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>"; ?> 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'"); ?> 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 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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