Not a group by Expression

Similar Questions

• Listagg function gives ORA-00979: not a GROUP BY expression error

  Hello

  I have an environment that supports the Listagg function. Suite works perfectly:

  Select task_code, LISTAGG(ename||) e ('| role |') (',',') WITHIN GROUP (ORDER BY ename) as employees

  team

  where task_code = '01.07.05'

  Task_code group

  However, this is just a test request, the real is a little wider and I can't get the function works, I get ORA-00979: not a GROUP BY expression error.

  Here's my query (the column names are in Dutch, but that shouldn't really be a problem):

  SELECT distinct op_sod.sod_code, op_sod.sod_omschr_lang, op_ood.ood_code, op_ood.ood_omschr_lang, op_activiteit.act_code, op_activiteit.act_omschr_lang, op_taak.taak_code, op_taak.taak_omschr_lang, op_afdeling___master.afd_code, op_taak.belmo, op_taak_j.planning_la,

  LISTAGG (op_ploeg.wn_naam |) ' ('| op_ploeg.rol |') (',',') WITHIN GROUP (ORDER BY op_ploeg.wn_naam) as employees.

  "Write subtaak/mijlpaal toe."

  op_teamplanning.subtaak_mijlpaal as subtaak_of_mijlpaal,

  op_teamplanning.subtaak_omschr | -case when op_teamplanning.subtaak_mijlpaal = 'BELMO Mijlpaal' then ' (BRUNO mijlpaalnr: ' | nvl (op_teamplanning.nummer, 0) |) ')' other ' ' end

  as subtaak_mijlpaal_omschrijving, op_teamplanning.deadline as date limit, op_teamplanning.status as status,

  trim (both

  case when op_teamplanning.vlag1 = 'Geen' and op_teamplanning.vlag2 = 'Geen' and op_teamplanning.vlag3 = 'Geen' and op_teamplanning.vlag4 = 'Geen'

  and op_teamplanning.vlag5 = 'Geen' or op_teamplanning.vlag1 is null and op_teamplanning.vlag2 is null and op_teamplanning.vlag3 is null

  op_teamplanning.vlag5 is null and op_teamplanning.vlag4 is null then 'Geen '.

  Another pad (both case when op_teamplanning.vlag1 = 'Geen' then "else trim (both from op_teamplanning.vlag1) end) |

  trim (both case when op_teamplanning.vlag2 = 'Geen' then "another ',' | trim (both from op_teamplanning.vlag2) end) |

  trim (both case when op_teamplanning.vlag3 = 'Geen' then "another ',' | trim (both from op_teamplanning.vlag3) end) |

  trim (both case when op_teamplanning.vlag4 = 'Geen' then "another ',' | trim (both from op_teamplanning.vlag4) end) |

  trim (both case when op_teamplanning.vlag5 = 'Geen' then "another ',' | trim (both from op_teamplanning.vlag5) end) end) as Beleidsvlaggen,

  op_teamplanning.naam_verantw as quotation, op_teamplanning.subtaak_id as subtaak_id, nvl (op_teamplanning.nummer, 0) as Nummer, nvl (op_teamplanning.regelgevingsagenda, 'Nee') as

  Regelgeving, op_teamplan_subsubtaak.subsubtaak_submijlpaal as subsubtaak_submijlpaal, op_teamplan_subsubtaak.subsubtaak_id as subsubtaak_id, op_teamplan_subsubtaak.subsubtaak_omschr

  like subsubtaak_omschr, op_teamplan_subsubtaak.deadline, sub_deadline, op_teamplan_subsubtaak.status as sub_status, op_teamplan_subsubtaak.naam_verantw as naam_verantw

  Of...

  WHERE THE...

  Group of op_sod.sod_code, op_sod.sod_omschr_lang, op_ood.ood_code, op_ood.ood_omschr_lang, op_activiteit.act_code, op_activiteit.act_omschr_lang, op_taak.taak_code, op_taak.taak_omschr_lang, op_afdeling___master.afd_code, op_taak.belmo, op_taak_j.planning_la

  
  

  Can someone tell me the reason for this error?

  Thanks in advance!

  Kind regards

  NDG

  NDG123 wrote:

  I don't really understand your other question:

  Is below for example, a group of expression?

  op_teamplanning.subtaak_mijlpaal

  After some additional research:

  The group by mistake is not appear anymore when I delete all fields after the listagg field, that might be a clue to the solution.

