On 10/22/2015 11:48 AM, Don Wieland wrote:
On Oct 20, 2015, at 1:24 PM, shawn l.green wrote:
Which release of MySQL are you using?
Version 5.5.45-cll
How many rows do you get if you remove the GROUP_CONCAT operator? We don't need
to see the results. (sometimes it is a good idea to look
> On Oct 20, 2015, at 1:24 PM, shawn l.green wrote:
>
> Which release of MySQL are you using?
Version 5.5.45-cll
> How many rows do you get if you remove the GROUP_CONCAT operator? We don't
> need to see the results. (sometimes it is a good idea to look at the raw,
> unprocessed results)
>
On 10/20/2015 1:54 PM, Don Wieland wrote:
Hi all,
Trying to get a query working:
SELECT
ht.*,
CONCAT(o.first_name, " ", o.last_name) AS orphan,
GROUP_CONCAT(DISTINCT hti.rec_code ORDER BY hti.rec_code ASC SEPARATOR ", ") AS
alloc
FROM hiv_transactions ht
LEFT JOIN tk_orphans o ON ht.orphan
On 2015-10-20 12:54 PM, Don Wieland wrote:
Hi all,
Trying to get a query working:
SELECT
ht.*,
CONCAT(o.first_name, " ", o.last_name) AS orphan,
GROUP_CONCAT(DISTINCT hti.rec_code ORDER BY hti.rec_code ASC SEPARATOR ", ") AS
alloc
FROM hiv_transactions ht
LEFT JOIN tk_orphans o ON ht.orphan_
Hi all,
Trying to get a query working:
SELECT
ht.*,
CONCAT(o.first_name, " ", o.last_name) AS orphan,
GROUP_CONCAT(DISTINCT hti.rec_code ORDER BY hti.rec_code ASC SEPARATOR ", ") AS
alloc
FROM hiv_transactions ht
LEFT JOIN tk_orphans o ON ht.orphan_id = o.orphan_id
LEFT JOIN hiv_trans_it
I'm using MySQL 5.0.92-log.
I'm trying to do a pivot-sort-of-thing. I've tried a few things from the
O'Reilly "SQL Cookbook," but I seem to be having a mental block.
I have a table of farm harvests. Each harvest has a date, quantity, and foreign
keys into product and harvester tables:
-
Hello All,
Happy Friday! I know how to do the following query:
>select count(*) from sales where WEEK(sale_date)=15 AND
YEAR(sale_date)=2013;
But can someone tell me I can do a query that will give me:
the count(*) for each week of 2013 so that I end up with:
WEEK | COUNT
1 | 22
2
Perfect! Thank you Larry et all.
Have a great weekend.
2013/4/19 Larry Martell
> On Fri, Apr 19, 2013 at 8:24 AM, Richard Reina
> wrote:
> > Hello All,
> >
> > Happy Friday! I know how to do the following query:
> >
> >>select count(*) from sales where WEEK(sale_date)=15 AND
> > YEAR(sale_dat
On Fri, Apr 19, 2013 at 8:24 AM, Richard Reina wrote:
> Hello All,
>
> Happy Friday! I know how to do the following query:
>
>>select count(*) from sales where WEEK(sale_date)=15 AND
> YEAR(sale_date)=2013;
>
> But can someone tell me I can do a query that will give me:
>
> the count(*) for each w
Hello Richard,
>the count(*) for each week of 2013 so that I end up with:
http://dev.mysql.com/doc/refman/5.1/en/group-by-functions.html
Ilya.
