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Birthday Mod --- Wie macht man folgendes?

Verfasst: 03.11.2003 18:35
von flomei
Wie muss ich das hier machen:
#
#-----[ SQL ]------------------------------------------
#
ALTER TABLE users ADD user_birthday INT DEFAULT '999999' not null
ALTER TABLE users ADD user_next_birthday_greeting INT DEFAULT '0' not null
INSERT INTO CONFIG (config_name, config_value) VALUES ("birthday_required", "0")
INSERT INTO config (config_name, config_value) VALUES ('birthday_greeting', '1')
INSERT INTO config (config_name, config_value) VALUES ('max_user_age', '100')
INSERT INTO config (config_name, config_value) VALUES ('min_user_age', '5')
INSERT INTO config (config_name, config_value) VALUES ('birthday_check_day', '7')
Das verstehe ihc nicht. Muss ich da was an der Datenbank ändern? Wie mach ich das?

Danke im Vorraus!

MfG Florian

Verfasst: 03.11.2003 18:44
von Henne
Entweder per phpMyAdmin oder so in der DB ausführen oder du führst ganz einfach die birthday_db_update.php aus. Da steht das nämlich auch drin.

Verfasst: 03.11.2003 18:50
von flomei
Aha.
Mal noch ne kleine Frage am Rand.
Ich hab keine deutsche Sprachdatei gefunden.
Kann ich das hier:
$lang['Birthday'] = 'Birthday';
$lang['No_birthday_specify'] = 'None Specified';
$lang['Age'] = 'Age';
$lang['Birthday_explain'] = 'The syntax used is %s, e.g. %s, remember prefixed zeros';
$lang['Wrong_birthday_format'] = 'The birthday format was entered incorrectly.';
$lang['Birthday_to_high'] = 'Sorry, this site, does not accept user older than %d years old';
$lang['Birthday_require'] = 'your Birthday are required on this site';
$lang['Birthday_to_low'] = 'Sorry, this site, does not accept user yonger than %d years old';
$lang['Submit_date_format'] = 'd-m-Y'; //php date() format - Note: ONLY d, m and Y may be used and SHALL ALL be used (different seperators are accepted)
$lang['Birthday_greeting_today'] = 'We would like to wish you congratulatons on reaching %s years old today.<br /><br /> The Management';//%s is substituted with the users age
$lang['Birthday_greeting_prev'] = 'We would like to give you a belated congratulatons for becoming %s years old on the %s.<br /><br /> The Management';//%s is substituted with the users age, and birthday
$lang['Greeting_Messaging'] = 'Congratulations';
$lang['Birthday_today'] = 'Users with a birthday today:';
$lang['Birthday_week'] = 'Users with a birthday within the next %d days:';
$lang['Nobirthday_week'] = 'No users are having a birthday in the upcoming %d days'; // %d is substitude with the number of days
$lang['Nobirthday_today'] = 'No users have a birthday today';
$lang['Year'] = 'Year';
$lang['Month'] = 'Month';
$lang['Day'] = 'Day';
wohl einfach ins Deutsche übersetzen oder geht das schief???

Danke!

MfG Florian

Verfasst: 03.11.2003 19:02
von Henne
Ich weiß nicht, ob du das kannst :)

Sonst guckst du hier: http://mods.db9.dk/viewtopic.php?t=156