  It is not difficult to understand. You can have other columns not aggregated in your projection on the side of those listed in the group by expression.

  The reason for ORA-01489 is that the resulting string will become long and exceeds 4000 bytes.

  You will find several solution in this forum.

  For example

  Re: Listagg function excdding 4000 characters

• error: not a GROUP BY expression

  Hello world
  
  Looks like I need some help here... :) don't know what that means, but im getting this message
  00979 00000 - "not a GROUP BY expression"
  * Cause:
  * Action:
  Error on line: column 137: 36
  
  It's the function:
  
  function get_path (materialID in number, materialTypeID number, inTopFolderID number default null, inMaterialFolderID in the number default null, webMode in default number 1 in)
  return varchar2;
  
  
  and im using it on my query
  
  , (select material_util.get_path (m.material_id, 8, 1021695, min (mf.)) (IP_MATERIALFOLDERID), 1).
  material m, materialfolder mf
  where mf.nfx_link = m.MATERIAL_ID
  and t.template_id = m.template_id) as path
  
  It seems that min (mf. IP_MATERIALFOLDERID) is having some problems here... don't know why...
  
  Let me know guys...
  
  Thank you...
  
  J

  Read below for informative post from Frank Kulash on ABC in the group by clause
  {message: id = 9969583}

• SQL - ORA-00979: not a GROUP BY expression

  Hi Expert,
  
  I received the error message: ORA-00979: not a GROUP BY expression at the bottom of the sql statement
  
  Please advise - display total number of point of sale and the total amount of all items. Only display results when the total quantity is greater than 2.
  
  SELECT o.Qty, sum(o.Qty*o.salesprice) AS TOTALDOLLARS, description and i.Itemcode
  OF ORDERLINE o
  INNER JOIN POINT I
  ON o.itemcode = i.itemcode
  GROUP BY i.Itemcode
  After having o.qty > 2
  
  > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
  
  Create table ORDERLINE
  (
  Qty number 4,
  Salesprice Number (6.2),
  OrderNo Number (3).
  ItemCode Varchar2 (6).
  Primary key (OrderNo, Itemcode)
  Foreign Key (Itemcode) refers to POINT.
  Foreign Key (OrderNo) made reference SALESORDER
  
  
  Create table POINT
  (
  ItemCode Varchar2 (6).
  Description Varchar2 (50).
  Retailprice Number (6.2),
  Number of weight (4.1),
  Purchaseprice Number (6.2),
  Category Varchar2 (20).
  Primary Key (Itemcode)
  );

  Hi 845720,

  Based on the sql code, the sum must be grouped itemcodewise, descriptionwise, qtywise.
  This means that you have all these in a group by clause too with sum().

  eg : group by  i.Itemcode, i.Description, o.Qty
  

  Is this wise sum qty is logical? I do not know...

  It will be useful,
  CKLP

  Published by: CKLP April 14, 2011 22:10

• Subquery in the select clause and not "a group by expression" error.

  Hi all
  I hope someone out there can help me out with this one. I have a sql query that performs a subquery in the select case combined with an aggregate function and gives me a "not a group by expression.
  
  I created a simplified example

   SELECT COUNT(*)                     ,
    ai.invoice_num invoice_number      ,
    ai.description invoice_description ,
    (SELECT poh.po_header_id
         FROM po_headers poh ,
        po_distributions pod
        WHERE pod.po_distribution_id = aid.po_distribution_id
      AND pod.po_header_id           = poh.po_header_id
      ) po_number
       FROM ap_invoice_distributions aid ,
      ap_invoices ai
      WHERE ai.invoice_id = 1476932
    AND aid.project_id    = 8608   
    AND aid.task_id       = 462202  
    AND ai.invoice_id     = aid.invoice_id
   GROUP BY ai.invoice_num,
                  ai.description;

  I'm sure he expected me to add the po_number to the group by clause, but this does not work and gives the error "invalid identifier".
  
  Can someone suggest a way for a new concert the query so that I can add po_number to my education by clause.
  