--
MySQL General Mailing List
For list archives: http://lists.mysql.com/mysql
To unsubscribe:http://lists.mysql.com/mysql
of course, "Group By"
bill
--
MySQL General Mailing List
For list archives: http://lists.mysql.com/mysql
To unsubscribe:http://lists.mysql.com/mysql
On 3/31/2013 7:32 AM, william drescher wrote:
I have a table, schedule:
CREATE TABLE `schedule` (
`schedule_id` mediumint(9) NOT NULL AUTO_INCREMENT,
`provider` varchar(15) NOT NULL,
`apptTime` datetime NOT NULL,
`location` varchar(10) NOT NULL,
`duration` smallint(5) unsigned NOT
I have a table, schedule:
CREATE TABLE `schedule` (
`schedule_id` mediumint(9) NOT NULL AUTO_INCREMENT,
`provider` varchar(15) NOT NULL,
`apptTime` datetime NOT NULL,
`location` varchar(10) NOT NULL,
`duration` smallint(5) unsigned NOT NULL,
`standing_script` mediumint(9) DEFAULT NULL,
> -Original Message-
> From: Rik Wasmus [mailto:r...@grib.nl]
> Sent: Monday, October 01, 2012 10:44 AM
> To: mysql@lists.mysql.com
> Subject: Re: NOT_REGEXP Query Help
>
> > I'm having an issue using regular expressions, and I was hoping
> > someon
> I'm having an issue using regular expressions, and I was hoping someone
> can see the syntax error that I'm missing (Monday morning, not enough
> coffee, etc). Running the following query returns the error:
>
> SELECT `a.custid`,`a.login` FROM ol.customers a WHERE `a.login` NOT_REGEXP
> '^ano
Hi all,
I'm having an issue using regular expressions, and I was hoping someone
can see the syntax error that I'm missing (Monday morning, not enough
coffee, etc). Running the following query returns the error:
SELECT `a.custid`,`a.login` FROM ol.customers a WHERE `a.login` NOT_REGEXP
'^anonym
9 AM
> To: 'Richard Reina'; mysql@lists.mysql.com
> Subject: RE: query help
>
> I think this will get you there:
>
> SELECT LEAD, COUNT(*) FROM ORDERS GROUP BY LEAD;
>
> It'll give you something more like:
>
> | LEAD | COUNT(*) |
> | F | 44
Reina [mailto:gatorre...@gmail.com]
Sent: Thursday, September 13, 2012 9:51 AM
To: mysql@lists.mysql.com
Subject: query help
I have a table like this:
|ORDERS|
|ID| DATE | QNT | LEAD |
|342 | 8-12-12 | 32 | F|
|345 | 8-15-12 | 12 | S|
|349 | 8-16-12 | 9 | R|
I have a table like this:
|ORDERS|
|ID| DATE | QNT | LEAD |
|342 | 8-12-12 | 32 | F|
|345 | 8-15-12 | 12 | S|
|349 | 8-16-12 | 9 | R|
I am looking for a way to query it with counts by the LEAD column in
order to tell what the number of each type lead is,
2012/07/31 15:02 -0700, Haluk Karamete
So, in a case like this
shop.orders.32442
shop.orders.82000
shop.orders.34442
It would be the record whose source_recid is shop.orders.82000. Why? Cause
82000 happens to be the largest integer.
Now, if they are always 5-digit-long integer
Hi,
Use LIMIT 1 to limit the number of output to single record.
Regards,
Vikas Shukla
On Wed, Aug 1, 2012 at 3:56 AM, Paul Halliday wrote:
> On Tue, Jul 31, 2012 at 7:02 PM, Haluk Karamete
> wrote:
> > I've got a text field called source_recid. It stores half string half
> > number like str
L
> Subject: Re: query help
>
> On Tue, Jul 31, 2012 at 7:02 PM, Haluk Karamete
> wrote:
> > I've got a text field called source_recid. It stores half string half
> > number like strings in it.
> >
> > Example
> >
> > shop.orders.32442
> >
On Tue, Jul 31, 2012 at 7:02 PM, Haluk Karamete wrote:
> I've got a text field called source_recid. It stores half string half
> number like strings in it.
>
> Example
>
> shop.orders.32442
>
> the syntax is DATABASENAME.TABLENAME.RECID
>
> My goal is to scan this col and find out the biggest RECI
I've got a text field called source_recid. It stores half string half
number like strings in it.
Example
shop.orders.32442
the syntax is DATABASENAME.TABLENAME.RECID
My goal is to scan this col and find out the biggest RECID ( the
integer) in it.
So, in a case like this
shop.orders.32442
shop
On 6/22/2012 12:18 AM, Anupam Karmarkar wrote:
Thanks Rick for your reply,
Here i am asking about logic to perpare query or whole query itself.
A set-based approach to doing the basic task is to convert your set of
start/stop times into duration values. The timediff() function mentioned
alr
Thanks Rick for your reply,
Here i am asking about logic to perpare query or whole query itself.
