Hello.
You're right. I haven't understood the problem properly.
Terence <[EMAIL PROTECTED]> wrote:
> Wouldnt that only work for the current year? For example I was born on
> 1970-08-25, "select id from members where birthday = now();" wouldnt
> return my birthday if it was today. Or a
Hi!
Thanks for all your answers. I'll go for a flexible solution, with one
column for day, one for month and if i want to compute the age, a year
column. This will also give more flexibility if someone does not want
to give his birthdate with precision (He would be able to only give
the month, and
Pooly wrote:
Hi,
I would like to display a list of members who have their birthday a
given day (today for instance).
For such an application I've used a single integer column to store a
number consisting of the month and day (day as 2 digits!!!) concatenated.
So dates range from 101 to 1231.
Try to avoid running functions against columns, unless you have very
little data, as it doesnt scale well (I'm talking 6 figure tables).
Two options I can suggest are:
1. Write a scheduled job using your MONTH() DAY() to replace the values
into a birthdays table on a daily basis
2. Use 2 sepera
Wouldnt that only work for the current year? For example I was born on
1970-08-25, "select id from members where birthday = now();" wouldnt
return my birthday if it was today. Or am I missing something new in
5.0.11?
Gleb Paharenko wrote:
Hello.
Why don't you want to use just:
select id
Hello.
Why don't you want to use just:
select id from members where birthday = now();
See:
mysql> show create table members\G;
*** 1. row ***
Table: members
Create Table: CREATE TABLE `members` (
`id` int(11) NOT NULL auto_increment,
"Pooly" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
Hi,
I would like to display a list of members who have their birthday a
given day (today for instance).
My idea is to store their birth date in a column, and then query the
table against the column. But the query would be like :
2005/8/24, Cummings, Shawn (GNAPs) <[EMAIL PROTECTED]>:
>
> The 29th wouldn't be an issue because if that is their birthday -- and
> today is 2/29 -- it will show up.
It will only happen once every 4 years... I'll go with the two fields
solution and make a special case for leap years.
thanks for
table bday
user int
mon int
day int
On Wed, 24 Aug 2005, Pooly wrote:
Hi,
I would like to display a list of members who have their birthday a
given day (today for instance).
My idea is to store their birth date in a column, and then query the
table against the column. But the query would be l
Hi,
I would like to display a list of members who have their birthday a
given day (today for instance).
My idea is to store their birth date in a column, and then query the
table against the column. But the query would be like :
select id from members where MONTH(birthday) = MONTH(NOW()) AND
DAY(b
10 matches
Mail list logo
