Display multiple paging

Question

hello all,
i am having problem to paging my result in multiple pages. i'm already searching in this forum on how to solve on paging pages but none of them can help me. here is my code..
// paging section
int rownumb = listof.countData(offer_type);
out.println("row : " +rownumb);
int perpage = 2;
int pageno = rownumb/perpage;
int balance = rownumb - (pageno * perpage);
if (balance != 0) {
pageno = pageno + 1;
out.println(" balance : " balance "<br>");
out.println(" rownumb pageno : " pageno "<br>");
int inc =0;
for (int i=1; i<=pageno ; i++) {
if ( i == pageno) { %>
<font color="#ff3366"><%=i%></font>
<% }
else { %>
<a href="ListOffer.jsp?offer_type=<%=offer_type%>&pageno=<%=i%>"><%
out.println("displaying page number : " i "<br></a>");
listof.queryCategory(offer_type);
col = listof.getColumn();
row = listof.getRow();
outVector = listof.getVectorRow();
++inc;
// end paging
//in my java bean file
function countData
String query = SELECT count(*) FROM offer
WHERE otype= '"+offer_type+"' AND ostatus = 'posted'";
function queryCategory
String query = SELECT offerid, otitle, odescription, oposted_date, oexpiry_date FROM offer WHERE otype='"+offer_type+"' AND ostatus='posted' AND (oposted_date <= '"+currentDate+"' AND oexpiry_date >= '"+currentDate+"')ORDER BY offerid ASC LIMIT 2 ";
the problem is i can't display 2 data in one pages eventhough the code already limit the data to display 2 rows in 1 pages.
can anybody solve for me?
</a>

Answer 1

I can offer my 2-cents but...
I'm assuming you are using a MySQL database. What does your query return when you run it from the MySQL prompt? It's good practice to execute your queries in some SQL client before hand.
From MySQL Reference Manual:
The LIMIT clause can be used to constrain the number of rows returned by the SELECT statement. LIMIT takes one or two numeric arguments.
If two arguments are given, the first specified the offset of the first row to return; the second specifies the maximum number of rows to return. The offset of the initial row is 0 (not 1):
mysql> select * from table LIMIT 5,10; # Retrieve rows 6-15
If one argument is given, it indicates the maximum number of rows to return:
mysql> select * from table LIMIT 5; # Retrieve first 5 rows
In other words, LIMIT n is equivalent to LIMIT 0,n.
Just a note, LIMIT performs much faster on large result sets, however, it is a non-standard MySQL extension, and therefore, it is suggested to avoid using it in queries.
First thing I can see is that the way you are using LIMIT, the first two records will always be returned. You would need to pass two parameters (or at least one -> start) to queryCategory, i.e.
public void queryCatetory(int start, int count)
String query = "... LIMIT " + start + "," + count;
Also, you don't need to execute another query (the countData function) to get the record count, simply move the ResultSet cursor to the last position and get the row number, i.e.
rs.last();
noRows = rs.getRow();
For the pageno value, this is what you should be getting as a parameter to the page.
int pageno = Integer.parseInt(request.getParameter("pageno"));
For the query paging part, I can offer this bit of code which might help:
final int ROW_COUNT = 10;
int currPage, rowStart, rowStop, totalPages, noRows;
String scriptName = request.getRequestURI();
String query = "SELECT offerid, otitle, odescription, oposted_date, oexpiry_date FROM offer WHERE otype='"+offer_type+"' AND ostatus='posted' AND (oposted_date <= '"+currentDate+"' AND oexpiry_date >= '"+currentDate+"') ORDER BY offerid";
// get the page parameter
try { currPage = Integer.parseInt(request.getParameter("pageno"); }
catch (NumberFormatException e) { currPage = 1; }
// connect to database and execute query
PreparedStatement stmt = conn.prepareStatement(query);
ResultSet rs = stmt.executeQuery();
// get the record count
rs.last();
noRows = rs.getRow();
if (noRows == 0)
out.println("Sorry, no records found!");
else
// calculate row start/stop and total pages
// if current page is 2, we want to display records 11-20
rowStart = (currPage - 1) * ROW_COUNT + 1;
// alternatively, you can do:
// rowStop = Math.min(noRows, currPage * ROW_COUNT);
rowStop = Math.min(noRows, rowStart + ROW_COUNT - 1);
// division by integer yeilds a truncated integer value so we
// have to check for a remainder, but you can also do:
// totalPages = (int)Math.ceil(ROW_TOTAL / (ROW_COUNT * 1.0));
totalPages = (noRows / ROW_COUNT) + (noRows % ROW_COUNT == 0 ? 0 : 1);
// set result set to start position and print the records
rs.absolute(rowStart);
for (int row = rowStart; row <= rowStop; row++)
// display current record data
rs.next();
// print query paging links
out.print("<p align=\"center\">");
if (currPage > 1)
out.print(" <a href=\"" + scriptName + "?pageno=" + (currPage - 1) + "\">< Prev</a>");
for (int i = 1; i <= totalPages; i++)
if (i == currPage)
out.print(" <b>" + i + "</b>");
else
out.print(" <a href=\"" + scriptName + "?pageno=" + i + "\">" + i + "</a>");
if (currPage < totalPages)
out.print(" <a href=\"" + scriptName + "?pageno=" + (currPage + 1) + "\">Next ></a>");
out.print("</p>");
I'm sure there is a more elegant solution out there. I believe that the standard taglib has some tags which can do this for you. Hopefully someone can confirm this can possibly provide further details.
Hope this helps :)