hallo
ik heb een zoekpagina maar daar mee kan ik maar op 1 rij zoeken. kunnen iemand mij helpen om met deze bestaande pagina op meerder rijen te zoeken. hij werkt wel verder maar dan op 1 rij.
<title>Search</title>
<center><h2>zoeken latijnse namen</h2></center>
<br>
<body bgcolor="#00FF33" text="#000000">
</body>
<center><form name="form" action="search.php" method="get">
<input type="text" name="q" />
<input type="submit" name="Submit" value="Search" />
</form>
<a href="searchnl.php">zoeken nederlandse namen </a></center><?php
$user="****";
$password="*****";
$database="*****";
$host="*****";
// 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($var))
{
echo "<p>We dont seem to have a search parameter!</p>";
exit;
}
//connect to your database
mysql_connect("$host","$user","$password"); //(host, username, password)
//specify database
mysql_select_db("$database") or die("Unable to select database"); //select which database we're using
// Build SQL Query
$query = "select * from plantdb where plantnaam like \"%$trimmed%\"
order by plantnaam";
$numresults=mysql_query($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: "" . $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";
$result = mysql_query($query) or die("Couldn't 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["plantnaam"];
$url = $row["link"];
echo "<br>$count.) <a href=\"$url\">$title</a><br><hr>" ;
$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>";
?>
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