  Kris

  You should try either:

  SELECT COUNT (*),
  AI.invoice_num invoice_number,
  have. Description invoice_description,
  (SELECT poh.po_header_id
  OF po_headers poh,.
  pod po_distributions
  WHERE pod.po_distribution_id = aid.po_distribution_id
  AND pod.po_header_id = poh.po_header_id
  ) po_number
  HELP ap_invoice_distributions.
  ap_invoices AI
  WHERE ai.invoice_id = 1476932
  AND aid.project_id = 8608
  AND aid.task_id = 462202
  AND ai.invoice_id = aid.invoice_id
  Ai.invoice_num GROUP,
  have. Description, aid.po_distribution_id;

  or

  SELECT COUNT (*),
  AI.invoice_num invoice_number,
  have. Description invoice_description,
  MAX ((SELECT poh.po_header_id
  OF po_headers poh,.
  pod po_distributions
  WHERE pod.po_distribution_id = aid.po_distribution_id
  AND pod.po_header_id = poh.po_header_id
  po_number))
  HELP ap_invoice_distributions.
  ap_invoices AI
  WHERE ai.invoice_id = 1476932
  AND aid.project_id = 8608
  AND aid.task_id = 462202
  AND ai.invoice_id = aid.invoice_id
  Ai.invoice_num GROUP,
  have. Description;

• Not a group by expression error

  Hello

  Can you please tell me why the query below is to launch one isn't a group by expression error

  Select count (*),

  TO_CHAR (date, "YYYYMM") table

  To_char (date, 'yyyymm') group;

  Basically I want a count by month.

  Help, please

  Thank you

  Check the status of Tom on this thread, which seems similar to your problem:

  Ask Tom & quot; Group By clause & quot;

• ORA-00979: not a GROUP BY expression

  Hello

  And if you want to include aggregate functions, you must make sure of all the other columns are included in the group by clause...

  SELECT THE CAG. SystemId,
  CAG. BankId,
  CR. ClearingNumber,
  CAG. AgreementId,
  CAs.timestampchanged AS ApprovedLGDDate,
  LGD. ApprovedLGDValue,
  CASE. ModelId,
  CASE. ModelVersion,
  CAG. CaseId,
  CASE. Creator,
  BOX WHEN CAG. SystemId = 'ASK' THEN CAG. SystemidProduct
  ANOTHER NULL
  END as SystemidProd,
  BOX WHEN CAG. SystemId = 'REQUEST' AND CAG.agreementid LIKE ' %: % ' THEN
  To_char (TO_NUMBER (SUBSTR (CAG.agreementid, INSTR (CAG.agreementid, ' :') + 1), "xxxxxxxxxxxxxxx")))
  WHEN CAG. SystemId = 'REQUEST' AND CAG.agreementid NOT LIKE ' %: % ' THEN
  CAG.agreementid
  ANOTHER NULL
  END as AgreementidUNP,
  case. DecisionDate
  OF CAG CustomerAgreement
  JOIN CreditLGD LGD ON (CAG. SystemId = LGD. SystemId
  AND CAG. BankId = LGD. BankId
  AND CAG. AgreementId = LGD. AgreementId
  AND CAG. AgreementSubId = LGD. AgreementSubId
  AND CAG. CaseId = LGD. CaseId
  )
  Credit to JOIN CR ON (CAG. SystemId = CR. SystemId
  AND CAG. BankId = CR. BankId
  AND CAG. AgreementId = CR. AgreementId
  AND CAG. AgreementSubId = CR. AgreementSubId
  AND CAG. CaseId = CR. CaseId
  )
  NO. CASE CASE OF JOIN ON (CAG. BankId = CASE. BankId
  AND CAG. CaseId = CASE. CaseId
  AND THE AC. State = 'approved '.
  )
  JOIN CU customer ON (CU.bankid = CAS.bankid
  AND CU.latestapprovedcaseid = CAS.caseid
  AND CU.customerid = CR.customerid
  )
  WHERE CAG. SYSTEMID IN ("ACC", "APL", "GOF", "GRK", "HPD", "KRF", "LIM", "PRO", "SPI", "SPS", "ASK")
  CAG group. SystemId
  CAG. BankId
  CR. ClearingNumber
  CAG. AgreementId
  cas.timestampchanged
  LGD. ApprovedLGDValue
  CASE. ModelId
  CASE. ModelVersion
  CAG. CaseId
  CASE. Creator
  CASE WHEN CAG. SystemId = 'ASK' THEN CAG. SystemidProduct
  ANOTHER NULL
  END
  CASE WHEN CAG. SystemId = 'REQUEST' AND CAG.agreementid LIKE ' %: % ' THEN
  To_char (TO_NUMBER (SUBSTR (CAG.agreementid, INSTR (CAG.agreementid, ' :') + 1), "xxxxxxxxxxxxxxx")))
  WHEN CAG. SystemId = 'REQUEST' AND CAG.agreementid NOT LIKE ' %: % ' THEN
  CAG.agreementid
  ANOTHER NULL
  END
  cas.decisiondate

