Saskaitīt stundas, minūtes.

[false]

DB

id | dep_time | dep_date    | arr_time | arr_date


1 | 18:00   	| 02.01.2012 | 02:42    | 03.01.2012
2 | 15:00   	| 03.01.2012 | 21:27 	| 03.01.2012
3 | 16:25   	| 04.01.2012 | 16:41 	| 04.01.2012
..

 

kods kas dabū cik stundas, minūtes pagāja no dep_time līdz arr_time

while($flight = mysql_fetch_array($flightsql)){


$date = $flight['dep_date'];
$dateA = $flight['arr_date'];
$beggin_hour = $flight['dep_time'];
$end_hour = $flight['arr_time'];

define("SECONDS_PER_HOUR", 60*60);
$start = strtotime($date." ".$beggin_hour);
$stop = strtotime($dateA." ".$end_hour);

$difference = $stop - $start;

$hours = round($difference / SECONDS_PER_HOUR, 0, PHP_ROUND_HALF_DOWN);

$minutes = ($difference % SECONDS_PER_HOUR) / 60;

echo $hours. " " .$minutes. "<br/>";
}

 

iznākums

id | stundas | minutes
1 | 8            | 42
2 | 6            | 27
3 | 0            | 36

un kāds var paskaidrot, kā dabūt kopējo stundu, minūšu skaitu? Ir jasanāk 15h 45min.

Edited January 5, 2012 by false

[daGrevis]

Kāpēc ir divas kolonas gan ielidošanas, gan izlidošanas laikiem (tjip, katram pa divām)? To visu varētu glabāt par divām kolonām mazāka izmēra shēmā.

[false]

Var, bet šādi arī var glabāt, ne?

Un lūgums nenovērsties no jautājuma..

Edited January 5, 2012 by false

[marrtins]

1) Tu nepareizi rēķini:

$hours = floor($difference / (60 * 60));
$mins = floor(($difference / 60) % 60);
$secs = $difference % 60;

2) kopsummas var dabūt summējot.

$hoursTotal += $hours;
utt.

 

3) kopsummas var dabūt summējot.

{
..
$differenceTotal += $difference;
..
}
$hoursTotal = floor($differenceTotal / (60 * 60));
$minsTotal = floor(($differenceTotal / 60) % 60);
$secsTotal = $differenceTotal % 60;

Edited January 5, 2012 by marrtins

[Val]

konstante ciklā, ā jē :)

[Morphius]

A kāpēc nevar glabāt visus laikus timestamp formāta? Vienkārši pieskaiti vai atņem noteiktu daudzumu sekundes un viss, un tad vajadzības gadījumā kovertē lai būtu lasāmāk?! date("Y-m-d", $timestamp);

[marrtins]

date("Y-m-d", 3); <- LOL

[Gints Plivna]

Visu vai gandrīz visu var izdarīt SQLā,

PHP ir lieks (tas tā fleimam ;) )

Paspēlējos ar MySQL datumu f-jām

BTW sākotnējā uzdevumā dati nesaskan ar rezultātu 3 rindiņā, bet tas sīkums.

Tātad:

 

 

mysql> create table t (id int, dep_time time, dep_date date, arr_time time, arr_
date date);
Query OK, 0 rows affected (0.03 sec)


mysql> insert into t values (1, '18:00', '2012.01.02', '02:42', '2012.01.03');
Query OK, 1 row affected (0.00 sec)

mysql> insert into t values (1, '15:00', '2012.01.03', '21:27', '2012.01.03');
Query OK, 1 row affected (0.00 sec)

mysql> insert into t values (1, '16:25', '2012.01.04', '16:41', '2012.01.04');
Query OK, 1 row affected (0.00 sec)

mysql> select * from t;
+------+----------+------------+----------+------------+
| id   | dep_time | dep_date   | arr_time | arr_date   |
+------+----------+------------+----------+------------+
|    1 | 18:00:00 | 2012-01-02 | 02:42:00 | 2012-01-03 |
|    1 | 15:00:00 | 2012-01-03 | 21:27:00 | 2012-01-03 |
|    1 | 16:25:00 | 2012-01-04 | 16:41:00 | 2012-01-04 |
+------+----------+------------+----------+------------+


mysql> SELECT SUM(HOUR(diff)) + FLOOR(SUM(MINUTE(diff))/60) as h,
   ->        MOD(SUM(MINUTE(diff)),60) as m
   -> FROM (
   ->   SELECT timediff(
   ->     date_add(arr_date, INTERVAL arr_time HOUR_SECOND),
   ->     date_add(dep_date, INTERVAL dep_time HOUR_SECOND)) AS diff
   ->   FROM t) AS t1;
+------+------+
| h    | m    |
+------+------+
|   15 |   25 |
+------+------+
1 row in set (0.00 sec)

 

 

Gints Plivna

http://datubazes.wordpress.com