Re: Query Help...

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.

Re: Query Help...

> 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,

Re: Query Help...

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

Re: Query Help...

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

RE: Query Help

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

Re: Query Help

On Fri, Apr 19, 2013 at 8:24 AM, Richard Reina gatorre...@gmail.com 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(*)

Re: Query Help

Perfect! Thank you Larry et all. Have a great weekend. 2013/4/19 Larry Martell larry.mart...@gmail.com On Fri, Apr 19, 2013 at 8:24 AM, Richard Reina gatorre...@gmail.com wrote: Hello All, Happy Friday! I know how to do the following query: select count(*) from sales where

Re: Query help -

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

Re: Query help - Solved

of course, Group By bill -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/mysql

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 | | S | 122 | | R | 32 | -Original Message- From: Richard Reina

RE: query help

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 | | S | 122 | | R | 32

Re: query help

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 integers, try ending

RE: query help

: MySQL Subject: Re: query help On Tue, Jul 31, 2012 at 7:02 PM, Haluk Karamete halukkaram...@gmail.com 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

Re: query help

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 paul.halli...@gmail.comwrote: On Tue, Jul 31, 2012 at 7:02 PM, Haluk Karamete halukkaram...@gmail.com wrote: I've got a text field called source_recid. It

Re: query help

On Tue, Jul 31, 2012 at 7:02 PM, Haluk Karamete halukkaram...@gmail.com 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

Re: Query help...

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) JOIN tl_clients

RE: Query help,,,

For (1), ponder: Group-by trick example: Find the most populous city in each state: SELECT state, city, population, COUNT(*) AS num_cities FROM ( SELECT state, city, population FROM us ORDER BY state, population DESC ) p GROUP BY state ORDER BY

Re: Query help,,,

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,

Re: Query help...

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

Re: Query help...

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

Re: Query help

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

RE: Query help

-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

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 Schwartz je...@gii.co.jp wrote: -Original Message- From: Tompkins Neil

RE: Query help

...@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

Re: Query Help

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`

Re: Query help

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

Re: Query help

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,

Re: Query help

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,

Re: Query help

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,

Re: query help

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

RE: Query Help

[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

Re: Query Help

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

Re: Query Help

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

Re: query help

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

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 joerg.bru...@sun.com wrote: Hi! Jay Blanchard wrote: [snip] I have a table similar to this: - |transactions | |ID |DATE

RE: query help

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

RE: query help

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

RE: query help

[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

Re: Query Help!

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

Re: Query Help!

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

RE: query help

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

RE: Query help

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:

Re: Query help

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|

Re: Query help

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

RE: Query Help

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

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,

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 logged into a different host. Basically find

RE: Query Help

: 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

Re: Query help, please..

On Dec 11, 2007 8:38 AM, Anders Norrbring [EMAIL PROTECTED] wrote: I'm looking at a situation I haven't run into before, and I'm a bit puzzled by it. I have this table structure: Table USERS: userid, class Table OBJECT: userid, class, result Now I want to query the database for a certain

Re: Query help, please..

On Dec 11, 2007, at 10:46 AM, Rob Wultsch wrote: On Dec 11, 2007 8:38 AM, Anders Norrbring [EMAIL PROTECTED] wrote: I'm looking at a situation I haven't run into before, and I'm a bit puzzled by it. I have this table structure: Table USERS: userid, class Table OBJECT: userid, class, result

Re: Query help, please..