• ORA-00979: not a GROUP BY expression error

  Hi all
  
  I added a GROUP BY clause that includes all expressions of individual column listed.
  I don't know why this error occurs. ?
  How could I solve this problem?
  
  
  

  SELECT site_to_data_source_lane_v.site_id,  
          site_to_data_source_lane_v.site_lane_id,
          traffic_record.record_id,
          traffic_record.datetime,
          NVL(traffic_class.v_class, 0),
          NVL(SUM(traffic_status.status_code), 0) vehicle_status_count,  
          NVL(COUNT(traffic_error.error_code), 0)  vehicle_error_count ,
          NVL(site_to_data_source_lane_v.site_lane_id, 0) ,
          NVL(traffic_status.status_code, 0),
          NVL(traffic_error.error_code, 0)
      FROM traffic_record
      LEFT OUTER JOIN traffic_class
        ON traffic_record.record_id = traffic_class.record_id    
      LEFT OUTER JOIN traffic_error
        ON traffic_record.record_id =traffic_error.record_id              
      LEFT OUTER JOIN traffic_status
        ON traffic_record.record_id =traffic_status.record_id         
      JOIN site_to_data_source_lane_v
        ON traffic_record.data_source_id = site_to_data_source_lane_v.data_source_id
       AND traffic_record.lane = site_to_data_source_lane_v.data_source_lane_id
      JOIN edr_rpt_tmp_report_lanes
        ON site_to_data_source_lane_v.site_lane_id = edr_rpt_tmp_report_lanes.site_lane_id
      JOIN edr_rpt_tmp_report_classes
        ON NVL(traffic_class.v_class, 0) = edr_rpt_tmp_report_classes.class_id
      JOIN edr_rpt_tmp_inclusion_table
        ON TRUNC(traffic_record.datetime) = TRUNC(edr_rpt_tmp_inclusion_table.date_time)
     WHERE site_to_data_source_lane_v.site_id = 32
       AND traffic_record.datetime >= to_timestamp('07/01/2008 00:00:00', 'mm/dd/yyyy hh24:mi:ss')
       AND traffic_record.datetime <  to_timestamp('07/31/2008 00:00:00', 'mm/dd/yyyy hh24:mi:ss') 
       GROUP BY site_id,                        
                            site_lane_id,
                            record_id,
                            date_time,
                            v_class,                
                            status_code,
                error_code;
                

  You choose: -.

  site_to_data_source_lane_v.Site_ID,
  site_to_data_source_lane_v.site_lane_id,
  traffic_record.record_id,
  traffic_record. DateTime,
  NVL (traffic_class.v_class, 0),
  NVL (Sum (traffic_status.status_code), 0) vehicle_status_count,
  NVL (Count (traffic_error. (Error_code), 0) vehicle_error_count,
  NVL (site_to_data_source_lane_v.site_lane_id, 0),
  NVL (traffic_status.status_code, 0),
  NVL (traffic_error.error_code, 0)

  But you are a grouping by

  Site_ID,
  site_lane_id,
  record_id,
  Date_Time,
  v_class,
  status_code,
  Error_code

  Quite a few differences...

• SQL error: ORA-00979: not a GROUP BY

  Hi team,

  I am trying to execute the query

  SELECT DISTINCT comp.company_name |' -' || COMP.contract_number AS 'company_name '.

  , UPPER (bh.first_name) AS 'budget_holder '.

  , INITCAP (sup.first_name) |' -' || SUP. Surname AS 'supervisor '.

  , INITCAP (emp.status) AS "Department".

  , INITCAP (emp.first_name |) » '|| EMP. Surname) AS "employee."

  r.employee_id

  , emp.paaf_number AS 'paaf_num '.

  , emp.employee_payroll_number AS 'payroll_num '.

  , emp.discipline AS 'discipline '.

