ger_mac74 Posted March 21, 2006 Share Posted March 21, 2006 Hi thereI am using this search script that I found online, in a web page that I am designing. It is searching a table in a database. When I open the file the search script finds and lists the first 10 results but the link to Next 10>> is not working. It does find enough results to have a next page as it is also printing "Showing results 1 to 10 of 35"" below the link. I am new to php and am having problems here. Can anyone help me out? Please!Thank you...Geraldine[code]<form name="form" action="search.php" method="get"> <input type="text" name="q" /> <input type="submit" name="Submit" value="Search" /></form><?php // Get the search variable from URL $var = @$_GET['q']; $trimmed = trim($var); //trim whitespace from the stored variable// rows to return$limit =10;// check for an empty string and display a message.if ($trimmed == "") { echo "<p>Please enter a search...</p>"; exit; }// check for a search parameterif (!isset($var)) { echo "<p>We dont seem to have a search parameter!</p>"; exit; }//connect to databasemysql_connect("cosmos.ucc.ie","gpmc1","itahkauf");//specify database ****mysql_select_db("2006_gpmc1") or die("Unable to select database");// Build SQL Query$query = "SELECT * FROM articles1 WHERE (description LIKE '%$trimmed%' ) OR (articlename LIKE '%$trimmed%' ) ORDER BY description";$numresults=mysql_query($query);$numrows=mysql_num_rows($numresults);// If we have no results, offer a google search as an alternativeif ($numrows == 0) { echo "Results"; echo "<p>Sorry, your search: "" . $trimmed . "" returned zero results</p>";// googleecho "<p><a href=\"http://www.google.com/search?q=" . $trimmed . "\" target=\"_blank\" title=\"Look up " . $trimmed . " on Google\">Click here</a> to try the search on google</p>"; }// next determine if s has been passed to script, if not use 0 if (empty($s)) { $s=0; }// get results $query .= " limit $s,$limit"; $result = mysql_query($query) or die("Couldn't execute query");// display what the person searched forecho "<p>You searched for: "" . $var . ""</p>";// begin to show results setecho "Results: <br>";$count = 1 + $s;// now you can display the results returned while ($row= mysql_fetch_array($result)) { $title = $row["articlename"]; $arturl = $row["articleurl"]; echo "$count.) <a href='$arturl'>$title</a><br>"; $count++;}$currPage = (($s/$limit) + 1);//break before paging echo "<br />"; // next we need to do the links to other results if ($s>=1) { // bypass PREV link if s is 0 $prevs=($s-$limit); print " <a href=\"$PHP_SELF?s=$prevs&q=$var\"><< Prev 10</a> "; }// calculate number of pages needing links $pages=intval($numrows/$limit);// $pages now contains int of pages needed unless there is a remainder from division if ($numrows%$limit) { // has remainder so add one page $pages++; }// check to see if last page if (!((($s+$limit)/$limit)==$pages) && $pages!=1) { // not last page so give NEXT link $news=$s+$limit; echo " <a href=\"$PHP_SELF?s=$news&q=$var\">Next 10 >></a>"; }$a = $s + ($limit); if ($a > $numrows) { $a = $numrows; } $b = $s + 1; echo "<p>Showing results $b to $a of $numrows</p>";?>[/code] Quote Link to comment Share on other sites More sharing options...
ger_mac74 Posted March 22, 2006 Author Share Posted March 22, 2006 does anyone have any ideas how i could fix one or two incorrect lines in this script?? Quote Link to comment Share on other sites More sharing options...
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