-Original Message-
From: Mihail Manolov [mailto:mihail.mano...@liquidation.com]
Sent: March 19, 2012 12:44 PM
To: Steven Staples
Cc: mysql@lists.mysql.com
Subject: Re: Group_Concat help...
Try this
SELECT `user_id`, `login_ip`,
COUNT(`id`) AS 'connections'
FROM `mysql_test
; 2012/03/19 12:06 -0400, Steven Staples
SELECT `user_id`, GROUP_CONCAT(DISTINCT `login_ip`) AS 'login_ips',
COUNT(`id`) AS 'connections'
FROM `mysql_test`
WHERE `login_datetime` BETWEEN '2012-03-19 00:00:00' AND '2012-03-19
23:59:59'
GROUP BY `user_id`
HAVING COUNT(`id`) 2
ORDER BY COUNT(`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 d...@pointmade.net 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
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
Check if this is in the [mysqldump] section of your my.cnf file(s). Of
course it might not be a valid option in mysqldump...I haven't checked...
On 13 Jun 2011 17:00, Brown, Charles cbr...@bmi.com wrote:
Hello All ~ I attempted to do a mysqldump when I got this message -- see
below:
mysqldump:
Subject: Re: need help with -- unknown variable
Check if this is in the [mysqldump] section of your my.cnf file(s). Of course
it might not be a valid option in mysqldump...I haven't checked...
On 13 Jun 2011 17:00, Brown, Charles cbr...@bmi.commailto:cbr...@bmi.com
wrote:
Hello All ~ I attempted
I could be wrong but I think your problem is the unique index.
--
João Cândido de Souza Neto
robert rottermann rob...@redcor.ch escreveu na mensagem
news:4dd967a8.5040...@redcor.ch...
Hi there,
I would like to create a table that optionally links to an other table.
The field company_id in
2011/03/15 17:51 -0500, LAMP
Let's say there is a table orders (simplified, of course)
CREATE TABLE orders (
item_id int,
org_id int,
) ENGINE=MyISAM
Need to select all (distinct) org_id they have item_id 34, 36, 58 and
63. All of them, not only some of them.
Result is org_id=2607 and
Hi!
I think that the query that you have proposed is the best possible for the
problem.
However, if there are duplicates in the orders table, then
HAVING COUNT(item_id) = 4
should be replaced with
HAVING COUNT(DISTINCT item_id) = 4
(I assume that you meant item_id and not org_id in the
On Mar 17, 2011, at 3:01 PM, Geert-Jan Brits wrote:
Indeed, I don't thing there is.
Just be sure that each record has an unique combination of org_id
and item_id, otherwise you might end up with an org_id that, for
example, references 4 times item_id 34 in 4 different records, but
no
2011/03/18 08:49 -0500, LAMP
Is here anybody from mysql development team, to suggest to build IN
ALL function?
There is a problem here: the basic operation is on the record, each record by
each record, all by itself. The solution to your problem entails acting on more
distinct records until
Yes, that was my question. Though, since English is not my first
language, let me try to post it again:
There is a list of all orgs and items org bough, from table called
orders
item_idorg_id
342607
342607
341520
362607
361520
368934
38
What I need is a list of orgs they bought all of items 34, 36, 58,
63. every of them.
Some solutions under What else did buyers of X buy at
http://www.artfulsoftware.com/infotree/queries.php.
PB
---
On 3/17/2011 12:00 PM, LAMP wrote:
Yes, that was my question. Though, since English is not
First I was thinking there is function IN ALL or something like that,
since there are functions IN and EXISTS. And I would be able to make a
query something like this
select distinct org_id, item_id
from orders
where item_id in all (34, 36, 58, 63)
order by org_id asc
But, there
On Mar 15, 2011, at 6:18 PM, Rhino wrote:
All you should need is this:
select distinct org_id
from orders
where item_id in (34, 36, 58, 63)
I'm assuming that the DISTINCT operator is available in the version
of MySQL that you are using. I don't currently have any version of
MySQL
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
-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
I want to extract all records from this table in which a user has
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 ?
