David,
>My goal is to create a report, that lists the Top 100 most expensive
BookNames, for every CategoryId in this table.
I think you can map the example under "Within-group quotas (Top N per
group)" at http://www.artfulsoftware.com/infotree/queries.php to your
requirement.
PB
David Per
Hi MySQL Users-
I have a query problem I have been working on for quite some time and I am
really at a loss to find a native function(s) to handle my task.
I have this table:
CREATE TABLE BookCategoryMetrics (
BookName VARCHAR(255),
CategoryId VARCHAR(128),
RatingSum DOUBLE,
Cost DOUBLE,
PRIMARY
EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
Sent: Friday, September 03, 2004 6:48 PM
Subject: Re: HELP ME WITH THIS
> Hi,
> The only way I can think of is to join the table to itself. It should look
> something like this:
>
> select unix_timestamp(concat(s.date,'
rji [mailto:[EMAIL PROTECTED]
> > Sent: Friday, September 03, 2004 5:06 AM
> > To: [EMAIL PROTECTED]
> > Subject: HELP ME WITH THIS
> >
> > Hello can anyone help me with this,
> >
> > i have a table, like the one below,
> > i
* Karma Dorji
> i have a table, like the one below,
> i need to find the time difference between the Start and Stop from a
> particular CallingStationId to particular CalledStationId.
>
> ++--++---+
> --
> +---+
nal Message-
> From: Karma Dorji [mailto:[EMAIL PROTECTED]
> Sent: Friday, September 03, 2004 5:06 AM
> To: [EMAIL PROTECTED]
> Subject: HELP ME WITH THIS
>
> Hello can anyone help me with this,
>
> i have a table, like the one below,
> i need to find the time differe
Hello can anyone help me with this,
i have a table, like the one below,
i need to find the time difference between the Start and Stop from a
particular CallingStationId to particular CalledStationId
Hi
I have 2 tables and I want a faster report like this.
| counter | user company name |
+-++
| 65537 | company1 |
| 65535 | company2 |
| 64563 | company3 |
| 52676 | company4 |
User table description
| Field
Hi!
I have 2 tables
Table 'candidate_index' holds information like
firstName,lastName,phoneNumber etc.
Example
id | firstName | lastName | phoneNumber
1 | Peter | Engström | 3236363
2 | John| Smith | 3773737
...
Table 'candidate_skills' holds the skills a candidate have
Exa
Hi.
IMHO, the self-join is the correct solution for the given table layout
(which looks reasonable to me).
Greetings,
Benjamin.
On Wed 2002-07-17 at 09:26:48 +0530, [EMAIL PROTECTED] wrote:
[...]
> > Select companyid, count(*) from CompanyServices where
> > serviceid = 1 or serviceid
ny) 1 services the three
service type.
Let me know if this is not what you wanted.
Bye and Good Luck.
> So how do we handle these situations
>
> -Arul
> - Original Message -
> From: "Francisco Reinaldo" <[EMAIL PROTECTED]>
> To: "Arul" <[E
"MySQL" <[EMAIL PROTECTED]>
Cc: "Arul" <[EMAIL PROTECTED]>
Sent: Wednesday, July 17, 2002 3:27 AM
Subject: Re: help me with this query
> Hi,
>
> Select companyid, count(*) from CompanyServices where
> serviceid = 1 or serviceid = 2 or serviceid = 3
>
Hi,
Select companyid, count(*) from CompanyServices where
serviceid = 1 or serviceid = 2 or serviceid = 3
group by companyid
having count(*) = 3
Bye and Good Luck.
--- Arul <[EMAIL PROTECTED]> wrote:
> Hi All
>
> The Table design goes like this
>
> Company Table
>
> companyid
Hi All
The Table design goes like this
Company Table
companyid integer
name varchar(100)
Services Table
ServiceIDinteger
Servicevarchar(50)
Company Services
---
companyidinteger - References Company(companyid
Hi All
The Table design goes like this
Company Table
companyid integer
name varchar(100)
Services Table
ServiceIDinteger
Servicevarchar(50)
Company Services
---
companyidinteger - References Company(companyid
select u.userid , c.companyname from
users u left join company c on (u.companyid = c.companyid)
where u.status = 'ACT'
Best regards,
Mikhail.
- Original Message -
From: "Arul" <[EMAIL PROTECTED]>
To: "MySQL" <[EMAIL PROTECTED]>
Sent: Tuesday, J
Hi All
I have two tables User and Company
User tables has userid , companyid and status
Company table has companyid and companyname
It's not necessary that all users should have company.There can be users
without company also and for them the companyid is null.
so i need all the users whose st
David Ayliffe writes:
> pc-xx-xx-190-118-du:~/mysql++-1.7.9/examples # g++ -I../sqlplusint/
> -I/usr/local/include/mysql -c simple1.cc -L../lib
> ../lib/libmysqlclient.a -lsqlplus
> g++: ../lib/libmysqlclient.a: linker input file unused since linking not
> done
> g++: -lsqlplus: linker input file
pc-xx-xx-190-118-du:~/mysql++-1.7.9/examples # g++ -I../sqlplusint/
-I/usr/local/include/mysql -c simple1.cc -L../lib
../lib/libmysqlclient.a -lsqlplus
g++: ../lib/libmysqlclient.a: linker input file unused since linking not
done
g++: -lsqlplus: linker input file unused since linking not done
Thi
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