--Anupam
From: Rick James
To: Anupam Karmarkar ; "mysql@lists.mysql.com"
Sent: Wednesday, 20 June 2012 10:52 PM
Subject: RE: Need Query H
http://dev.mysql.com/doc/refman/5.1/en/date-and-time-functions.html#function_timediff
and SEC_TO_TIME()/3600
> -Original Message-
> From: Anupam Karmarkar [mailto:sb_akarmar...@yahoo.com]
> Sent: Wednesday, June 20, 2012 2:39 AM
> To: mysql@lists.mysql.com
> Subject:
Hi All,
I need query help for following table struture, where we need to calculate
login duration of that employee for give period.
Example table
EmployeeID LoginTime LogoutTIme
101 2012-05-01 10:00:00 2012-05-01 12:30:00
102 2012
2012/05/28 08:03 -0700, Don Wieland
Any assistance would be appreciated. Thanks!
Maybe something like this:
SELECT usie.client_id, first_name, last_name, COUNT(anie.client_id)
FROM
(SELECT client_id, first_name, last_name, time_start
FROM tl_appt
JOIN tl_rooms USING(room_id)
I have been working with a query but need to add a few more
conditions. I was a to do a query that contains a few more selects but
want to know if there is a more efficient way to get results I need:
This query is to find NEW or FORMER CLIENT within a moving window
(Date Range) of time.
I
I would work from the inside out. What you're doing is grouping scenes
by DVD and throwing away the ones that have no scenes. If you start
with DVDs and do a subquery for each row, you'll process DVDs without
scenes and then filter them out. If you start with a subquery that's
grouped by DVD ID, al
elly [mailto:my...@wastedtimes.net]
Sent: Saturday, May 19, 2012 3:34 PM
To: mysql@lists.mysql.com
Subject: Re: SQL query help. Retrieve all DVDs that have at least one scene
of a certain encoding format
Hi.
On Friday 18 May 2012 18:21:07 Daevid Vincent wrote:
> Actually, I may have figured
Hi.
On Friday 18 May 2012 18:21:07 Daevid Vincent wrote:
> Actually, I may have figured it out. Is there a better way to do this?
I don't see why you need the dvds table when the dvd_id is in the scene table:
SELECT a.dvd_id
FROM scenes_list a, moviefiles b
WHERE a.scene_id = b.scene_id
AND
> -Original Message-
> Sent: Friday, May 18, 2012 5:34 PM
>
> I have a table of DVDs, another of scenes and a last one of encoding
> formats/files...
>
> I want to find in one query all the dvd_id that have > 0 scene_id that's
> encoded in format_id = 13.
> In other words all DVDs that ar
I have a table of DVDs, another of scenes and a last one of encoding
formats/files...
I want to find in one query all the dvd_id that have > 0 scene_id that's
encoded in format_id = 13.
In other words all DVDs that are format_id = 13 despite not having a direct
link.
CREATE TABLE `dvds` (
`dvd_
On 2012-05-17 9:37 AM, Don Wieland wrote:
Hi folks,
I am trying to compile a query that does statistics on appointments
based on specific criteria. Here is my starting query:
SELECT
u.user_id,
c.client_id,
c.first_name,
c.last_name,
a.time_start AS stime,
FROM_UNIXTIME
| 40 |
| AZ| Phoenix |1428509 | 51 |
| CA| Los Angeles |3877129 |447 |
...
> -Original Message-
> From: Don Wieland [mailto:d...@pointmade.net]
> Sent: Thursday, May 17, 2012 7:37 AM
> To: mysql@lists.mysql.com
> Subject: Query help,,,
>
Hi folks,
I am trying to compile a query that does statistics on appointments
based on specific criteria. Here is my starting query:
SELECT
u.user_id,
c.client_id,
c.first_name,
c.last_name,
a.time_start AS stime,
FROM_UNIXTIME(a.time_start,'%Y-%m-%d') AS formatted
F
Hi Rafael,
You can try using correlated subquery instead of outer join. This can be slow
with big tables though:
SELECT * FROM users WHERE accept_email = 1 and email not in (SELECT email FROM
sent_emails WHERE sent_emails
.email = users.email AND messageID NOT LIKE = ‘XX’)
OR OUTER JOIN as
Dear Friends,
I m new on this list, and I m trying to learn more about mysql.
After perform a lot of searchs in the Internet, I have no answer to my
question and would like to ask your help.
