Having a problem with mysql_fetch_array()...

Penguin · 01-04-2007, 11:17 AM

So, here it is:
On my computer I have Apache2 with PHP5..
I'm writing all my codes on this system.

So I built a simple code with mysql_fetch_array().. On my Comp. it works great,
but when i uploaded it to some hosting company (who uses php5 too) it dosent work.

Here is my code:

PHP Code:


			
<?php

    $link = mysql_connect ($dbhost, $dbusername, $dbpass)

        or die ("Can not connect to MySQL");

    if (mysql_select_db ($dbname)) {

        $result = mysql_query ("Select * from GuestBook order by gb_index desc");

        if (mysql_affected_rows() == 0) {

            echo "<font face='Arial' size='2'>àéï ëøâò øòéåðåú.</font>\n";

        } else {

            while ($row = mysql_fetch_array ($result)) {



                echo "<html dir='rtl'>\n";

                echo "<p>\n";

                echo "<table dir='rtl' align='center' cellspacing='5' bgcolor='#e9e9e9' border=0 width='70%'style='font: 8pt verdana, geneva, lucida, arial, helvetica, sans-serif;'>\n";

                echo "<tr><td align=right>\n";

                echo "<b>îàú:</b>\n";

                echo "<br>".$row["name"]." <br>\n";

                echo "<b> ôåøñí á: </b>\n";

                echo "<br>".$row["date"]." <br><br>\n";

                echo "<b> äøòéåï: </b>\n";

                echo "<br>".$row["message"]."\n";

                echo "</tr></td></table>\n";

                echo "</html>\n";



            }

        }

    } else {

        echo "<font face='Arial' size='2'>An error occured will trying to select the database.</font>\n";

    }

    mysql_close ($link);

?>

Returns the next error:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home2/yazamim/webapps/htdocs/ideas/admin1.php on line 43

Line 43:

PHP Code:


			
while ($row = mysql_fetch_array ($result)) {

Can anyone help me?

MRMAN · 01-04-2007, 11:21 AM

change this line

PHP Code:


			
$result = mysql_query ("Select * from GuestBook order by gb_index desc");

to

PHP Code:


			
$result = mysql_query ("Select * from GuestBook order by gb_index desc") or die (mysqlL_error());

That error tends to mean there is no data in $result so you could check that my putting this line

PHP Code:


			
print mysql_num_rows($result);

just before the while loop

that won't fix the problem but might help find out what the problem is

Penguin · 01-04-2007, 11:33 AM

Thank You!
I have managed to find the problem..
The table name at the database wasn't correct (case-sensitive)

Thanks Again

CFMaBiSmAd · 01-04-2007, 02:38 PM

Quote:

That error tends to mean there is no data in $result

The error means the query failed (syntax error, typo's or incorrect table/column names, no database selected, no connection to database server...) and returned a FALSE value. Executing a mysql function using the FALSE value instead of a resource result is what causes the error. A query that succeeds returns a valid resource result even if the number of rows in the result is zero.

Executing a mysql_fetch_array($result) on a resource result with zero rows returns a FALSE value. Executing a mysql_num_rows($result) on a resource result with zero rows returns a ZERO value.

Executing a mysql_fetch_array($result) or a mysql_num_rows($result)following a failed query, results in the error.