s4salman Posted April 24, 2011 Share Posted April 24, 2011 i want to change this code : $result = mysql_query("select count(*) from 3gp where category='Bollywood'"); with this $result = mysql_query("select count(*) from 3gp where category='$category'"); but i am getting error message "Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/funxy/public_html/salmobile.info/3gp-videos.php on line 151". I am calling the page using http://abc.com/3gp-videos.php?category=Bollywood Please help. <?php $db="funxy_db"; $conn = mysql_connect("localhost","funxy_saba","myasco001"); @mysql_select_db($db) or die( "Unable to select database, Please contact your administrator"); class Pager { function getPagerData($numHits, $limit, $page) { $numHits = (int) $numHits; $limit = max((int) $limit, 1); $page = (int) $page; $numPages = ceil($numHits / $limit); $page = max($page, 1); $page = min($page, $numPages); $offset = ($page - 1) * $limit; $ret = new stdClass; $ret->offset = $offset; $ret->limit = $limit; $ret->numPages = $numPages; $ret->page = $page; return $ret; } } // get the pager input values $page = $_GET['page']; $limit = 3; $result = mysql_query("select count(*) from 3gp where category='Bollywood'"); $total = mysql_result($result, 0, 0); // work out the pager values $pager = Pager::getPagerData($total, $limit, $page); $offset = $pager->offset; $limit = $pager->limit; $page = $pager->page; // use pager values to fetch data $query = "select * from 3gp where category='Bollywood' order by id DESC limit $offset, $limit"; $result = mysql_query($query); // use $result here to output page content //my addition //grab all the content //Custom Table Stsrt// $cols = 4; //number of coloms $i =1; echo "<table border=\"0\" cellpadding=\"2\" cellspacing=\"2\" width=\"100%\" id=\"table1\" bordercolor=\"#FFFFFF\" bgcolor=\"#FFFFFF\">" ."<tr>"; while($r=mysql_fetch_array($result)) { //the format is $variable = $r["nameofmysqlcolumn"]; //modify these to match your mysql table columns $id=$r["id"]; $name=$r["name"]; $views=$r["views"]; $url=$r["url"]; $image=$r["image"]; $category=$r["category"]; //display the row $mybox = " <br> <a href='http://salmobile.info/3gp-videos-download.php?id=$id' class=\"classd\"><b><u> $name</u></b></a> <BR> <span class=\"text\"><img src=\"$image\" width=\"60\" height=\"60\"></span> <BR> <span class=\"text2\">Download This 3GP Video</span> <br>"; if (is_int($i / $cols)){ echo "<td width='336' align='center' style=\"border-style: dotted; border-width: 0\">$mybox</td></tr><tr>"; }else{ echo "<td width='336' align='center' style=\"border-style: dotted; border-width: 0\">$mybox</td>"; } if ( $i / $cols == 3) echo "<td colspan='3' align=\"center\"> </td></tr><tr>"; if ( $i / $cols == echo "<td colspan='3' align=\"left\"> </td></tr><tr>"; $i++; //end if }//end while echo "</tr></table>"; //Custom Table End// //ends my addition // output paging system (could also do it before we output the page content) if ($page == 1) // this is the first page - there is no previous page echo "<font color=\"#FFB300\">Previous</font>"; else // not the first page, link to the previous page echo "<a href=\"http://salmobile.info/index-" . ($page - 1) . ".html\" id=\"navigationURL\"><font color=\"#FFB300\">Previous</font></a>"; for ($i = 1; $i <= $pager->numPages; $i++) { echo " <font color=\"#FFFFFF\">|</font> "; if ($i == $pager->page) echo "<font color=\"#FFB300\"> $i</font>"; else echo "<a href=\"http://salmobile.info/index-$i.html\" id=\"navigationURL\"> <font color=\"#FFB300\">$i</font></a>"; } if ($page == $pager->numPages) // this is the last page - there is no next page echo "Next"; else // not the last page, link to the next page echo " <a href=\"http://salmobile.info/index-" . ($page + 1) . ".html\" id=\"navigationURL\"><font color=\"#FFB300\">Next</font></a>"; ?> Quote Link to comment https://forums.phpfreaks.com/topic/234591-query-string-causing-problem/ Share on other sites More sharing options...
Fadion Posted April 24, 2011 Share Posted April 24, 2011 <?php $category = mysql_real_escape_string($_GET['category']); $result = mysql_query("SELECT COUNT(*) FROM 3gp WHERE category='$category'"); ?> Quote Link to comment https://forums.phpfreaks.com/topic/234591-query-string-causing-problem/#findComment-1205583 Share on other sites More sharing options...
s4salman Posted April 24, 2011 Author Share Posted April 24, 2011 thank you for the help. matter resolved Quote Link to comment https://forums.phpfreaks.com/topic/234591-query-string-causing-problem/#findComment-1205591 Share on other sites More sharing options...
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