I wanna a perform a query that depends of the result from another (query)
table inside the same
2012/03/01 19:56 -0800, Don Wieland
I do not get the same results. Am I missing something? Hopefully
something simple ;-)
O, you are. You do not want GROUP_CONCAT in the subquery. It gives you the
comma-separated string whereto you referred, which, as far as the IN goes, is
o
Appreciate a little guidance here:
Background: I have an invoicing system. Invoices are generated and
(invoice and Invoice Items) and Payments are generated (Payments and
Payment Items). Payment items are amount of the Payment Total
allocated to payoff open invoices. So I may have 3 open in
On 2/29/2012 1:15 PM, Don Wieland wrote:
Little help...
In my mySQL query editor, I am trying to return a value of 0 when
there is no related rows from this query:
(select if(count(ip.payment_amount) IS NOT NULL,
count(ip.payment_amount) , 0) FROM tl_trans_pmt_items ip WHERE
t.transaction_id =
http://dev.mysql.com/doc/refman/5.0/en/control-flow-functions.html#function_ifnull
On Wed, Feb 29, 2012 at 13:15, Don Wieland wrote:
> Little help...
>
> In my mySQL query editor, I am trying to return a value of 0 when there is
> no related rows from this query:
>
> (select if(count(ip.payment_
Little help...
In my mySQL query editor, I am trying to return a value of 0 when
there is no related rows from this query:
(select if(count(ip.payment_amount) IS NOT NULL,
count(ip.payment_amount) , 0) FROM tl_trans_pmt_items ip WHERE
t.transaction_id = ip.inv_id GROUP BY ip.inv_id) as d
2011/10/24 16:31 -0700, Daevid Vincent
WHERE cs.customer_id = 7
GROUP BY customer_id
Well, the latter line is now redundant.
How will you make the '7' into a parameter?
--
MySQL General Mailing List
For list archives: http://l
October 24, 2011 4:06 PM
> To: mysql@lists.mysql.com
> Subject: RE: Within-group aggregate query help please - customers and
latest
> subscription row
>
> A kind (and shy) soul replied to me off list and suggested this solution,
> however,
> this takes 28 seconds (that's f
-
> From: Daevid Vincent [mailto:dae...@daevid.com]
> Sent: Monday, October 24, 2011 1:46 PM
> To: mysql@lists.mysql.com
> Subject: Within-group aggregate query help please - customers and latest
> subscription row
>
> I know this is a common problem, and I've been struggling
I know this is a common problem, and I've been struggling with it for a full
day now but I can't get it.
I also tried a few sites for examples:
http://www.artfulsoftware.com/infotree/queries.php#101
http://forums.devarticles.com/general-sql-development-47/select-max-datetime
-problem-10210.html
The OR conditions require a full table scan everytime this is called.
You didn't say how many rows you had, nor if there were indexes on
your various phone_xxx fields. If you do, you should get some value
by approaching it as a UNION
select count(id)from (
select id from leads where phone_work
Dear all,
I have a query that takes a rather long time and was wondering if there is
anyway to optimize it.
Normally we removing duplicate records by phone number. This query takes about
a
second and
it really slows down the process when we are importing several 1000 records a
day.
Here is
lto:neil.tompk...@googlemail.com]
>Sent: Wednesday, March 02, 2011 10:12 AM
>To: Jerry Schwartz
>Cc: [MySQL]
>Subject: Re: Query help
>
>Thanks for the response. This is what I was after. Although, I am looking
>to find out the email addresses used to login from the same IP
il.com]
> >Sent: Wednesday, March 02, 2011 6:00 AM
> >To: [MySQL]
> >Subject: Query help
> >
> >Hi
> >
> >I've the following basic table
> >
> >login_id
> >email_address
> >ip_address
> >
> >I want to extract all reco
>-Original Message-
>From: Tompkins Neil [mailto:neil.tompk...@googlemail.com]
>Sent: Wednesday, March 02, 2011 6:00 AM
>To: [MySQL]
>Subject: Query help
>
>Hi
>
>I've the following basic table
>
>login_id
>email_address
>ip_address
>
&g
Hi Neil,
select
login_id,
ip_address
from
basic_table
group by
login_id,ip_address
having
count(login_id,ip_address)>1
this should work
in case you want to see also the list of emails add:
group_concat(email_address,',') as list_of_used_emails
to the select fields.
Claudio
Hi
I've the following basic table
login_id
email_address
ip_address
I want to extract all records from this table in which a user has used the
same IP address but different email address to login ?
