$q=mysql_querry('SELECT * from table')
$qf = mysql_fetch_array;
while($data =$qf){echo $qf['email']."" $qf['time']; }

так цикл получается бесконечным, а нужно вывести все значения в построенную таблицу,
как сделать проще?

Глупость какая то.

$q=mysql_query('SELECT * from table');
if(mysql_num_rows($q)){
while($row = mysql_fetch_assoc($q)){echo $row['email']."" $row['time']; }
}

_____________
Mysql, Postgresql, Redis, Memcached, Unit Testing, CI, Kohana, Yii, Phalcon, Zend Framework, Joomla, Open Cart, Ymaps, VK Api

1. не использовать mysql_query , есть mysqli расширение
2. посмотреть документацию по php на php.net

<?php
$mysqli = new mysqli("localhost", "my_user", "my_password", "world");

/* check connection */
if ($mysqli->connect_errno) {
    printf("Connect failed: %s\n", $mysqli->connect_error);
    exit();
}

/* Create table doesn't return a resultset */
if ($mysqli->query("CREATE TEMPORARY TABLE myCity LIKE City") === TRUE) {
    printf("Table myCity successfully created.\n");
}

/* Select queries return a resultset */
if ($result = $mysqli->query("SELECT Name FROM City LIMIT 10")) {
    printf("Select returned %d rows.\n", $result->num_rows);

    /* free result set */
    $result->close();
}

/* If we have to retrieve large amount of data we use MYSQLI_USE_RESULT */
if ($result = $mysqli->query("SELECT * FROM City", MYSQLI_USE_RESULT)) {

    /* Note, that we can't execute any functions which interact with the
       server until result set was closed. All calls will return an
       'out of sync' error */
    if (!$mysqli->query("SET @a:='this will not work'")) {
        printf("Error: %s\n", $mysqli->error);
    }
    $result->close();
}

$mysqli->close();
?>