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Re: newbie wanting advice on search form

Posted by wgerry on 09/26/05 21:07

Thanks for all your kind help guys. I have tried what you have
suggested but I am still having trouble. I have made the following
changes but it still comes up with the warning mysql_num_rows is not a
valid SQL resource and my query does not return a result and it comes
up with couldnt execute query,
<html>
<head>
<title>search</title>

</head>

<body>

<form name="form" action="search09.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 ","usename",password"); //(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);
$numrows= mysql_num_rows($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: &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";
$results = mysql_query($query) or die ("Couldnt execute query");
}
// now you can display the results returned
while ($row = mysql_fetch_array($results)){
$title =$row["surname"];

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>";
?>

any help give me with the above greatfully recieved and thanks again
for the help already given.

Kind Regards
Gerry

 

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