  , NVL (SUM (tg.week_day1), 0) AS "day 1".

  , NVL (SUM (tg.week_day2), 0) AS "day 2".

  , NVL (SUM (tg.week_day3), 0) AS day '3 '.

  , NVL (SUM (tg.week_day4), 0) AS "day4.

  , NVL (SUM (tg.week_day5), 0) AS "day5".

  , NVL (SUM (tg.week_day6), 0) AS "day6.

  , NVL (SUM (tg.week_day7), 0) AS "day 7".

  r.valid_from

  r.valid_to

  Tw_roster_data r

  JOIN tw_employee emp

  ON emp.employee_id = r.employee_id

  JOIN tg tw_timesheet_stage

  ON tg.employee_id = r.employee_id

  AND tg.week_commencing = r.valid_from

  JOIN the tw_contracting_company comp

  ON comp.contracting_company_id = r.contracting_company_id

  JOIN tw_supervisor sup

  ON sup.supervisor_id = tg.supervisor_id

  JOIN tw_budget_holder bh

  ON bh.budget_holder_id = sup.budget_holder_id

  WHERE r.contracting_company_id = 101

  AND r.VALID_FROM = January 31, 2015.

  AND r.VALID_TO = February 6, 2015"

  Group of r.employee_id, comp.company_name, bh.first_name,

  EMP.paaf_number, emp.employee_payroll_number, emp.discipline,

  r.valid_from, r.valid_to;

  But still the error below

  Error in the command line: 1 column: 17

  Error report-

  SQL error: ORA-00979: not a GROUP BY expression

  00979 00000 - "not a GROUP BY expression"

  Can help.

  Hello

  When to use analytical function oracle other columns used based on the analysis, must be included in the GROUP BY Clause

  SELECT expression1, expression2, ... expression_n,
    aggregate_function (expression)
  FROM tables
  WHERE conditions
  GROUP BY expression1, expression2, ... expression_n;
  

  SQL 101: Seen are medium and other aggregate data

    SELECT DISTINCT
           comp.company_name || ' - ' || comp.contract_number AS "company_name",
           UPPER (bh.first_name) AS "budget_holder",
           INITCAP (sup.first_name) || ' - ' || sup.surname AS "supervisor",
           INITCAP (emp.status) AS "department",
           INITCAP (emp.first_name || ' ' || emp.surname) AS "employee",
           r.employee_id,
           emp.paaf_number AS "paaf_num",
           emp.employee_payroll_number AS "payroll_num",
           emp.discipline AS "discipline",
           NVL (SUM (tg.week_day1), 0) AS "day1",
           NVL (SUM (tg.week_day2), 0) AS "day2",
           NVL (SUM (tg.week_day3), 0) AS "day3",
           NVL (SUM (tg.week_day4), 0) AS "day4",
           NVL (SUM (tg.week_day5), 0) AS "day5",
           NVL (SUM (tg.week_day6), 0) AS "day6",
           NVL (SUM (tg.week_day7), 0) AS "day7",
           r.valid_from,
           r.valid_to
      FROM tw_roster_data r
           JOIN tw_employee emp
              ON emp.employee_id = r.employee_id
           JOIN tw_timesheet_stage tg
              ON tg.employee_id = r.employee_id
                 AND tg.week_commencing = r.valid_from
           JOIN tw_contracting_company comp
              ON comp.contracting_company_id = r.contracting_company_id
           JOIN tw_supervisor sup
              ON sup.supervisor_id = tg.supervisor_id
           JOIN tw_budget_holder bh
              ON bh.budget_holder_id = sup.budget_holder_id
     WHERE     r.contracting_company_id = 101
           AND r.VALID_FROM = '31-JAN-2015'
           AND r.VALID_TO = '06-FEB-2015'
  GROUP BY
  comp.contract_number
  ,comp.company_name
  ,bh.first_name
  ,sup.first_name
  ,sup.surname
  , emp.status
  ,emp.first_name
  ,emp.surname
  r.employee_id,
  emp.paaf_number,
  emp.employee_payroll_number,
  emp.discipline,
   r.valid_from,
   r.valid_to;
  

• in Outlook Express, how can I send the same message to a group important (in the address book already), but have to each recipient can only see his name (not the Group)

  In Outlook Express, how can I send the same message to the large group (already in the address book), but have to each recipient can only see his name (not the Group)?

