PHP Issue :[ please help?

This is the key to your problem.

I'd be checking for errors.

The script excutes fine when I test it. The only error I receive is a warning "Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/gclunis/public_html/moodtunez.com/test-site/search.php on line 52" but i'm not sure what that means. I didn't think that would cause all html below the script to not display though. The html is outside of the closing php tag so that error should be read seperately no?

alright i'll look through it. Thank's so much for the help. I'll let you all know how it works out.

Hey guys, so I can't quite figure out what's wrong with the script. Can you guys check it out and get back to me?

[php]

<?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 tell us how you are feeling...</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","username","password"); //(host, username, password)

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

// Build SQL Query
$query = "select mood from userupload where 1st_field like \"%$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 please try again.</p>";

// google
echo "<p><a href=\"http://moodtunez.com"
. $trimmed . "\" target=\"_blank\" title=\"Try
" . $trimmed . " again\">Click here</a> to try the
search again</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>Your mood is: &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]

*The line in question has been put in bold for demonstration purposes
*username, password, and database name has been blanked

mood does exist...i didn't know that 1st_field was a value that it was trying to pull. Is 1st_field the primary field?

OOOO that was it. thank you sooo much :]

I had the same. but it runs sometimes well and sometimes not. might be problem with database

Are you developing on a local machine?