On Wed, Mar 2, 2011 at 2:49 PM, Jerry Schwartz je...@gii.co.jp wrote:
-Original Message-
From: Tompkins Neil
...@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 ?
On Wed, Mar 2, 2011 at 2:49 PM, Jerry
Hi Travis,
That query kind of gives me the desired result. However, if is showing
me 1, 18, 11, 23, 3, 2010-11-14 17:18:17 record and not 2, 11, 10, 3, 6,
2010-12-20 22:17:13, which is when they changed teams. Any thoughts ?
Cheers
Neil
On Thu, Feb 3, 2011 at 10:32 PM, Travis Ard
Something like this might help you find all of the times where your user_id
switched to a different team_id:
select team_id, user_id, min(last_changed)
from
(select home_team_id as team_id, home_user_id as user_id, last_changed
from data
union all
select away_team_id as team_id, away_user_id as
Hello,
Thank for your help
I just try mysqldump with --quick or --opt option ... to avoid
out of memory problem but
-- dump fails with --max_allowed_packet=2048M and --quick :
r...@pcjahia01:/# /usr/bin/mysqldump -A --max_allowed_packet=2048M
--quick --default-character-set=UTF8 -u
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`
Tompkins,
Which field stores the result of matches.
regards
anandkl
On Mon, Sep 6, 2010 at 4:45 PM, Tompkins Neil
neil.tompk...@googlemail.comwrote:
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
These two fields
home_goals and away_goals
Cheers
Neil
On Mon, Sep 6, 2010 at 12:58 PM, Ananda Kumar anan...@gmail.com 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,
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
neil.tompk...@googlemail.comwrote:
These two fields
home_goals and away_goals
Cheers
Neil
On Mon, Sep 6, 2010 at 12:58 PM,
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,
Force all traffic to Node A
Break/Stop replication
backup Node using mysqldump
Restore backup file on Node B
Resume replication
Install Maatkit
Should do the trick...
From: ext Norman Khine [nor...@khine.net]
Sent: 30 August 2010 15:50
To:
Something like this might work:
insert into domains
select a.accountid, reverse(a.domainid), a.mailname
from domains a
left outer join domains b on b.domainid = reverse(a.domainid) and
b.accountid = a.accountid and b.mailname = a.mailname
where b.domainid is null;
-Travis
On Sun, Jul 25, 2010 at 8:49 PM, Nguyen Manh Cuong
cuong.m...@vienthongso.com wrote:
Hi Mark,
Please test this query:
select test1.*, (select name from test2 where test2.id=test1.`v_id` limit
1) as name_1,
(select name from test2 where test2.id=test1.`h_id` limit 1) as name_2
from test1;
Hi Mark,
Please test this query:
select test1.*, (select name from test2 where test2.id=test1.`v_id` limit 1) as
name_1,
(select name from test2 where test2.id=test1.`h_id` limit 1) as name_2
from test1;
- test1 table:
col1v_idh_id
America 1 2
- test2 table:
id name
2
On 07/25/2010 09:29 PM, Mark Phillips wrote:
I have been away from sql for awhile, and can't seem to figure out how to
write a simple query for two tables.
Table 1 has many columns, two of which are hID and vID. Table 2 has two
columns, ID and name. The hID and vID in table 1 correspond to the
[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
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
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
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 give the
Thank you very much for all the insightful replies. I think I can get it to
work with a join.
Joerg Bruehe joerg.bru...@sun.com wrote:
Hi!
Jay Blanchard wrote:
[snip]
I have a table similar to this:
-
|transactions |
|ID |DATE
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
CC: jblanch
Hi Richard,
This is a LEFT JOIN, but with only one table you may be wondering what to join.
Typically you'll have to create a new table that contains all the dates in the
range you want to check. Then you left join your transaction table, and all
the rows from your dates table that don't
[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
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
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
It may only be returning 51 rows but its having to read significantly more.
