Ben Galin wrote:

I want to define() TABLE as a constant instead of using the variable $table

[PHP code]
  // This runs without an error but does *not* insert a new row
  define("TABLE", "mytable");
  $mysql_connect("localhost","","") or die("Error: ".mysql_error());
  $mysql_select_db("mydb") or die("Error: ".mysql_error());
  $mysql_query("INSERT INTO TABLE (`id`,`name`) VALUES ('','$name')");
[/PHP code]

You can't have a constant in a string. You'd do it like this:

mysql_query("INSERT INTO " . TABLE . " (`id`,`name`) VALUES ('','$name')");

The other error you have, and I don't know how you think this code runs, are the dollar signs before mysql_connect(), mysql_select_db() and mysql_query() are not needed.


---John Holmes...

Amazon Wishlist:

php|architect: The Magazine for PHP Professionals –

PHP Database Mailing List (
To unsubscribe, visit:

Reply via email to