Warning: mysql_num_rows() Error

BryantA · June 4, 2012

Hi,

I'm very new to PHP and MySQL. But I'm pretty sure there's a simple solution to my problem because there usually is.

What I'm trying to do is create a search engine for a website that retrieves links from pages that I've placed in a table in my database according to their designated keywords. But every time I enter a keyword into the search box this error comes up:

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/xxxxxx/public_html/search.php on line 35

No results found for "keyword"

I underlined line 35 with a row of asterisks.

If you think you have an idea to what my problem is I would really, really would appreciate your help.

Thank you!

Here is the PHP code:

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">

<head>

<title>Search Engine</title>

</head>

<body>

<h1>Search</h1>

</center>

</form>

<hr />

<?php

$search = $_GET['search'];

$terms = explode(" ", $search);

$query = "SELECT * FROM search WHERE ";

foreach ($terms as $each){

$i++;

if ($i == i)

$query .= "keywords LIKE '%$each%' ";

else

$query .= "OR keywords LIKE '%$each%' ";

}

// connect

mysql_connect("localhost", "username", "password");

mysql_select_db("databasename");

$query = mysql_query($query);

$numrows = mysql_num_rows($query);

**************************************

if ($numrows > 0) {

while ($row = mysql_fetch_assoc($query)) {

$id = $row['id'];

$title = $row['title'];

$description = $row['description'];

$keywords = $row['keywords'];

$url = $row['url'];

echo "<h2><a href='$url'>$title</a></h2>

$description<br /><br />";

}

else {

echo "No results found for \"<b>$search</b>\"";

//disconnect

mysql_close();

}

?>

</body>

</html>

Psycho · June 4, 2012

This forum has [ code ] tags that you can put around code so it displays in a readable format - please use them. You don't even have line breaks in what you posted.

But, from what I see the code is a mess. The problem when you see these errors is that the query failed - therefore any function that tries to use the results of the query will produce errors. You need to implement error handling into your code. Below is a rewrite of your code. I changed a lot and I'm not going to take the time to explain why I changed what I did. But, you can ask specific questions if you want.

<?php

//Process the input (if exists)
$searchTermsStr = isset($_POST['search']) ? trim($_POST['search']) : '';
$searchOutput = ''; //Var to hold the output

//Verify there was a valid input
if (!empty($searchTermsStr))
{
    // connect
    mysql_connect("localhost", "username", "password");
    mysql_select_db("databasename");

    //Create an array of search terms (filter out empty values)
    $searchTermsAry = array_filter(explode(" ", $searchTermsStr));

    //Process terms array into Query strings
    foreach($searchTermsAry as $key => $value)
    {
        $value = mysql_real_escape_string($value);
        $searchTermsAry[$key] = "keywords LIKE '%$value%'"
    }

    //Create and run query
    $query = "SELECT url, title, description FROM search WHERE " . implode(' OR ', $searchTermsAry);
    $result = mysql_query($query);

    //Verify query results
    if(!$result)
    {
        $searchOutput .= "There was a problem running the query:<br><br>";
        $searchOutput .= "Query: $query<br><br>";
        $searchOutput .= "Error: " . mysql_error();
    }
    elseif(!mysql_num_rows($result))
    {
        $searchOutput .= "No results found for '<b>{$searchTermsStr}</b>'";
    }
    else
    {
        while ($row = mysql_fetch_assoc($query))
        {
            $searchOutput .= "<h2><a href='{$row['url']}'>{$row['title']}</a></h2> {$row['description']}<br /><br />";
        }
    }

    //disconnect
    mysql_close();
}

?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
    <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
    <title>Search Engine</title>
</head>
<body>
    <center>
    <h1>Search</h1>
    <form action='search.php' method='post'>
    <input type='text' name='search' size='90' value='<?php echo $searchTermsStr; ?>' />
    <input type='submit' name='submit' value='Search' >
    </center>
</form>
<hr />
<?php echo $searchOutput; ?>
</body>
</html>

waynew · June 4, 2012

Your query failed.

Change:

$query = mysql_query($query);

to

$query = mysql_query($query) or die(mysql_error());

and post whatever errors are shown on your screen.

BryantA · June 5, 2012

Psycho,

Thank you so much for the feedback. I thought I did use the PHP code feature for this post, I guess not :-[ . Sorry though, I'll make sure to preview my post next time.

But I tried to use your revised code but when I copy and paste it into Dreamweaver it shows an error on line 30 that I can't figure out. I'm sure it works.

I'm pretty sure it would've been much easier to read if I would've of used the tags, silly me! But here is my original code with the code tags :

Thanks again.

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Search Engine</title>
</head>
<body>

<center>
<h1>Search</h1>
<form action='search.php' method='GET'>
<input type='text' name='search' size='90' value='<?php echo $_GET['search']; ?>'  /> 
<input type='submit' name='submit' value='Search' >
</center>
</form>
<hr />
<?php
$search = $_GET['search'];
$terms = explode(" ", $search);
$query = "SELECT * FROM search WHERE ";

foreach ($terms as $each){
	$i++;
	if ($i == i)
		$query .= "keywords LIKE '%$each%' ";
	else
		$query .= "OR keywords LIKE '%$each%' ";
}

// connect
mysql_connect("localhost", "username", "password");
mysql_select_db("databasename");

$query = mysql_query($query) or die(mysql_error());
$numrows = mysql_num_rows($query);
*****************************************	

if ($numrows > 0) {

	while ($row = mysql_fetch_assoc($query)) {
		$id = $row['id'];
		$title = $row['title'];
		$description = $row['description'];
		$keywords = $row['keywords'];
		$url = $row['url'];

		echo "<h2><a href='$url'>$title</a></h2>
		$description<br /><br />";
}
}
else {
		echo "No results found for \"<b>$search</b>\"";
		//disconnect 
		mysql_close();
	 }
?>
</body>
</html>

BryantA · June 5, 2012

Waynewex,

Thank you for the advice. I added the wonderful line of code that you suggested by replacing it with that original line. And the great news is I'm no longer getting that "Warning: mysql_num_rows()..." error message. But now it's showing me this error message:

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'OR keywords LIKE '%keyword%'' at line 1

I guess there might be something wrong with my MySQL database or table I'm not sure though. I know I have my collation set to "utf8_unicode_ci" because I heard that that was the best to use.

Do you have any suggestions as to why it might be saying this and what I can do to fix it?

Thank you so much for your help

Side Note: I followed a YouTube video that gave instructions on how to code a search engine. That's where I got the code from. And as he was testing it when he was done it worked perfect for him. And all my original coding was exactly like his. I quadruple checked it!

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