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

Re: query help

I only want to view the product per product name which has the lowest price. SELECT product_name,MIN(price) FROM tbl GROUP BY product_name; PB uYe wrote: Hi, I have a table with product_name, price field. The product name may be exist in more than one record. I only want to view the product

Re: Query Help plss

Thank u so much. It woked now! - Original Message - From: Visolve DB Team [EMAIL PROTECTED] To: Visolve DB Team [EMAIL PROTECTED]; Renish [EMAIL PROTECTED]; mysql@lists.mysql.com Sent: Tuesday, October 10, 2006 1:42 PM Subject: Re: Query Help plss Hi, Hope this link will you. http

Re: Query Help plss

Renish koshy wrote: Hi all, I have installed MySql 4.1 in my system. when I tried to run Mysql , I always gets this error. Could not start MySql service on a local computer Error 1067: the process terminated unexpectedly. Stop sending this message! We get the idea! The more you send it,

Re: Query Help plss

Xp - Original Message - From: Visolve DB Team [EMAIL PROTECTED] To: Renish koshy [EMAIL PROTECTED] Sent: Tuesday, October 10, 2006 11:42 AM Subject: Re: Query Help plss Hi, On which platform? Thanks ViSolve DB Team. - Original Message - From: Renish koshy [EMAIL PROTECTED

Re: Query Help plss

@lists.mysql.com Sent: Tuesday, October 10, 2006 12:35 PM Subject: Fw: Query Help plss - Original Message - From: Renish To: mysql@lists.mysql.com ; Visolve DB Team Sent: Tuesday, October 10, 2006 12:34 PM Subject: Re: Query Help plss b) And what happend when you open

Re: Query Help plss

:03 AM Subject: Re: Query Help plss Hi The .err file shows that the service is stopped Normally and no error was found. sure the service is properly shutdown. Try, a) net stop mysql mysqld-nt remove mysqld-nt install net start mysql b) Also run 'services.msc' and double click on the mysql

RE: Query Help for Loosely Couple Properties

generally simplify this? R. -Original Message- From: Robert DiFalco [mailto:[EMAIL PROTECTED] Sent: Friday, August 04, 2006 4:12 PM To: [EMAIL PROTECTED] Cc: mysql@lists.mysql.com Subject: RE: Query Help for Loosely Couple Properties I think this only works if I know that Orange

RE: Query Help for Loosely Couple Properties

, 2006 9:37 AM To: Robert DiFalco Cc: mysql@lists.mysql.com Subject: Re: Query Help for Loosely Couple Properties On Tue, 2006-08-01 at 16:23 -0700, Robert DiFalco wrote: The question is, how do I query this? Say I want all records from table T whose COLOR property value is ORANGE. The only

RE: Query Help for Loosely Couple Properties

They are user defined properties. -Original Message- From: Jay Pipes [mailto:[EMAIL PROTECTED] Sent: Tuesday, August 01, 2006 8:11 PM To: Robert DiFalco Cc: mysql@lists.mysql.com Subject: Re: Query Help for Loosely Couple Properties On Tue, 2006-08-01 at 16:23 -0700, Robert DiFalco

Re: Query Help for Loosely Couple Properties

On Tue, 2006-08-01 at 16:23 -0700, Robert DiFalco wrote: The question is, how do I query this? Say I want all records from table T whose COLOR property value is ORANGE. The only thing I can come up with (and I'm no SQL expert and this looks wrong to me) is the following: SELECT * FROM T

Re: Query Help for Loosely Couple Properties

On Tue, 2006-08-01 at 16:23 -0700, Robert DiFalco wrote: I have a table that contains properties that can be associated with any table whose primary key is a LONG. Lets say that there is just one kind of property. The table looks something like this: TABLE StringVal REF_ID BIGINT

Re: query help-multiple joins

Hi, I guess you should be able to do something like: SELECT t1.term_id, t1.name, r.type_id, t2.term_id, t2.name FROM term t1 LEFT JOIN relationTerm r ON r.term_id1 = t1.term_id LEFT JOIN term t2 ON r.term_id2 = t2.term_id /Johan mel list_php skrev: Hi! I'm stuck with a join query 2

Re: query help-multiple joins

Perfect! I tried aliasing the field names but didn't think about the table, and was just stuck looking at that query without any idea... Thanks a lot for your help. melanie From: Johan Höök [EMAIL PROTECTED] To: mel list_php [EMAIL PROTECTED] CC: mysql@lists.mysql.com Subject: Re: query

