Michael McFadden <[EMAIL PROTECTED]> wrote on 11/10/2005 08:55:13 AM:
> Hi Jerry.
>
> I'm new to the list, so don't take this as the final
> answer. Wait for a guru to pounce with a better
> solution!
>
> But, here is my idea:
>
> Before we start, note that "order" is a reserved word.
> So we
item_status table can have more than one status. I need to get the latest
status from the table.
Thanks
On 11/10/05, ISC Edwin Cruz <[EMAIL PROTECTED]> wrote:
>
> Try it:
>
> select distinct a.*
> from order a,
> item b,
> item_status c,
> status d
> where a.order_id = b.order_id
> and b.item_i
Try it:
select distinct a.*
from order a,
item b,
item_status c,
status d
where a.order_id = b.order_id
and b.item_id=c.item_id
and c.item_status_id = d.item_status_id
where d.status = 'completed'
It isn´t "tunned" but I think that it works for that you want
-Mensaje original-
De: Jerry
Hi Jerry.
I'm new to the list, so don't take this as the final
answer. Wait for a guru to pounce with a better
solution!
But, here is my idea:
Before we start, note that "order" is a reserved word.
So we must backtick `order` to reference the table in
SQL (or the interpreter will think we're u
0/2000
| 2 | 01/10/2000 |
+--++---+---+---
-+---++
10 rows in set (0.00 sec)
-Original Message-
From: Mike Rains [mailto:[EMAIL PROTECTED]
Sent: Monday, February 21, 2005 9:33 AM
To: mysql@lists.mysql.com
Subject: Re: how to write this query?
SELECT
start_date,
end_date,
DATEDIFF(end_date,
SELECT
start_date,
end_date,
DATEDIFF(end_date, start_date) -
(WEEK(end_date) - WEEK(start_date)) * 2
AS business_days
FROM DateDiffs
ORDER BY start_date;
+-+-+---+
| start_date | end_date| business
It's not precisely correct.
When time difference is less than 7, the time is calcualted wrong
end_time 2005-01-10 17:53:33
end_time 2005-01-04 16:44:57
Result: days 6
Result: bussiness_days 6
On Sat, 19 Feb 2005 09:50:06 -0500, Mike Rains <[EMAIL PROTECTED]> wrote:
> On Sat, 19 Feb 2005
On Sat, 19 Feb 2005 14:01:05 +, Jerry Swanson <[EMAIL PROTECTED]> wrote:
> I have two dates (start_date, end_date). Datediff() function returns
> difference in days.
> I need the difference but not including Satuday and Sunday.
>
> Any ideas?
C:\Program Files\MySQL\MySQL Server 4.1\bin>mysql
s for B.*. Then, the second left join gives you C.* for that
A_ID; it doesn't matter that the B.* part contains nulls.
Bill
> From: sean peters <[EMAIL PROTECTED]>
> To: "Kevin Fries" <[EMAIL PROTECTED]>, <[EMAIL PROTECTED]>
> Subject: Re: How to write th
ssage-
> From: sean peters [mailto:[EMAIL PROTECTED]
> Sent: Wednesday, October 01, 2003 2:23 PM
> To: Kevin Fries; [EMAIL PROTECTED]
> Subject: Re: How to write this query
>
>
> Unfortunately that wont always work either.
>
> For instance, assume that there is an
Unfortunately that wont always work either.
For instance, assume that there is an A record with A_ID = 4
And that there is a C record where A_ID = 4,
but NO B record where A_ID = 4
So, executing the query:
> SELECT A_data, B_data, C_data
> FROM A LEFT JOIN B ON A.A_ID = B.B_ID LEFT JOIN C ON A.A_
You're on the right track with LEFT JOIN. Just continue the thought...
Try:
SELECT A_data, B_data, C_data
FROM A LEFT JOIN B ON A.A_ID = B.B_ID LEFT JOIN C ON A.A_ID = C.C_ID
WHERE A.A_ID = 4;
> -Original Message-
> From: sean peters [mailto:[EMAIL PROTECTED]
> Sent: Wednesday, October
>Ways around inner select statments
>Select ID, Sum(CASE WHEN phone.PHN = NULL THEN 1 ELSE 0 END) as PHNCount
>from person left outer join phone on ID
>where PHNCount = 0 GROUP BY phone.ID;
The alias in the WHERE clause is illegal; it would have to be
SELECT persons.ID,
Sum( CASE WHE
If you have the option to change the table structure, just replace
the date and time columns with a timestamp column.
If you must keep the current structure then the following wil
work, but it will not use indexes in the search:
select ...
where concat(today,' ',heure1)
be
14 matches
Mail list logo
