问输入开始日期和结束日期，然后将日期插入db中。

EN

function dateRange( $first, $last, $step = '+1 day', $format = 'd/m/Y' ) {

    $dates = array();
    $current = strtotime( $first );
    $last = strtotime( $last );

    while( $current <= $last ) {

        $dates[] = date( $format, $current );
        $current = strtotime( $step, $current );
    }

    return $dates;
}

$dateRange = dateRange("03/20/2012", "03/20/2013);

$stmt = $dbh->prepare("INSERT INTO table (from, to, date_from, date_to) VALUES (:from, :to, :datefrom, :dateto)");
$stmt->bindParam(':from', "Dar Es Salaam");
$stmt->bindParam(':to', "zanzibar");
$stmt->bindParam(':from', $datefrom);
$stmt->bindParam(':to', $dateto);
for ($i = 0; $i < count($dateRange); $i++){
   $datefrom  = strototime($dateRange[$i]);
   $dateto  = strototime($dateRange[$i+1]);
   $stmt->execute();
}

$int_date_from = strtotime($str_date_from); // Will be used often
$int_days = (strtotime($str_date_to) - $int_date_from); // Calculate difference in seconds
$int_days /= floor(60 * 60 * 24); // Seconds to days

$arr_values = array();
for($i = 0; $i < $int_days; $i++) {
    $str_new_date_from = date('m/d/Y', strtotime("+{$i} Days", $int_date_from));
    $str_new_date_to = date('m/d/Y', strtotime('+'.($i + 1).' Days', $int_date_from));

    $arr_values[] = "('{$str_from}', '{$str_to}', '{$str_new_date_from}', '{$str_date_to}')";
}

$str_values = implode(', ', $arr_values);

mysql_query("
INSERT INTO table
VALUES {$str_values}
");