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Thread: PHP MySQL problem

  1. #1
    Join Date
    Jan 2009

    Exclamation PHP MySQL problem

    Hello guys,
    I am having a search script which suppose to take results out of my mysql
    database and display on a page 10 results each
    if the results are more then 10 it should create another page
    or pages with Next >> and Prev << link so anyone can move
    forward and backward on those results page.

    This script do everything fine counting total records display's 1 to 10
    records on first page but at second location on same page i want it to daisplay
    Showing results 1 to 10 of total records found it show 5 to 14, ant it
    displays a link to Next which is not working.

    Here is the code to generate pages please have a look and let me
    know where I am wrong.

    PHP Code:
    // begin to show results set
    echo " found  $numrows results <p>";
    $count $s;

    // now you can display the results returned
    while ($rowmysql_fetch_array($result)) {

    $title $row["sname"];
    $url $row["url"];
    $message $row["massage"];
    //&nbsp;<a href=\"$row['url']\">"$row['sname']</a><br /> 

      echo "$count.)&nbsp";
      echo "<b><a href=\"$url\" target=\"_blank\">$title</a></b>";
      echo "<br>";
      echo "$message<br><hr height=12 color=lightblue><br>";


    $currPage = (($s/$limit) + 1);

    //break before paging
      echo "<br />";

      // next we need to do the links to other results &lt;&lt;
      if ($s>=1) { // bypass PREV link if s is 0
      print "&nbsp;<a href='
    {$_SERVER['PHP_SELF']}?s=$prevs&q=$var'> <<<
      Prev 10</a>&nbsp&nbsp;";

    // calculate number of pages needing links
    //  $s=intval($numrows/$limit);

    // $s now contains int of pages needed unless there is a remainder from 

      if ($numrows%$limit) {
      // has remainder so add one page

    // check to see if last page
      if (!((($s+$limit)/$limit)==$s) && $s!=1) {

      // not last page so give NEXT link  $PHP_SELF &gt;&gt;

      echo "&nbsp;<a href='
    {$_SERVER['PHP_SELF']}?s=$news&q=$var'>Next 10 >>> 


    $a = $s+$limit;
      if ($a > $numrows) { $a = $numrows; }
    $b = $s+1;
      echo "<p>Showing results $b to $a of $numrows</p>";
    include ('');
    Thank You
    Last edited by PeejAvery; January 26th, 2009 at 06:55 PM. Reason: Added PHP tags.

  2. #2
    PeejAvery's Avatar
    PeejAvery is offline Super Moderator Power Poster
    Join Date
    May 2002

    Re: PHP MySQL problem

    Pagination is a very simple task. The following should get you through it.

    PHP Code:
    $page = (isset($_GET['page'])) ? $_GET['page'] : 1;
    $page--; // because databases start at zero, humans start at one
    $start $page $show;

    $query "SELECT * FROM blog ORDER BY date DESC LIMIT $start$show";
    $sql mysql_query($query);
    $rows mysql_fetch_object(mysql_query("SELECT FOUND_ROWS() AS `rows`"))->rows;
    while (
    $row mysql_fetch_object($sql)) {
    // do your outputting from the database here

    if (
    $page 0) {echo '<a href="?page=' $page '">&lt;&lt; Newer Posts</a>';}
    $total ceil(mysql_num_rows(mysql_query("SELECT id FROM blog")) / $show);
    if (
    $page $total 1) {echo '<a href="?page=' . ($page 2) . '">Older Posts &gt;&gt;</a>';}
    Last edited by PeejAvery; March 28th, 2012 at 05:29 PM. Reason: Fixed the row count
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