Get rid of the derived table join if possible. Something like
SELECT TAP.ID http://tap.id/, M.UID, TAP.FirstName, TAP.MI, TAP.LastName,
TAP.State,
TAP.SchoolName, TAP.StateApproved, TAP.DiplomatApproved, C.ChapterType,
Can you please show us the indexes on both the tables.
regards
anandkl
On Tue, Mar 16, 2010 at 11:47 PM, Jesse j...@msdlg.com wrote:
I have the following query:
SELECT TAP.ID http://tap.id/, M.UID, TAP.FirstName, TAP.MI,
TAP.LastName, TAP.State,
TAP.SchoolName, TAP.StateApproved,
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
What error are you getting?
===
John Daisley
MySQL 5.0 Certified Database Administrator (CMDBA)
MySQL 5.0 Certified Developer
Cognos BI Developer
Telephone: +44(0)1283 537111
Mobile: +44(0)7812 451238
Email: john.dais...@butterflysystems.co.uk
===
Sent via HP
Please paste the error and don't give the spave between -p and password.
On Tue, Feb 2, 2010 at 12:05 PM, muralikrishna g
muralikrishn...@gmail.comwrote:
hi to all..
i was in need to take backup of database..
i am using sql server version:5.0.27-coomunity-nt
i have tried by using
mysqldump
hi to all..
i was in need to take backup of database..
i am using sql server version:5.0.27-coomunity-nt
i have tried by using
mysqldump -u name -p password database_name backup.sql;
but i am getting error.. i am using windows xp system..
please help me
mysql mysqldump -u root -p dbadmin murali
@lists.mysql.com
Subject: Re: hi help to take backup-mysql-windows-xp
hi to all..
i was in need to take backup of database..
i am using sql server version:5.0.27-coomunity-nt
i have tried by using
mysqldump -u name -p password database_name backup.sql;
but i am getting error.. i am using windows xp
Message-
From: sureshkumar...@gmail.com [mailto:sureshkumar...@gmail.com]
Sent: Tuesday, February 02, 2010 12:22 PM
To: muralikrishna g
Cc: mysql@lists.mysql.com
Subject: Re: hi help to take backup-mysql-windows-xp
Hi Murali,
You have to execute it outside mysql prompt i.e command prompt.
Thanks
where
b.lieferant is null
-Ursprüngliche Nachricht-
Von: Carlos Proal [mailto:carlos.pr...@gmail.com]
Bereitgestellt: Dienstag, 19. Jänner 2010 20:15
Bereitgestellt in: gmane.comp.db.mysql.general
Unterhaltung: Quick help with Insert
Betreff: Re: Quick help with Insert
Hi
Hey
Not really quick ; - But nobody knows an answer?
THX
-Ursprüngliche Nachricht-
Von: Intell! Soft [mailto:intellis...@fachoptiker.net]
Bereitgestellt: Donnerstag, 14. Jänner 2010 17:40
Bereitgestellt in: gmane.comp.db.mysql.general
Unterhaltung: Quick help with Insert
Betreff:
Hi !!
You need a left join and then an insert.
Please read: http://dev.mysql.com/doc/refman/5.1/en/join.html or google
for tutorials on left join
And tell me if you have further questions
Carlos
On 1/19/2010 1:00 PM, Intell! Soft wrote:
Hey
Not really quick ; - But nobody knows an
...@gmail.com]
Bereitgestellt: Dienstag, 19. Jänner 2010 20:15
Bereitgestellt in: gmane.comp.db.mysql.general
Unterhaltung: Quick help with Insert
Betreff: Re: Quick help with Insert
Hi !!
You need a left join and then an insert.
Please read: http://dev.mysql.com/doc/refman/5.1/en/join.html or google
Depending on whether you just need to count or the transaction numbers, one of
the following three should get you where you want/need to be:
To identify the count for comp_id = 675:
select count(distinct trans_no) from trans where comp_id = 675 and result = 'o';
To identify the transactions:
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| 629|
SELECT count(distinct trans_no) from SEARCHES WHERE comp_id=675 and
result='o';
- Original Message -
From: Richard Reina rich...@rushlogistics.com
To: mysql@lists.mysql.com
Cc: rich...@rushlogistics.com
Sent: Sunday, December 13, 2009 11:36 AM
Subject: Query help
I was wondering if
Is your table innodb? If so i think the 'STRICT_TRANS_TABLES' sql_mode 'may' be
causing the problem.