Thanks,
Neil
On 10/27/2010 6:55 AM, Nuno Mendes wrote:
I have 3 tables: (1) Companies, (2) locations and (3) employees:
CREATE TABLE `companies` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`name` varchar(75) NOT NULL,
UNIQUE KEY `id` (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8
CREATE TABLE `locations ` (
`id` i
I have 3 tables: (1) Companies, (2) locations and (3) employees:
CREATE TABLE `companies` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`name` varchar(75) NOT NULL,
UNIQUE KEY `id` (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8
CREATE TABLE `locations ` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`
Then you'll probably need to define it with a separate select before using
it. I'm half-guessing here, really, but that sounds like it makes sense :-)
On Mon, Sep 27, 2010 at 11:49 AM, Tompkins Neil <
neil.tompk...@googlemail.com> wrote:
> Hi,
>
> I did try defining it before the IF statement, bu
Hi,
I did try defining it before the IF statement, but still the same ?
Cheers
Neil
On Mon, Sep 27, 2010 at 7:58 AM, Johan De Meersman wrote:
> At a guess, because you use @team in an if statement before you actually
> define it.
>
>
> On Sun, Sep 26, 2010 at 12:35 AM, Tompkins Neil <
> neil.to
At a guess, because you use @team in an if statement before you actually
define it.
On Sun, Sep 26, 2010 at 12:35 AM, Tompkins Neil <
neil.tompk...@googlemail.com> wrote:
> Hi,
>
> I've the following query
>
> SELECT teams_id AS teams_id ,SUM(rating) AS total_team_rating FROM (SELECT
> teams_id
Hi,
I've the following query
SELECT teams_id AS teams_id ,SUM(rating) AS total_team_rating FROM (SELECT
teams_id ,players_id ,rating ,IF(@team <> teams_id, @row := 1, @row := @row
+ 1) AS rank ,@team := teams_id FROM ( SELECT players.teams_id
,players.players_id ,players_master.rating FROM player
I wondered if anyone can help me ? Do you need any further information ?
Cheers
Neil
-- Forwarded message --
From: Tompkins Neil
Date: Thu, Sep 23, 2010 at 9:49 AM
Subject: Query help please
To: "[MySQL]"
Hi all,
I've the following
Hi all,
I've the following query :
SELECT fixtures_results.seasons_id ,
home_teams_id AS teams_id ,
1 AS home ,0 AS away ,
(SELECT SUM(goals) FROM players_appearances WHERE
fixtures_results.fixtures_results_id =
players_appearances.fixtures_results_id AND players_appearances.teams_id =
home_teams
For sure here is some sample data
home_teams_id,away_teams_id,home_goals,away_goals,home_users_id,away_users_id
100,200,2,1,5,6
200,100,1,1,6,5
Here is two rows of data for the same fixture both home and away
Let me know if you need any more info.
Cheers
Neil
On Mon, Sep 6, 2010 at 1:08 PM,
Also, can u please lets u know the value's in this table.
Just one row, an example would do.
regards
anandkl
On Mon, Sep 6, 2010 at 5:35 PM, Tompkins Neil
wrote:
> These two fields
>
> home_goals and away_goals
>
> Cheers
> Neil
>
>
> On Mon, Sep 6, 2010 at 12:58 PM, Ananda Kumar wrote:
>
>>
These two fields
home_goals and away_goals
Cheers
Neil
On Mon, Sep 6, 2010 at 12:58 PM, Ananda Kumar wrote:
> Tompkins,
> Which field stores the result of matches.
>
> regards
> anandkl
>
> On Mon, Sep 6, 2010 at 4:45 PM, Tompkins Neil <
> neil.tompk...@googlemail.com> wrote:
>
>> Hi,
>>
>> I
Tompkins,
Which field stores the result of matches.
regards
anandkl
On Mon, Sep 6, 2010 at 4:45 PM, Tompkins Neil
wrote:
> Hi,
>
> I've the following fields within a table :
>
> fixtures_results_id
> home_teams_id
> away_teams_id
> home_goals
> away_goals
> home_users_id
> away_users_id
>
> From
Hi,
I've the following fields within a table :
fixtures_results_id
home_teams_id
away_teams_id
home_goals
away_goals
home_users_id
away_users_id
>From this I want to extract the following type of information if
the home_users_id or away_users_id = 1 :
total number of games games
number of games
--
From: "Steven Buehler"
Sent: Friday, August 20, 2010 3:30 PM
To:
Subject: query help
I am hoping that I can do this with one query, I have a table, "Domains"
with 3 columns
accountID, domainID, mailname
what I am trying to do is find
I am hoping that I can do this with one query, I have a table, "Domains"
with 3 columns
accountID, domainID, mailname
what I am trying to do is find all accountID's for "domainID" of 12345 and
see if a second row with "domainID" of 54321 exists for that
"accountID,mailname". If it doesn't exis
Table 1
Product_id | Product_Name
1| Product A
2| Product B
3| Product C
Table 2
Category_id | Category_Name
1 | Admin
2 | Marketing
3 | Support
4 | IT
Table 3
Product_id
[snip]
For the life of me I cannot remember how to make a query like this and
what
it is called.