Re: Query Help

SELECT id, count(*) AS cnt FROM `table_name` GROUP BY id ORDER BY cnt DESC [ LIMIT 1 ] -- Gabriel PREDA Senior Web Developer

Re: Query Help

Thanks that got it. Robert Gehrig Webmaster at www.gdbarri.com e-mail: [EMAIL PROTECTED] -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]

Re: Query help with count and join on same table I think

2wsxdr5 wrote: I have a table of people. Some of the people in this table are related. You can find out who is related by comparing a familyID number. I have a query to select a certain group of people from the table and I want to also select anyone who is related to them, even though

Re: Query help

OKAN ARI wrote: I have 3 tables Table 1: user(id, name, surname) Table 2: crime(id, detail) Table 3: user_crime(id, user_id, crime_id) Table 1 1, OKAN, ARI Table 2 1, detail 1 2, Detail 2 Table 3 1, 1, 1 1, 1, 2 So user 1 takes 2 crime from crime table... I want to receive info with 1

Re: query help?

I am a novice when it come to queries such as this and was hoping someone could help me write a query that tells me how many records have the same ID and vendor number. |ID| vendor_no| date| |2354 | 578 | 2005-12-23| |2355 | 334 |

RE: query help?

Richard, If you mean with _both_ the same id _and_ vendor id, try this: Select id, vendor_id, count(*) from tablename group by id, vendor_id; If you just want separate counts for id and vendor_id, use: Select id, count(*) from tablename group by id; Select vendor_id, count(*) from tablename

Re: query help?

I's so sorry. You are very correct. The sample data is bad. ID should be unique. Here it is corrected. |ID| vendor_no| date| |2354 | 578 | 2005-12-23| |2355 | 334 | 2005-12-24| |2356 | 339 | 2005-12-26| |2357 | 339

Re: query help?

If you are looking just for duplicate (ID,vendort_no) combinations, this will find them: SELECT ID, vendor_no, count(1) as dupes FROM table_name_here GROUP BY ID, vendor_no HAVING dupes 1; Shawn Green Database Administrator Unimin Corporation - Spruce Pine Richard Reina [EMAIL PROTECTED]

Re: query help?

Actually I am looking for duplicates (vedor_no, date), but I think I can hopefully adapt the solution you have given me. [EMAIL PROTECTED] wrote: If you are looking just for duplicate (ID,vendort_no) combinations, this will find them: SELECT ID, vendor_no, count(1) as dupes FROM

Re: Query Help

Ian Barnes wrote: Hi, This is my current query which works in mysql 4, but not in 5. Its from mambo, but im trying to modify it because they don't officially support mysql5 yet. The original query: SELECT c.*, g.name AS groupname, cc.name, u.name AS editor, f.content_id AS frontpage, s.title

RE: Query Help

, cc.title, c.ordering LIMIT 0,10; Thanks, Ian -Original Message- From: gerald_clark [mailto:[EMAIL PROTECTED] Sent: 24 January 2006 09:50 PM To: Ian Barnes Cc: mysql@lists.mysql.com Subject: Re: Query Help Ian Barnes wrote: Hi, This is my current query which works in mysql 4

RE: Query Help

PM To: Ian Barnes Cc: mysql@lists.mysql.com Subject: Re: Query Help Ian Barnes wrote: Hi, This is my current query which works in mysql 4, but not in 5. Its from mambo, but im trying to modify it because they don't officially support mysql5 yet. The original query: SELECT c

Re: Query Help

Subject: Re: Query Help Ian Barnes wrote: Hi, This is my current query which works in mysql 4, but not in 5. Its from mambo, but im trying to modify it because they don't officially support mysql5 yet. The original query: SELECT c.*, g.name AS groupname, cc.name, u.name AS editor