Try inserting the value as
'2008-03-09 02:56:34.737'
Do you get the same error?
What mysql version is your server?
Regards
John Daisley
Mobile +44(0)7812 451238
Email
I had already tried that, actually. Produces the same error. I
should have mentioned that as well, sorry!
The version is 5.1.34
thanks for the help, btw!
Martin
On Fri, Sep 4, 2009 at 11:14 AM, John
Daisleyj...@butterflysystems.co.uk wrote:
Is your table innodb? If so i think the
Mysql doesn't store sub-second values.
try
2008-03-09 02:56:34
Instead of
2008-03-09 02:56:34.737
Regards,
Gavin Towey
-Original Message-
From: Proemial [mailto:proem...@gmail.com]
Sent: Friday, September 04, 2009 8:37 AM
To: John Daisley
Cc: mysql@lists.mysql.com
Subject: Re: Fwd: Help
AndrewJames schrieb:
Hey,
i have a table called users which has my users in it, each have a uid
field.
I also have a stories table which has stories in it each with a sid field
for each story but also a uid field so i know which user the story
belongs
to.
i want to write a query that
AndrewJames schrieb:
Hey,
i have a table called users which has my users in it, each have a uid
field.
I also have a stories table which has stories in it each with a sid field
for each story but also a uid field so i know which user the story
belongs
to.
i want to write a query that
AndrewJames schrieb:
Hey,
i have a table called users which has my users in it, each have a uid
field.
I also have a stories table which has stories in it each with a sid field
for each story but also a uid field so i know which user the story
belongs
to.
i want to write a query that
I actually get the feeling you are not connecting as root.
Try mysql -uroot -p test instead of just mysql test
Have a nice day,
- Martijn
On Wed, Aug 26, 2009 at 03:02, Joemysql@bluepolka.net wrote:
OK, thanks, that got me in. But upon inspection, the user.host
values do not look fouled
Step # 1 : Stop mysql service
# /etc/init.d/mysql stop
Step # 2: Start to MySQL server w/o password:
# mysqld_safe --skip-grant-tables
Step # 3: Connect to mysql server using mysql client:
# mysql -u root
Step # 4: Setup new MySQL root user password
mysql use mysql;
mysql update user set
Hey Joe,
stop the server, start it with --skip-grant-tables, change the root
entry in mysql.user to your liking, and then restart the server
without --skip-grant-tables.
viola!
Walter
On Wed, Aug 26, 2009 at 02:12, Joemysql@bluepolka.net wrote:
We have an inaccessible MySQL v5.0.45 DB
You have to reset the permissions.
http://dev.mysql.com/doc/refman/5.0/en/resetting-permissions.html
Carlos
On 8/25/2009 7:12 PM, Joe wrote:
We have an inaccessible MySQL v5.0.45 DB (w/Innodb) we really
need some help regaining access to. While attempting to
adjust/add remote user access,
OK, thanks, that got me in. But upon inspection, the user.host
values do not look fouled up as I thought they were (it appears
the bogus update may have aborted). But my access problem
remains
If I start with --skip-grant-tables, 'show databases' shows all
DBs. But without that flag, I
On Wed, Aug 26, 2009 at 02:12, Joemysql@bluepolka.net
wrote:
We have an inaccessible MySQL v5.0.45 DB (w/Innodb) we
really need some help regaining access to. While attempting
to adjust/add remote user access, we accidentally did the
following:
use mysql;
update user set
When you are in without the flag , issue the following:
Select current_user();
It should return root.
Then do this:
Grant all privileges on *.* 'root'@'%' identified by 'letmein'
It should work If you did not mess too much with grant tables.
