radiations3 Posted October 25, 2011 Share Posted October 25, 2011 HI, I am trying to use pagination on the record getting displayed into the table and there's restriction on record that at a time total 15 records can be displayed and others will be displayed by clicking on NEXT (on which i did the following coding). In my SQL query 'where' clause there is select * from db where status = $query <?php if ($pageNum_Recordset1 < $totalPages_Recordset1) { // Show if not last page ?> <a href="<?php printf("%s?pageNum_Recordset1=%d%s", $currentPage, min($totalPages_Recordset1, $pageNum_Recordset1 + 1), $queryString_Recordset1); ?>?id=<?php echo $yes;?>">Next</a> <?php } // Show if not last page ?> The problem that i am facing is when i click on next then the $query gets equal to null which results to showing no record into the table kindly help how can i modify the above code so the previous value present in $query doesn't get NULL by clicking on next and show next 15 records sucessfully??? NOTE: The main purpose of the above mentioned code is to show next 15 records (which works fine if $query != NULL). Link to comment https://forums.phpfreaks.com/topic/249793-problems-with-sending-values-from-one-form-to-another-page/ Share on other sites More sharing options...
radiations3 Posted October 25, 2011 Author Share Posted October 25, 2011 SOLVED IT MYSELF <?php if ($pageNum_Recordset1 < $totalPages_Recordset1) { // Show if not last page ?> <a href="<?php printf("%s?pageNum_Recordset1=%d%s&id=$yes", $currentPage, min($totalPages_Recordset1, $pageNum_Recordset1 + 1), $queryString_Recordset1) ?>">Next</a> <?php } // Show if not last page ?> Link to comment https://forums.phpfreaks.com/topic/249793-problems-with-sending-values-from-one-form-to-another-page/#findComment-1282180 Share on other sites More sharing options...
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