retrieving query php errors

halebop

I am new to using php and Mysql - wow is all I can say.

I have been spending the day setting up my results.php and ran into the following errors.

If you want to see the error first hand visit the site and enter a search. www.plasticsdirectory.com

This is what I get on my results page -

Results found matching your query:

Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in (leaving out source)results.php on line 78

Company Name:
Category:
Contact Person:
Address1:
Address2:
City, State:
Postal Code:
Country
E mail:
Web Address:
Description:

I have been taking the code straight from a book and it will not take the code for displaying the results.

Here is the orginally code - using this I don't get the Company Name... listing and the error mysql_num_rows():

Here is the complete code as given by the book:

<?php
// create short variable names
$state=$HTTP_POST_VARS['searchtype'];
$keyword=$HTTP_POST_VARS['keyword'];
$keyword= trim($keyword);

						if (!$searchtype || !$keyword)
						{
						echo 'You have not entered search details. Please go back and try again.';
						exit;
						}
						$searchtype = addslashes($searchtype);
						$keyword = addslashes($keyword);

						// connect to database
						@ $db = mysql_pconnect('localhost', 'user', 'pw');

						if (!$db)
						{
						echo 'Error: Could not connect to database. Please try again later.';
						exit;
						}

						mysql_select_db('plas27');
						$query = "select * from plas27 where ".$searchtype." like '%".$keyword."%'";
						$result = mysql_query($query);
						$num_results = mysql_num_rows($result);										
						echo '<p>Number of results found matching your query: '.$num_results.'</p>';
						for ($i <$num_results; $i++)	
						{
						 $row = mysql_fetch_array($result);
						echo '<p><strong>'.($i+1).'. Company Name: ';
						echo htmlspecialchars(stripslashes($row['Company_Name']));
			echo '<br />Category: ';
			echo stripslashes($row['Category']);
			echo '<br />Contact Person: ';
						echo stripslashes($row['Contact_Person']);
						echo '<br />Address1: ';
						echo stripslashes($row['Address1']);
						echo '<br />Address2: ';
						echo stripslashes($row['Address2']);
						echo '<br />City, State: ';
						echo stripslashes($row['City, State']);
						echo '<br />Postal Code: ';
						echo stripslashes($row['Postal_Code']);
						echo '<br />Country ';
						echo stripslashes($row['Country']);
						echo '<br />E mail: ';
						echo stripslashes($row['E_mail']);
						echo '<br />Web Address: ';
						echo stripslashes($row['Web_Address']);
						echo '<br />Description: ';
						echo stripslashes($row['Description']);
						echo '</p>';
						}
						?>

Any help to get this error fixed would be great.

Thanks
Mel

Mucello

I'm not seeing the mysql_fetch_row() function even being used in the code you posted. Are you sure you're posting the right section of code?

clandestine555

hrm, what is line 75 - 80 in results.php?

HalfaBee

Your for loop construction is wrong.

for( $i=0;$i<$numrows;+i++ )

HalfaBee

halebop

Thanks for the replies -

the coding that I put up is the orginal code that was in the book - I changed somethings according to the book, which resulted in the error listed first.

The script as you see it prompts the error for the
mysql_fetch_row() it says that it doesn't recognize the structure of the command.

Melissa

halebop

$num_results = mysql_num_rows($result);

laserlight

Could you indent your code and post it again, enclosing it within [php] and [/php] bbcode tags?

Also, use an inline comment to mark out the line that the parser said produced an error.

Judging from what you have posted, it is likely that the query was constructed poorly by the script.

halebop

HaffaBee - I entered the script you suggested and got the following message.

Parse error: parse error, expecting `')'' in /usr/shared/web/plas27/results.php on line 79

Adopted Daughter I will do my best to show you how it appears. I am new this type of communication also.

Thanks for the help.

halebop

I have attached the code and highlighted the area in red that prompts the error message.

Melissa

halebop

<?php
// create short variable names
	$state=$HTTP_POST_VARS['searchtype'];							$keyword=$HTTP_POST_VARS['keyword'];
	$keyword= trim($keyword);

if (!$searchtype || !$keyword)
{
echo 'You have not entered search details. Please go back and try again.';
exit;
}
$searchtype = addslashes($searchtype);
$keyword = addslashes($keyword);

// connect to database
@ $db = mysql_pconnect('localhost', 'user', 'password');

if (!$db)
{
echo 'Error: Could not connect to database. Please try again later.';
exit;
}

mysql_select_db(' database');
$query = "select * from database where ".$searchtype." like '%".$keyword."%'";
$result = mysql_query($query);

(error is here)  $num_results = mysql_num_rows($result);

echo '<p>Results found matching your query: '.$num_results.'</p>';

for ($i=0; $i <$num_results; $i++) 
{
 $row = mysql_fetch_row($result);
echo '<p><strong>'.($i+1).'. Company Name: ';
echo htmlspecialchars(stripslashes($row['Company_Name']));
echo '<br />Category: ';
echo stripslashes($row['Category']);
echo '<br />Contact Person: ';
echo stripslashes($row['Contact_Person']);
echo '<br />Address1: ';
echo stripslashes($row['Address1']);
echo '<br />Address2: ';
echo stripslashes($row['Address2']);
echo '<br />City, State: ';
echo stripslashes($row['City, State']);
echo '<br />Postal Code: ';
echo stripslashes($row['Postal_Code']);
echo '<br />Country ';
echo stripslashes($row['Country']);
echo '<br />E mail: ';
echo stripslashes($row['E_mail']);
echo '<br />Web Address: ';
echo stripslashes($row['Web_Address']);
echo '<br />Description: ';
echo stripslashes($row['Description']);
echo '</p>';
}
?>

$num_results = mysql_num_rows($result);

HalfaBee

Change this line so you can see what the error is.

$result = mysql_query($query) or die( mysql_error()."SQL= $sql" );

Sorry about the +i++ should have been $i++ 🙁

HalfaBee

halebop

I inserted the new code so that I can see the error - here is what is is telling me

first is said: Table 'plas27.plas27' doesn't existSQL=

plas27 is the name of the database - so I then went in and entered all the table names in one of the areas where I had plas27

here is what I got back

Table 'plas27.Category' doesn't existSQL=

I am not for sure the order that it needs to be in - I may be able to figure it out on my own at this point - but would gladly take more help.

Thanks

halebop

I figured it out. I know have it posting the number of results.

I didn't my tables listed right - THANK YOU.

NEXT problem - now it doesn't want to show my results. I will try to fumble through.

any thoughts?

HalfaBee

You need to use mysql_fetch_assoc() or mysql_fetch_array()
instead of mysql_fetch_row(), this onlky returns a simple array not an assoc array.
HalfaBee