i have search page to search by title, author, isbn from the dropdown list.

But when they display the result for my search, it unable to retrieve my other data from my table (product4) like the price. it shows this error below

Notice: Undefined variable: price in D:\xampp\htdocs\book\search.php on line 96

Notice: Undefined index: in D:\xampp\htdocs\book\search.php on line 96

this is line 96, n 98

$price = $row["$price"];

  echo "$count.) $title,$price" ;

this is the full code

<div class="searchmenu"></div>
<li class="searchli" style="padding-bottom: 8px">
         <form name="form" action="search.php" method="get">

         <span class="slogan" style="padding-top: 12px"><b>Search for :</b></span>

            <input type="text" size="10" maxlength="40"  name="q"/> <div class="button"></div> 
<SELECT NAME="searchtype">
<OPTION VALUE="">Search Type
<OPTION VALUE="title">Title</option>
<OPTION VALUE="author">Author</option>
<OPTION VALUE="isbn">ISBN</option>
<OPTION VALUE="publisher">Publisher</option>
           <input type="submit" name="Submit" value="Search" />

  <span class="slogan" style="padding-top: 12px">Advanced Search</span>
 <br><br><br> </div>

  // Get the search variable from URL
  $var = @$_GET['q'] ;
    $searchtype = @$_GET['searchtype'] ;
  $trimmed = trim($var); //trim whitespace from the stored variable

// rows to return

// check for an empty string and display a message.
if ($trimmed == "")
  echo "<p>Please enter a search...</p>";

// check for a search parameter
if (!isset($var))
  echo "<p>We dont seem to have a search parameter!</p>";

//connect to your database ** EDIT REQUIRED HERE **
mysql_connect("localhost","root","boon84"); //(host, username, password)

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

// Build SQL Query  
$query = "select * from products4 where   $searchtype like \"%$trimmed%\"  
  order by   $searchtype"; // EDIT HERE and specify your table and field names for the SQL query


// 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)) {

// 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 <br/>";
$count = 1 + $s ;

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

  echo "$count.)&nbsp;$title,$price" ;

  $count++ ;

$currPage = (($s/$limit) + 1);

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

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

// calculate number of pages needing links

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

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

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

  // not last page so give NEXT link

  echo "&nbsp;<a href=\"search.php?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>";
5 Years
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Last Post by cereal
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