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View Full Version : PHP Search Engine



joeiscoolone
06-22-2006, 10:10 AM
Im having trouble with a search script, I get an error when I try to search for something. The error is

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in c:\websites\sdeinterfacenet91\sde.interface-net.com\sde\jherbert\test\searchone.php on line 30

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in c:\websites\sdeinterfacenet91\sde.interface-net.com\sde\jherbert\test\searchone.php on line 42
Sorry, but we can not find an entry to match your query

Here is the script

<?
//This is only displayed if they have submitted the form
if ($searching =="yes")
{
echo "<h2>Results</h2><p>";

//If they did not enter a search term we give them an error
if ($find == "")
{
echo "<p>You forgot to enter a search term";
exit;
}

// Otherwise we connect to our Database
mysql_connect("mysql.yourhost.com", "user_name", "password") or die(mysql_error());
mysql_select_db("database_name") or die(mysql_error());

// We preform a bit of filtering
$find = strtoupper($find);
$find = strip_tags($find);
$find = trim ($find);

//Now we search for our search term, in the field the user specified
$data = mysql_query("SELECT * FROM users WHERE upper($field) LIKE'%$find%'");

//And we display the results
LINE 30.while($result = mysql_fetch_array( $data ))
{
echo $result['fname'];
echo " ";
echo $result['lname'];
echo "<br>";
echo $result['info'];
echo "<br>";
echo "<br>";
}

//This counts the number or results - and if there wasn't any it gives them a little message explaining that
LINE 42.$anymatches=mysql_num_rows($data);
if ($anymatches == 0)
{
echo "Sorry, but we can not find an entry to match your query<br><br>";
}

//And we remind them what they searched for
echo "<b>Searched For:</b> " .$find;
}
?>
Can someone tell me how to fix those errors.

NOTE:I did not put the host,user name,password and database name in the code in this post for obvious reasons.

keissfootball
06-22-2006, 10:19 AM
don't have much time, but I already see one mistake, change if ($find == "") to elseif ($find == "")

vinyl-junkie
06-22-2006, 02:24 PM
It looks like the problem may be because you don't have a space after the word LIKE.


$data = mysql_query("SELECT * FROM users WHERE upper($field) LIKE '%$find%'");



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