php script to find mysql

I am trying to implment php script to find search in mysql database there are two files search.html and search.php i am pasting them simulatenously.

this is returning Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in C:\xamp\xampp\htdocs\search.php on line 36

<?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 parameter
if (!isset($var))
  {
  echo "<p>We dont seem to have a search parameter!</p>";
  exit;
  }

//connect to your database ** EDIT REQUIRED HERE **
mysql_connect("localhost","aaa","aaa"); //(host, username, password)

//specify database ** EDIT REQUIRED HERE **
mysql_select_db("shopcart") or die("Unable to select database"); //select which database we're using

// Build SQL Query  
$query = "select productdetails.* from shopcart.productdetails where 1st_field like \"%$trimmed%\"  
  order by 1st_field"; // EDIT HERE and specify your table and field names for the SQL query

 $numresults=mysql_query($query);
 $numrows=mysql_num_rows($numresults);

// If we have no results, offer a google search as an alternative

if ($numrows == 0)
  {
  echo "<h4>Results</h4>";
  echo "<p>Sorry, your search: &quot;" . $trimmed . "&quot; returned zero results</p>";

// google
 echo "<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 for
echo "<p>You searched for: &quot;" . $var . "&quot;</p>";

// begin to show results set
echo "Results";
$count = 1 + $s ;

// now you can display the results returned
  while ($row= mysql_fetch_array($result)) {
  $title = $row["1st_field"];

  echo "$count.)&nbsp;$title" ;
  $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-$limit);
  print "&nbsp;<a href=\"$PHP_SELF?s=$prevs&q=$var\">&lt;&lt; 
  Prev 10</a>&nbsp&nbsp;";
  }

// 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 "&nbsp;<a href=\"$PHP_SELF?s=$news&q=$var\">Next 10 &gt;&gt;</a>";
  }

$a = $s + ($limit) ;
  if ($a > $numrows) { $a = $numrows ; }
  $b = $s + 1 ;
  echo "<p>Showing results $b to $a of $numrows</p>";
  
?>

PHP:

 

I'm assuming the line is this code:
$numrows=mysql_num_rows($numresults);

What I would is change this...
$numresults=mysql_query($query);
into..
$numresults=mysql_query($query) or die(mysql_error());

This will give you a better idea on what is wrong, whatever is happening in that query is incorrect (and this debug will print out what's wrong). Then you can work from there.