mysql_select_db("filterseveuk");
$query = ( " SHOW COLUMNS FROM user ");
$result = mysql_query ( $query ) or die( mysql_error () );
while ( $row = mysql_fetch_array ($result) ){
$x = 0 ;
echo $row[$x] ;
$x++ ;
}
I would like to replace the table name "user" with a variable $table, that
wil automatically take the value of the selected table. I have tried many
times but cannot get the correct sql syntax, it keeps saying
"you have an error in your sql syntax near 'user' line 1" when i try it with
$table.
any ideas??
thank's,
dave rice (haloplayer@hotmail.com)
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