----- Original Message ----- From: "Ray Cantwell" <malbojia@xxxxxxx> To: <php-general@xxxxxxxxxxxxx> Sent: Tuesday, February 21, 2006 1:27 PM Subject: Parse Error > Hi all, > I am a noob and super confused right now. I have some really simple code > and i am getting an error that reads: > *Parse error*: syntax error, unexpected T_VARIABLE in > */var/www/mysql_up.php* on line > > here is the code: > > <html> > <head><title>Test MySQL</title></head> > <body> > <!-- mysql_up.php --> > <?php > $host="localhost" > $user="ray" > $password="*****" > > mysql_connect($host,$user,$password) ; > $sql="show status"; > $result = mysql_query($sql); > if ($result == 0) > echo "<b>Error " . mysql_errno() . ": " > . mysql_error() . "</b>"; > else > { > ?> > <!-- Table That Displays the results --> > <table border="1"> > <tr><td><b>Variable_name</b></td><td><b>Value</b> > </td></tr> > <?php > for ($i = 0; $i < mysql_num_rows($result); $i++) { > echo "<TR>"; > $row_array = mysql_fetch_row($result); > for ($j = 0; $j < mysql_num_fields(result); $j++) > {/ > echo "<TD>" . $row_array[$j] . "</td>"; > } > echo "</tr>"; > } > ?> > </table> > <?php } ?> > </body></html> > > I am really confused because i cannot see any obvious errors. > any help would be appreciated. > > Ray. > > > > you need to add a ; to every variable you assign... you had: $host="localhost" $user="ray" $password="*****" it should be: $host="localhost"; $user="ray"; $password="*****"; also i belive a better way to get the result of your query would be mysql_connect($host, $user, $pass) or die(mysql_error()); -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php