I know it is fairly basic though.
Table 1
Product_id Product_Name
Table 2
Category_id, Category_name
Table 3
Product_id, Category_id
Each product can have one or more categories.
So I want a result
I believe you're describing a crosstab query. This should help you put it
together:
http://rpbouman.blogspot.com/2005/10/creating-crosstabs-in-mysql.html
---Michael
On Friday, July 09, 2010 07:37:41 pm Phillip Baker wrote:
> Hello All,
>
>
> For the life of me I cannot remember how to make a
Hello All,
For the life of me I cannot remember how to make a query like this and what
it is called.
I know it is fairly basic though.
Table 1
Product_id Product_Name
Table 2
Category_id, Category_name
Table 3
Product_id, Category_id
Each product can have one or more categories.
So I want a
Informationen und entfaltet keine rechtliche Bindungswirkung.
Aufgrund der leichten Manipulierbarkeit von E-Mails koennen wir keine Haftung
fuer den Inhalt uebernehmen.
> From: rich...@rushlogistics.com
> To: joerg.bru...@sun.com; mysql@lists.mysql.com
> Subject: Re: query help
>
Thank you very much for all the insightful replies. I think I can get it to
work with a join.
Joerg Bruehe wrote:
>
> Hi!
>
>
> Jay Blanchard wrote:
> > [snip]
> > I have a table similar to this:
> >
> > -
> > |transactions |
> > |ID |DATE |
Hi!
Jay Blanchard wrote:
> [snip]
> I have a table similar to this:
>
> -
> |transactions |
> |ID |DATE |EMPLOYEE|
> |234 |2010-01-05| 345|
> |328 |2010-04-05| 344|
> |239 |2010-01-10| 344|
>
> Is there a way to query such a table to gi
[snip]
I have a table similar to this:
-
|transactions |
|ID |DATE |EMPLOYEE|
|234 |2010-01-05| 345|
|328 |2010-04-05| 344|
|239 |2010-01-10| 344|
Is there a way to query such a table to give the days of the year that
employee 344 did not
don't have a match is your answer.
Regards,
Gavin Towey
-Original Message-
From: Richard Reina [mailto:rich...@rushlogistics.com]
Sent: Tuesday, June 15, 2010 11:30 AM
To: mysql@lists.mysql.com
Subject: query help
I have a table similar to this:
-
|tr
I have a table similar to this:
-
|transactions |
|ID |DATE |EMPLOYEE|
|234 |2010-01-05| 345|
|328 |2010-04-05| 344|
|239 |2010-01-10| 344|
Is there a way to query such a table to give the days of the year that employee
344 did not have
Not tested, but I think it can help you or at least give you an ideia on how
to do it.
select
EndDateTime + INTERVAL 1 SECOND as startLazy,
(select StartDateTime - INTERVAL 1 SECOND from table t2 where
t2.StartDateTime > t1.EndDateTime limit 1) as endLazy
from
table t1
where
(select Star
Hmm. You seem to have overlap, too. I suspect this would be easiest to do in
code - the data you're looking for doesn't exist in the data you have, only
the opposite of that data does.
You could try populating a table with a full day, using the resolution you
need (1 minute resolution means 1440 r
Hi All,
I have a query I need to run but can't think how to get this working so I am
hoping someone can advise.
