MySQL- select -date range weekly results?
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MySQL- select -date range weekly results?
zdarveite
imam BIG problem kakto se kazva
imam edna tablica s edno pole text i edno date pylna s danni
ta iskam da nparavia select * from table ot data do data
do tuk ok
oba4e - iskam si4ki resulti da mi doidat na edin page kato resultite sa listnati po sedmica, sortirani ASC
demek neshto takova
result:
------------------------
week 1
report1 - 02-01-2004
report2 - 03-01-2004
...
week 2
report3 - 11-01-2004
report4 - 12-01-2004
...
niakakvi idei?
niakakvi genialni funkcii?:)
predvaritelno Vi blagodaria
-----------------
signatura
imam BIG problem kakto se kazva
imam edna tablica s edno pole text i edno date pylna s danni
ta iskam da nparavia select * from table ot data do data
do tuk ok
oba4e - iskam si4ki resulti da mi doidat na edin page kato resultite sa listnati po sedmica, sortirani ASC
demek neshto takova
result:
------------------------
week 1
report1 - 02-01-2004
report2 - 03-01-2004
...
week 2
report3 - 11-01-2004
report4 - 12-01-2004
...
niakakvi idei?
niakakvi genialni funkcii?:)
predvaritelno Vi blagodaria
-----------------
signatura
Re: MySQL- select -date range weekly results?
ами аз лично бих направил
select * from table where date_field between date1 and date2 order by date_field
и после с php-то хмммм май date('Yw', strtotime($row['date_field']));
нещо такова
нещо такова не е много чист код, ама се надявам да е ясно кво имам предвид
select * from table where date_field between date1 and date2 order by date_field
и после с php-то хмммм май date('Yw', strtotime($row['date_field']));
нещо такова
$prev_week = '';
while ($row = mysql_fetch_row($res)) {
$current_week = date('Yw', strtotime($row['date_field']));
if ($prev_week != $current_week) {
$prev_week = $current_week;
echo "Седмица ето таз";
}
.....
}
нещо такова не е много чист код, ама се надявам да е ясно кво имам предвид
Re: MySQL- select -date range weekly results?
aaaaaaa - mojelo:):)
tarikatlyshtinata e tuka bratia
MySQL function - DATE_FORMAT(day, '%v'),
ama niakoi triabva da 4ete - nali se seshtas:)
polzva se i taka
$sql = "SELECT day, report, work_time, client, cat, DATE_FORMAT(day, '%v'), DATE_FORMAT(day, '%a') FROM ".$prefix."_reports ";
$sql .= " WHERE (day >= '$start' and day <= '$end') and uid = '$rest[user]' order by day ASC ";
kato ti vryshta resulta v otdelen row - mnogo iako
-----------------
signatura
tarikatlyshtinata e tuka bratia
MySQL function - DATE_FORMAT(day, '%v'),
ama niakoi triabva da 4ete - nali se seshtas:)
polzva se i taka
$sql = "SELECT day, report, work_time, client, cat, DATE_FORMAT(day, '%v'), DATE_FORMAT(day, '%a') FROM ".$prefix."_reports ";
$sql .= " WHERE (day >= '$start' and day <= '$end') and uid = '$rest[user]' order by day ASC ";
kato ti vryshta resulta v otdelen row - mnogo iako
-----------------
signatura
Редактирано от an_ на
DATE_FORMAT ne ti vyrshi rabota ;)
Tva ne ti vyrshi koi znae kakva rabota ... shtoto ti iskash da otdelish select resulta spored sedmicata :) Taka che dali s DATE_FORMAT ili v string vse shte triabva da sravniavash sedmicata na tekushtia zapis s tazi na prdhodnia, za da proverish dali triabva da zapochnesh nova sekcia za novata sedmica ili triabva da go printnesh v starata sekcia :)) Triabva da se chete ... spor niama ;) Ta postinga na phpguru shte ti svyrshi rabota, a ne DATE_FORMAT ...
Re: MySQL- select -date range weekly results?
Между другото, нямаше ли един клуб за бази данни?
Само да добавя ...
Аз предпочитам да изполвам за подобни ситуации функциите in_array и Array_push
Break The UnBreakable
Break The UnBreakable