<?PHP
    $dbconn = sqlite_open('combadd.sqlite');

    if ($dbconn) {
        $result = sqlite_query($dbconn,"SELECT * FROM combo_calcs WHERE options='easy'");
        var_dump(sqlite_fetch_array($result,sqlITE_ASSOC));
    } else {
        print "Connection to database Failed!\n";
    }
?>

但是,我得到这个错误：

Warning: sqlite_open() [function.sqlite-open]: file is encrypted or is not a database in C:\xampp\htdocs\deepthi\combadd\combadd_db.PHP on line 4
Connection to database Failed!

怎么了,怎么解决？