nacunis

To responder 1, thank you. It didn't work because the actual code had a backtick. Oh well. 8)8)8) As for responder 2, that code came from a PHP programmer (not me) and it worked. So the question about why I did it that way and not another way was not helpful for me. :(:(:(:( I wouldn't call myself a PHP programmer at all. If the forum results in people debating code rather than being helpful, I'll drop my association with this forum. Again, thanks to responder 1. At least you tried to help rather than debate. 8)8)

I get this message with the code at the bottom: session_start(); $connection=mysql_connect("localhost","database username","db user password"); $db=mysql_select_db("database name",$connection ); $users_table = 'table name'; $username = $_POST['myusername']; $result3 = mysql_query("SELECT `nickname` from `databasename`.`tablename` WHERE `loginsv4_md5`.`username` = '$myusername'"); $result4 = mysql_query("SELECT `lastlogin` from `databasename'.'tablename` WHERE `tablename`.`username` = '$myusername'"); $result5 = mysql_query("SELECT `announce` from `databasename`.`tablename` WHERE `tablename`.`username` = '$myusername'"); $num_rows = mysql_num_rows($result3); $row3 = mysql_fetch_assoc($result3); while ($row3 = mysql_fetch_assoc($result3)) { echo " \n <br>"; echo $row3['nickname']; echo " "; echo " \n"; } $row5 = mysql_fetch_assoc($result5); while ($row5 = mysql_fetch_assoc($result5)) { echo $row5['announce']; echo " "; } $row4 = mysql_fetch_assoc($result4); while ($row4 = mysql_fetch_assoc($result4)) { echo $row4['lastlogin']; echo " "; } ..... code not displayed Error message seems to point to mysql_fetch_assoc. Other pieces of this code are being executed, so I know the module is called and executing, and of course, there is the PHP warning message from the log. Please help