I have a table which logs start and end times of Scheduled jobs. It includes
for simplicity a `DayID`, `StartDateTime` and `EndDateTime` column. Both
`StartDateTime` and `EndDateTime` a
SELECT ID, check_no, amount FROM payables UNION SELECT ID, check_no, amount
FROM paychecks;
Regards,
Gavin Towey
-Original Message-
From: Richard Reina [mailto:rich...@rushlogistics.com]
Sent: Tuesday, February 09, 2010 9:23 AM
To: mysql@lists.mysql.com
Subject: query help
I am trying
I am trying to write a query that merges 2 columns from different tables and
show them as one column of data. Something like the following.
payables
ID |check_no| amount|
3 |3478| 67.00 |
4 |3489| 98.00 |
8 |3476| 56.00 |
paychecks
ID |check_no| amount
23 |3469|498.00 |
SELECT count(distinct trans_no) from SEARCHES WHERE comp_id=675 and
result='o';
- Original Message -
From: "Richard Reina"
To:
Cc:
Sent: Sunday, December 13, 2009 11:36 AM
Subject: Query help
I was wondering if someone could lend a hand with the following q
On December 13, 2009 01:36:41 pm Richard Reina wrote:
> I was wondering if someone could lend a hand with the following query. I
> have table.
>
> SEARCHES
>
> |ID |trans_no|comp_id|result
>
> 13 | 455| 675| o
> 15 | 302| 675| o
> 16 | 455| 675| o
> 12 | 225|
gt; From: Richard Reina [mailto:rich...@rushlogistics.com]
> Sent: Sunday, December 13, 2009 12:37 PM
> To: mysql@lists.mysql.com
> Cc: rich...@rushlogistics.com
> Subject: Query help
>
> I was wondering if someone could lend a hand with the following query.
> I have table.
>
> SEARC
I was wondering if someone could lend a hand with the following query. I have
table.
SEARCHES
|ID |trans_no|comp_id|result
13 | 455| 675| o
15 | 302| 675| o
16 | 455| 675| o
12 | 225| 629| y
SELECT count(*) FROM SEARCHES WHERE comp_id=675 AND result='o' G
Hello,
Currently, I have four tables (Items, UpdatePrice, UpdateStatus and
UpdateRelease). All the Update tables are linked to Items.ItemID via
Update(Price|Status|Release)ItemKey. Personally, I don't feel that
this is the best database design I could have, but I can't seem to
come up with one tha
: mysql@lists.mysql.com
> From: c...@hosting4days.com
> Subject: Basic SQL Query Help Needed
> Date: Tue, 25 Aug 2009 16:21:45 -0700
>
> I have a basic invoice table with related line items table
>
> Goal :I'd like to get ALL the related line items - for ALL the
> &#x
I have a basic invoice table with related line items table
Goal :I'd like to get ALL the related line items - for ALL the
'open' invoices...
-- this should get a list of open (unpaid) invoices
$query_invoice = "SELECT DISTINCT ID from invoices where status =
'open'"
-
--
: mysql@lists.mysql.com
Subject: RE: Query Help
"Ben Wiechman" wrote on 02/10/2009 01:30:14 PM:
> Thanks for the input! That is close to what I need, however not exactly.
It
> will give me the last time a user logged into the host in question but I
> want to prune users who hav
"Ben Wiechman" wrote on 02/10/2009 01:30:14 PM:
> Thanks for the input! That is close to what I need, however not exactly.
It
> will give me the last time a user logged into the host in question but I
> want to prune users who have since logged into a different host.
Basically
> find out how man
00 sec)
-Original Message-
From: Ben Wiechman [mailto:b...@meltel.com]
Sent: Tuesday, February 10, 2009 11:32 AM
To: mysql@lists.mysql.com
Subject: Query Help
I keep hacking at this but haven't been able to get it right yet.
I have two tables
Userinfo contains a login, User's Na
he host in question the information is
returned. This produces too many results as some of those users have since
migrated to a different access point.
-Original Message-
From: Andrew Wallo [mailto:theme...@microneil.com]
Sent: Tuesday, February 10, 2009 12:05 PM
To: Ben Wiechman
Subject: Re:
I keep hacking at this but haven't been able to get it right yet.
I have two tables
Userinfo contains a login, User's Name, Group Name
Log contains login, host, datetime of last login
What I need to do is return user information (userinfo.name/groupname) of
users that have logged into
Dear All,
I am interested in performing a sub query that removes duplicate records from
a temporary table prior to pushing the data to the main table.
I am not sure if it is possible and thought I would ask prior to the endeavor.
I currently use php to perform this operation but is really bogs
Anders,
>I also want to find out the user's position relative to others
depending on the result.
For a given pUserID, something like this?
SELECT userid,result,rank
FROM (
SELECT o1.userid,o1.result,COUNT(o2.result) AS rank
FROM object o1
JOIN object o2 ON o1.result < o2.result OR (o1.resu
1 - 100 of 560 matches
Mail list logo