Claudio
Il giorno 26 ago, 2009 4:36 m., Todd Lyons
Hi,
abdulazeez alugo wrote:
Date: Sun, 19 Apr 2009 23:19:56 +0100
From: andy-li...@networkmail.eu
To: defati...@hotmail.com
CC: mysql@lists.mysql.com
Subject: Re: Need help with mysql prob
Hi Alugo,
Hi Andy,
Thanks for your prompt response. However, since tbl1_id has
Date: Mon, 20 Apr 2009 08:06:05 +0100
From: andy-li...@networkmail.eu
To: defati...@hotmail.com
CC: mysql@lists.mysql.com
Subject: Re: Need help with mysql prob
Hi,
abdulazeez alugo wrote:
Date: Sun, 19 Apr 2009 23:19:56 +0100
From: andy-li...@networkmail.eu
To: defati
Hi Alugo,
abdulazeez alugo wrote:
Hi Andy,
Thank you very much you have been really very helpful. All those
mistakes you pointed at in the script about the apostrophe and others,
are simple mistakes I just made in the rush of typing the message; and
yes $conn is the result from
Date: Mon, 20 Apr 2009 13:03:14 +0100
From: andy-li...@networkmail.eu
To: defati...@hotmail.com
CC: mysql@lists.mysql.com
Subject: Re: Need help with mysql prob
Hi Alugo,
abdulazeez alugo wrote:
Hi Andy,
Thank you very much you have been really very helpful. All those
mistakes
Hi,
Now I have successfully created a relationship between the two tables but how
do I make sure the value of tbl1_id in tbl1 is equal to the value of tbl1_id in
tbl2???
MySQL handles this for you. Simply INSERT the value into tbl1, then
INSERT the value of tbl1_id in tbl2. Then try
Date: Sun, 19 Apr 2009 22:50:20 +0100
From: andy-li...@networkmail.eu
To: defati...@hotmail.com
CC: mysql@lists.mysql.com
Subject: Re: Need help with mysql prob
Hi,
Now I have successfully created a relationship between the two tables but
how do I make sure the value of tbl1_id
Hi Alugo,
Hi Andy,
Thanks for your prompt response. However, since tbl1_id has an auto_increment value in tbl1, mysql is actually generating the values for it automatically.
Ah, I see your point. I'm guessing by your code you're using PHP? If
so call mysql_insert_id() after you've
Hi Abdul-
you may want to check this thread out regarding the same issue...
http://www.webmasterworld.com/php/3565843.htm
On Apr 19, 2009, at 3:23 PM, abdulazeez alugo wrote:
Hi guys,
I'm having a really terrible problem with my mysql coding and I'ld
appreciate any help I can get on
Date: Sun, 19 Apr 2009 23:19:56 +0100
From: andy-li...@networkmail.eu
To: defati...@hotmail.com
CC: mysql@lists.mysql.com
Subject: Re: Need help with mysql prob
Hi Alugo,
Hi Andy,
Thanks for your prompt response. However, since tbl1_id has an
auto_increment value in tbl1
; mysql@lists.mysql.com
Cc: wi...@lists.mysql.com; mysql-h...@lists.mysql.com
Subject: RE: Please help me.
Velentin,
http://dev.mysql.com/doc/refman/5.1/en/innodb-foreign-key-constraints.html
Note the section for the droping of foreign keys used the contraint name,
not the key name. Try
Velentin,
http://dev.mysql.com/doc/refman/5.1/en/innodb-foreign-key-constraints.html
Note the section for the droping of foreign keys used the contraint name, not
the key name. Try this and see if it solves the first problem (of removing the
constraint). Then you should be able to drop the
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: Query Help
Alright to reply to myself I can return the information but have been unable
to return the last row... instead it always returns the first row. Tried
max, tried min, tried converting the datetime to a timestamp with the same
results...
mysql SELECT da_userinfo.UserName, da_userinfo.Name,
Ben Wiechman b...@meltel.com 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
: mysql@lists.mysql.com
Subject: RE: Query Help
Ben Wiechman b...@meltel.com 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
On Dec 17, 2008, at 2:56 PM, Lamp Lists wrote:
I hate when somebody put in Subject line something like I just did
but after 15 minutes to try to be specific just with one short
sentence - I gave up. So, you can hate me - I understand (though,
help with my problem too) :-)
I have let
Hi Afan
Why not prefix your field names with the table name?
select
p.first_name AS person_first_name,
p.last_name AS person_last_name,
p.status AS person_status,
p.date_registered AS person_date_registered,
o.org_id AS organization_org_id,
o.org_name AS organization_org_name,
-Original Message-
From: Lamp Lists [mailto:lamp.li...@yahoo.com]
Sent: Wednesday, December 17, 2008 2:57 PM
To: mysql@lists.mysql.com
Subject: need help with query...
