The warning means the query failed in mysql. Perhaps it failed in the
connect. You do not need the number of rows.
To determine what the problem is, ALWAYS include a die() clause:
$db = mysql_connect("localhost", "", "") or die("Error: ".mysql_error());
mysql_select_db("employee",$db) or die("Error: ".mysql_error());
$result = mysql_query("SELECT * FROM employees",$db) or die("Error:
".mysql_error());
-----Original Message-----
From: Todor Stoyanov [mailto:[EMAIL PROTECTED]]
Sent: Friday, February 22, 2002 10:13 AM
To: [EMAIL PROTECTED]
Subject: [PHP] Re: difference between php3 and php4
You must first get the number of record set rows.
The warning is because you trying to fetch a non existing row.
"Toni Baker" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> Warning: Supplied argument is not a valid MySQL result
> resource in c:\program files\apache
> group\apache\cgi-bin\emp.php3 on line 10
>
> $db = mysql_connect("localhost", "", "");
> mysql_select_db("employee",$db);
> $result = mysql_query("SELECT * FROM employees",$db);
> echo "<table border=0>";
> echo "<tr>";
> echo "<td bgcolor=#FAFOE6 colspan=2><B>Name</B></td>";
> echo "<td bgcolor=#FAFOE6><B>Address</B></td>";
> echo "<td bgcolor=#FAFOE6><B>Position</B></td>";
> echo "</tr>";
> while($row = mysql_fetch_row($result))
> {
> $fname = $row[1];
> $lname = $row[2];
> $address = $row[3];
> $position = $row[4];
> echo "<tr bgcolor=#C6E7DE>";
> echo "<td>$fname</td>";
> echo "<td>$lname</td>";
> echo "<td>$address</td>";
> echo "<td>$position</td>";
> echo "</tr>";
> }
>
> This script works in php3 but not in windows php4.
> The error message above points to the mysql_fetch_row
> function. Does this function work differently in
> php4? If not, what else might cause this script to
> work in php3 and not in php4?
>
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