[Resolved] search form giving strange error

sher

Dear coders,

I'm getting an error when searching for a string in the table of a database. The error is as follows: Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in C:\apachefriends\xampp\htdocs\xampp\cmdb\Categorieen_zoeken_resultaten.php on line 24

What could be possibly wrong?

Thanks in advance.

Categorieen_zoeken_resultaten.php(results)

<?php
trim($categorienaam);
if (!$categorienaam)
{
echo "Er is geen zoekterm ingevoerd. Graag terug en opnieuw proberen.";
exit;
}

$categorienaam = addslashes($categorienaam);

$link = mysql_connect("localhost", "root", "password")
or die("Kon geen verbinding tot stand brengen: " . mysql_error());
mysql_select_db('cmdb', $link) or die ( mysql_error());
$query = "select * from categorie where like '%".$categorienaam."%'";
$result = mysql_query($query);
$num_results = mysql_num_rows($result); //line 24

echo "<p>Aantal gevonden categorienamen: ".$num_results."</p>";

for ($i=0; $i < $num_results; $i++)
{
$row = mysql_fetch_array($result);
echo "<p>".($i+1).". Categorienaam: ";
echo htmlspecialchars( stripslashes($row["categorienaam"]));
echo "</p>";
}
?>

Categorieen_zoeken.php(searchform)

<html>
<head>
<link rel="stylesheet" type="text/css" href="stylesheet.css">
<title></title>
</head>
<body>
<H1>Categorieën zoeken</H1>
<form action="Categorieen_zoeken_resultaten.php" method="post">
<table border=0 width="571" cellspacing=1 cellpadding=3 bgcolor="#353535" align="center">
<tr>
<td bgcolor="#ffffff" width="31%">
<P><b>Voer een categorienaam in:</b></P>
</td>
<td bgcolor="#ffffff" width="70%">
<input type="text" name="categorienaam">
</td>
</tr>
<tr>
<td bgcolor="#ffffff" colspan=2 align="center">
<input type="submit" name="submit" value="Zoek">
</td>
</tr>
</table>
</form>
</body>
</html>

wilku

I think there's an error in your query.

$query = "select * from categorie where like '%".$categorienaam."%'";//what should be like ...
$result = mysql_query($query)or die ( mysql_error());//see what comes out of it.

sher

Dear wilku,

Thanks very much for your post. I'm still getting an error. 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 'LIKE '%test1%'' at line 1

Originally posted by wilku
I think there's an error in your query.

$query = "select * from categorie where like '%".$categorienaam."%'";//what should be like ...
$result = mysql_query($query)or die ( mysql_error());//see what comes out of it.

[/B]

sher

Originally posted by sher
Dear wilku,

Thanks very much for your post. I'm still getting an error. 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 'LIKE '%test1%'' at line 1

I got the problem 🙂 The sql statement should have been in the following format: SELECT * FROM tablename WHERE columnname LIKE '%name%'

I was missing a columnname.

Greetz,

Sher

wilku

Originally posted by sher
I got the problem 🙂 The sql statement should have been in the following format: SELECT * FROM tablename WHERE columnname LIKE '%name%'

That's what I meant by

//what should be like ...

Glad you solved that one.