|
|
Posted by wgerry on 09/30/04 11:27
Hi
having trouble with the following code. when I try it comes up with the
message "could'nt execute query and returns zero results. It also gives
a warning about the mysql_num_rows line that it is not a valid mysql
resource. Any help given greatfully recieved as I have been struggling
to make this work for about a month now.
<html>
<head>
<title>search08</title>
<
</head>
<body>
<form name="form" action="search08.php" method="get">
<input type="text" name="q" />
<input type="submit" name="Submit" value="Search" />
</form>
</body>
</html>
<?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($var))
{
echo "<p>We dont seem to have a search parameter!</p>";
exit ;
}
//connect to your database ** EDIT REQUIRED HERE **
mysql_connect("host","username","passord"); //(host, username,
password)
//specify database ** EDIT REQUIRED HERE **
mysql_select_db("online") or die("Unable to select database"); //select
which database we're using
// Build SQL Query
//$query = 'SELECT * FROM `arrow` WHERE `Surname`LIKE \"%trimmed%\" ;
$results= mysql_query($query);
$numrows= mysql_numrows($results);
// If we have no results, offer a google search as an alternative
if ($numrows == 0)
{
echo "<h4>Results</h4>";
echo "<p>Sorry, your search: "" . $trimmed . "" 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";
$results = mysql_query($query) or die ("Couldnt execute query");
// display what the person searched for
echo "<p>You searched for: "" . $var. ""</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["surname"];
echo "$count.) $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-$limit);
print " <a href=\"$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>";
?>
[Back to original message]
|