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 $name="name"; 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.
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