Re: query help

[EMAIL PROTECTED] wrote on 10/18/2005 06:28:03 AM: Hello, CREATE TABLE t ( dt datetime ) Please assume this is the table structure and contains thousands of records. And I want to list them only last 12 months (that mean last 1 year) records exactly. For that I tried using the

Re: query help

R, I tried using the following query, but it list sometimes 13 months when the current date is in the middle of the month. SELECT * FROM t WHERE DATE_ADD(dt, INTERVAL 1 YEAR) = NOW(); Try... DATE_ADD( dt, INTERVAL IF(YEAR(NOW())%40 AND YEAR(NOW())%100=0,366,365) DAY ) = NOW()

Re: query help

grKumaran wrote: Hello, CREATE TABLE t ( dt datetime ) Please assume this is the table structure and contains thousands of records. And I want to list them only last 12 months (that mean last 1 year) records exactly. For that I tried using the following query, but it list sometimes 13

Re: query help

[EMAIL PROTECTED] wrote: lets say i have two tables: module - VARCHAR module_name INTEGER module_id module_config INTEGER module_id VARCHAR config_name VARCHAR config_value config item names and values are rather arbitrary and depend on the module. each module can

Re: query help

[EMAIL PROTECTED] wrote on 10/11/2005 01:27:12 PM: lets say i have two tables: module - VARCHAR module_name INTEGER module_id module_config INTEGER module_id VARCHAR config_name VARCHAR config_value config item names and values are rather arbitrary

Re: Query help

Balazs Bagi [EMAIL PROTECTED] wrote on 08/24/2005 05:10:55 AM: Hi there, I'm kind of new to joining two tables. Please bear with me with this simple example. I have three tables. Guests is a list of guests that are coming to the party. The primary key of this table is 'id' and the

Re: Query Help

Jason Chan wrote: Jason Chan wrote: I have a student Table and a SubjectGrade table Create Table Student( StudentID INT NOT NULL, StudentName VARCHAR(30) ) Create Table SubjectGrade( StudentID INT NOT NULL, Subject VARCHAR(30) NOT NULL, Grade CHAR(1) ) let's say have following record in

Re: Query Help

I am using mysql 4.0.25 with no subquery support : ( Jason Chan [EMAIL PROTECTED] ¦b¶l¥ó news:[EMAIL PROTECTED] ¤¤¼¶¼g... I have a student Table and a SubjectGrade table Create Table Student( StudentID INT NOT NULL, StudentName VARCHAR(30) ) Create Table SubjectGrade( StudentID INT NOT NULL,

Re: Query Help

Select s.StudentID, s.StudentName from Student s, SubjectGrade sg where s.StudentID = sg.StudentID and sg.Subject = 'Maths' and sg.Subject = 'Chem' and sg.Grade = 'A' The key here is using the aliases for the tables. The linking of the tables comes in the s.StudentID = sg.StudentID statement.

Re: Query Help

Jason Chan wrote: I have a student Table and a SubjectGrade table Create Table Student( StudentID INT NOT NULL, StudentName VARCHAR(30) ) Create Table SubjectGrade( StudentID INT NOT NULL, Subject VARCHAR(30) NOT NULL, Grade CHAR(1) ) let's say have following record in SubjectGrade 1

Re: Query Help

Select s.StudentID, s.StudentName from Student s, SubjectGrade sg where s.StudentID = sg.StudentID and sg.Subject = 'Maths' and sg.Subject = 'Chem' and sg.Grade = 'A' Take a close look at the WHERE clause: sg.Subject = 'Maths' and sg.Subject = 'Chem' will never return a result - how could the

Re: Query Help

Jason Chan wrote: I have a student Table and a SubjectGrade table Create Table Student( StudentID INT NOT NULL, StudentName VARCHAR(30) ) Create Table SubjectGrade( StudentID INT NOT NULL, Subject VARCHAR(30) NOT NULL, Grade CHAR(1) ) let's say have following record in SubjectGrade 1

