$select = "mysql_select_db ($database)"; << like Mike said you don't need to have this in " and no need for a var to be assigned...
Then when you want to print out the data
print_r( $row );
you can use $row['table_row_name']; :) Adam
At 08:26 PM 3/22/2003 -0500, you wrote:
I have been having this problem for a while now, and I cant figure it out. the code is:
<?
$host = "localhost";
$uname = "sniper";
$pass = "starcraft";
$database = "sha";
$tablename = "members";
$connect = mysql_connect ($host, $uname, $pass);
$select = "mysql_select_db ($database)";
$query = "SELECT * from $tablename";
$result = mysql_query ($query, $connect);
while ($row = mysql_fetch_array($result)) { print_r( $row ); echo mysql_error(); } ?>
mysql has an error on line 19. Line 19 is while ($row = mysql_fetch_array($result)). What am I doing wrong?
Thanks for your help, Mike
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