=?");

s.setString(1,value);

s.executeQuery();

if (rs.)//want to know is resultset have records

{

PreparedStatement st=con.prepareStatement("update check set ans1=? where name Like ?");

st.setString(1,ans1);

st.setString(2,value);

st.executeUpdate();

out.println(value);

}

else

{

out.println("ALREADY REGISTERED");

}

table have cols

1 name

2 emailid

3 ans1 answer of questions 1(its online test website)

4ans2

and so on

regards

rachna

rachna_arora82a at 2007-7-7 22:15:11 >

# 4

Here is the simple way to do this.

String sql = "Select 1 From check Where name = '" + submitted_name_goes_here + "'";

Execute the above query and iterate the resultset.

if(rs.next()){

//Say the username is already used.....

}

else{

//Insert statement will goes here

}

Regards

Vinay

cybervink2000a at 2007-7-7 22:15:11 >

# 5

hi vinay

thanx for answering.

now my code is

String ans1=(request.getParameter("R1"));

//String name=(request.getParameter("T2"));

Class.forName("sun.jdbc.odbc.JdbcOdbcDriver");

Connection con=DriverManager.getConnection("jdbc:odbc:exam","","");

PreparedStatement s;

ResultSet rs;

String value = (session.getValue("s1")).toString();

String name=(request.getParameter("T1"));

s=con.prepareStatement("select emailid from check where emailid=?");

s.setString(1,value);

rs=s.executeQuery();

rs.next();

//int ResultCount = rs.getInt(rowcount) ;

if (rs.next())

{

out.println("already reg");

}

else

{

PreparedStatement st=con.prepareStatement("update check set ans1=? where name Like ?");

st.setString(1,ans1);

st.setString(2,value);

st.executeUpdate();

out.println(value);

}

and now its showing error

javax.servlet.ServletException

at org.apache.jasper.runtime.PageContextImpl.handlePageException(PageContextImpl.java:384)

at _0005conline_0002ejsponline_jsp_0._jspService(_0005conline_0002ejsponline_jsp_0.java:101)

at org.apache.jasper.runtime.HttpJspBase.service(HttpJspBase.java:126)

at javax.servlet.http.HttpServlet.service(HttpServlet.java:865)

at org.apache.jasper.runtime.JspServlet$JspServletWrapper.service(JspServlet.java:161)

at org.apache.jasper.runtime.JspServlet.serviceJspFile(JspServlet.java:247)

at org.apache.jasper.runtime.JspServlet.service(JspServlet.java:352)

at javax.servlet.http.HttpServlet.service(HttpServlet.java:865)

at org.apache.tomcat.core.ServiceInvocationHandler.method(ServletWrapper.java:626)

at org.apache.tomcat.core.ServletWrapper.handleInvocation(ServletWrapper.java:534)

at org.apache.tomcat.core.ServletWrapper.handleRequest(ServletWrapper.java:378)

at org.apache.tomcat.core.Context.handleRequest(Context.java:644)

at org.apache.tomcat.core.ContextManager.service(ContextManager.java:440)

at org.apache.tomcat.service.http.HttpConnectionHandler.processConnection(HttpConnectionHandler.java:144)

at org.apache.tomcat.service.TcpConnectionThread.run(TcpEndpoint.java:310)

at java.lang.Thread.run(Thread.java:484)

Root cause:

java.lang.NullPointerException

at _0005conline_0002ejsponline_jsp_0._jspService(_0005conline_0002ejsponline_jsp_0.java:70)