Re: Query Help

Jason Chan wrote: I am using mysql 4.0.25 with no subquery support : ( Jason Chan [EMAIL PROTECTED] ¦b¶l¥ó news:[EMAIL PROTECTED] ¤¤¼¶¼g... I have a student Table and a SubjectGrade table Create Table Student( StudentID INT NOT NULL, StudentName VARCHAR(30) ) Create Table SubjectGrade(

Re: Query Help

Alternatively, use an IN list for the subject, then use a HAVING clause to limit the results to students with 2 matching rows, like this: SELECT StudentID, StudentName FROM Student s JOIN SubjectGrade sg ON s.StudentID = sg.StudentID WHERE sg.Subject IN ('Maths', 'Chem') AND

Re: Query Help

Jason Chan wrote: I have a student Table and a SubjectGrade table Create Table Student( StudentID INT NOT NULL, StudentName VARCHAR(30) ) Create Table SubjectGrade( StudentID INT NOT NULL, Subject VARCHAR(30) NOT NULL, Grade CHAR(1) ) let's say have following record in SubjectGrade 1Maths

Re: Query Help

Arno Coetzee wrote: Jason Chan wrote: I have a student Table and a SubjectGrade table snip I want to find out students who have got A in both Maths and Chem How the SQL look like? select s.StudentID , s.StudentName from Student as s , SubjectGrade as sj where s.studentID = sj.studentID

Re: Query Help

I havn't write my schema clearly , (StudentID, Subject) is the key of SubjectGrade Philippe Poelvoorde [EMAIL PROTECTED] ??? news:[EMAIL PROTECTED] ???... Alternatively, use an IN list for the subject, then use a HAVING clause to limit the results to students with 2 matching rows, like this:

Re: Query Help

So i have to write 3 join if I have 3 conditions and so on, right? Scott Noyes [EMAIL PROTECTED] ??? news:[EMAIL PROTECTED] ???... Select s.StudentID, s.StudentName from Student s, SubjectGrade sg where s.StudentID = sg.StudentID and sg.Subject = 'Maths' and sg.Subject = 'Chem' and sg.Grade =

Re: Query Help

SELECT s.StudentID, StudentName FROM Student s JOIN StudentGrade sg1 USING (StudentID) JOIN StudentGrade sg2 USING (StudentID) WHERE sg1.Subject = 'Maths' AND sg1.Grade = 'A' AND sg2.Subject = 'Chem' AND sg2.Grade = 'A' So i have to write 3 join if I have 3 conditions and so on,

Re: Query Help

Jason Chan wrote: I have a student Table and a SubjectGrade table Create Table Student( StudentID INT NOT NULL, StudentName VARCHAR(30) ) Create Table SubjectGrade( StudentID INT NOT NULL, Subject VARCHAR(30) NOT NULL, Grade CHAR(1) ) let's say have following record in SubjectGrade

Re: Query Help

Thanks Scott! Scott Noyes [EMAIL PROTECTED] ??? news:[EMAIL PROTECTED] ???... SELECT s.StudentID, StudentName FROM Student s JOIN StudentGrade sg1 USING (StudentID) JOIN StudentGrade sg2 USING (StudentID) WHERE sg1.Subject = 'Maths' AND sg1.Grade = 'A' AND sg2.Subject = 'Chem' AND

Re: Query HELP!

m i l e s [EMAIL PROTECTED] wrote on 08/08/2005 03:10:21 PM: Hi, IS the following query counting cumulative (see below **) pHITS or is it counting individual counts for each user for a particular day. Im not skilled enough to answer this question myself. My instinct says that its

Re: Query HELP!

Miles, IS the following query counting cumulative (see below **) pHITS or is it counting individual counts for each user for a particular day. Im not skilled enough to answer this question myself. My instinct says that its counting cumulative values and NOT individual counts for each

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