mysql_fetch_array

cliffmar3024

I get this message when trying to display my page

Warning: mysql_fetch_array(): supplied argument is not a
valid MySQL result resource in /home/fsd-web/public_html/testdb.php on line 8

my code is below PLEASE HELP

<HTML>
<?php
$db = mysql_connect("localhost", "fsdweb_cliff", "god") or die("could not connect");
mysql_select_db("fsdweb_future",$db);
$result = mysql_query("SELECT * FROM personnel",$db) ;
echo "<TABLE>";
echo"<TR><TD><B>Full Name</B><TD>&nbsp<B>Nick Name</B><TD>&nbsp<B>Salary</B></TR>";
while ($myrow = mysql_fetch_array ($result))
{
echo "<TR><TD>";
echo $myrow["firstname"];
echo " ";
echo $myrow["lastname"];
echo "<TD>";
echo $myrow["nick"];
echo "<TD>";
echo $myrow["salary"];
}
echo "</TABLE>";
?>
</HTML>

laserlight

Does the personnel table contain any rows?

You might want to check with mysql_num_rows($result) first.

ahundiak

Even if personnel had no rows you would still get a valid $result.
Echo out any errors.
$result = mysql_query("SELECT * FROM personnel",$db) ;
if (mysql_errno()) die(mysql_error());

laserlight

In retrospect yeah, it should be a problem with the query itself or your selection of a DB, not so much a null result query affecting the later operations.

John_Canyon

$result = mysql_query("SELECT * FROM personnel",$db) ;

Your syntax is incorrect, thus the result ID is returning negative, or no results (ERROR).

Your query would be correct if you were using something like the following which has been deprecated.

$result = mysql_db_query($db,"SELECT * FROM personnel") ;

The Correct syntax for the query you are trying to make would be..

$result = mysql_query("SELECT * FROM personnel");
if(mysql_num_rows($result) == "0")
{
die(mysql_error());
}

$result = mysql_query("SELECT * FROM personnel") or die(mysql_error());

Cheers.