...snip...
I have let say 3 tables people, organization, addresses. and they are
linked to each other with column
-Original Message-
From: Andy Shellam [mailto:andy-li...@networkmail.eu]
Sent: Wednesday, December 17, 2008 3:29 PM
To: Lamp Lists
Cc: mysql@lists.mysql.com
Subject: Re: need help with query...
Hi Afan
Why not prefix your field names with the table name?
select
p.first_name
From: Andy Shellam andy-li...@networkmail.eu
To: Lamp Lists lamp.li...@yahoo.com
Cc: mysql@lists.mysql.com
Sent: Wednesday, December 17, 2008 2:29:08 PM
Subject: Re: need help with query...
Hi Afan
Why not prefix your field names with the table name?
select
Jerry Schwartz wrote:
-Original Message-
From: Andy Shellam [mailto:andy-li...@networkmail.eu]
Sent: Wednesday, December 17, 2008 3:29 PM
To: Lamp Lists
Cc: mysql@lists.mysql.com
Subject: Re: need help with query...
Hi Afan
Why not prefix your field names with the table name?
select
Hi,
Hi Andy,
the reason I can't use this because fields (columns) in select
statement (p.first_name, p.last_name,...) are actually dynamically
created. In my project different client will select different fields
to be shown. 99% will select first_name, and last_name, but some don't
care
From: Andy Shellam andy-li...@networkmail.eu
To: Lamp Lists lamp.li...@yahoo.com
Cc: mysql@lists.mysql.com
Sent: Wednesday, December 17, 2008 2:48:31 PM
Subject: Re: need help with query...
Hi,
Hi Andy,
the reason I can't use this because fields (columns
This seems to do it:
SELECT phone_work FROM leads WHERE phone_work REGEXP '[(]{1}([0-9]){3}[)]{1}[
]?([^0-1]){1}([0-9]){2}[ ]?[-]?[ ]?([0-9]){4}'
- Original Message
From: Paul Nowosielski [EMAIL PROTECTED]
To: mysql@lists.mysql.com
Sent: Wednesday, December 3, 2008 2:39:54 PM
On Wednesday 03 December 2008 08:39:54 Paul Nowosielski wrote:
Hi,
Please, can anyone lend a hand in helping pullout
phone numbers from the DB that only match
the format (nnn) nnn- ?
([0-9]{3}) [0-9]{3}-[0-9]{4}
I think
HTH
W
SELECT phone_work FROM leads WHERE phone_work REGEXP
what rate of pay are you offering?
On Mon, Nov 24, 2008 at 5:13 AM, Tanveer Bhurani
[EMAIL PROTECTED]wrote:
Dear All,
I am looking for help as I want to make a website like orkut.
can u plz help me in designing the Data Table and Queries
--
Thanks Regards,
Tanveer Bhurani
The Bend on
Shaun Adams schrieb:
When I perform a dump in mysql5 to mysql 4 DB, I get the error (below).
Does anyone know how I can resolve this?
QUERY (windows server from the cmd prompt)
mysqldump --lock-tables --user=root [SOURCE DB] | mysql --user=[USERNAME]
--password=[PASSWORD]
Hi Saul,
I need to use C++ and I'm not writing a web application.
Thanks anyway.
Kandy
I have done queries to the database in PHP with variables like month but
easily can select from a range of time and data to produce the same
results, the output goes directly to the web so if that is what
Hi Kandy,
this could be the query you are looking for. It should return record
with the closest timestamp to your required time:
(SELECT TIMEDIFF('20080815091907', timestamp_column) AS diff, t.* FROM
table1 t
WHERE timestamp_column = '20080815091907'
ORDER BY timestamp_column DESC LIMIT 1
)
Kandy Wong wrote:
Hi Saul,
I need to use C++ and I'm not writing a web application.
Thanks anyway.
you can do something like:
select min(abs(timediff(targettime,timestamp))) from table where
condition ;
if you use the libmysql you can get the result as strings back (the method i
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