  To use the functionality of the BCC (Blind Carbon Copy), in a new message window, click View | All headers and type the addresses separated by a; Or: Click the button to. You will see your list of contacts and the choice to put them in the To, CC, or BCC field. Click on an address, and then click BCC to add it. Repeat for all of the contacts you want. (Note: most of the providers have limits as to how many messages you can send both a. 25 is common, but it varies).

  Put your own address in the box to. That's the only name/address that any recipient will see. If you leave the field empty, some people may not receive e-mail because of message rules they put in place, or restrictions imposed by their ISP.

  You can also make an entry in the address book using your e-mail address, but put something like the mailing list in the area of the display.

  *********************

  You can also create groups if you send e-mail to the same people often.

  To create a group of contacts:

  You can create a single group name (or alias) to use when sending a message to several contacts at the same time. Just create a group name and add individual contacts to the group. Then just type the name of the group in the box for when you send mail.

  1. in the address book, select the folder in which you want to create a group. Click New in the toolbar, and then click New Group.
  2. the Properties dialog box opens. In the group name box, type the name of the group.
  3. There are several ways to add people to the Group:
  a. to add a person from your address book list, click Select members, and then click a name in the address book list.
  (b) to add a person directly to the group without adding the name to your address book, type the person's name and e-mail address at the bottom of the Properties dialog box and then click Add.
  to add a person to both the Group and your address book, click New Contact and enter the appropriate information.
  (d) to use a directory service, click Select members, and then click search. Select a directory service in the drop-down list at the end of the text box. After finding and selecting an address, it is automatically added to your address book.
  4. Repeat for each addition until your group is defined.

  Note

  a. to view a list of your groups separately from the list of address book in the address book on the view menu, make sure that folders and groups is selected.
   
  b. you can create multiple groups, and contacts can belong to several groups.

  If you want to send to a group without addresses are displayed for each recipient, open the address book and make a new entry. Enter the name of the group in the area of the display, but use your address. (Some Internet service providers require a legitimate address in the line to and recipients will know it's from the Group).

  Click on the button to. Enter the name of the group using your address. Enter the group in the BCC field.

• Apple Watch: can I receive all messages from whatsapp on my Apple Watch, but not the Group whatsapp messages, don't know why... Help, please. Besides, my watch is custom in Spanish, but only answer messages Whatsapp understands English... someone kn

  I can receive all messages from whatsapp on my Apple Watch, but not the Group whatsapp messages, don't know why... Help, please.

  Besides, my watch is custom in Spanish, but only answer messages Whatsapp understands English... anyone know why? Thanks to you all!

  Hello

  To change the language for a response on your watch: Press answer > firmly press the screen > press on choose language > select Spanish.

• Linking New Extreme old Extreme and Express. New Airport utility does not recognize old Extreme/Express

  I had a linear new extreme connection (connected to the wireless model Cat5) related to the old extreme then Express (newish).

  I would have put in place using airport Utility (for example) 4 years ago and has worked well.

  I changed my wireless modem (formerly broken) but cannot now reconnect the old extreme and express to the extreme again.

  The extreme new connects to internet just fine as long as it is in Bridge mode (does not work in DHCP/NAT mode).

  My real question is the utility software Airport now seems not to recognize the Express extreme or old old to get re-connected.

  The manual of Apple still shows configurations using the old utilitarian approach Airport (with each unit sitting on LHS page)

  If you're still using OS X (10.5.8)... as, you enter in your profile information... then the AirPort Utility in this operating system must be able to recognize the old AirPort Extreme and Express.

  However, if you have updated to a new operating system such as the Mountain Lion (10.8.x), Mavericks (10.9.x), Yosemite (10.10.x) or El Capitan (10.11.x)... and... you have more "round" or version 'dome' in the form of AirPort Extreme, and an old version of the AirPort Express A1084... so, the new version of AirPort Utility is not compatible with older airports.

  So, this would help a lot if we could get clarification from your part...

  (1) what operating system you're using now?  If you are not sure about this, click on the Apple icon in the upper left corner of the screen, then click on about this Mac and post back with the Version of OS X number that you see there.

  (2) what version of the AirPort Extreme you may have.  Is it a 'rounder', or 'dome' in the form of version... or... The "flat" or "square" version? If you have flat or square version, the number of model of the device is required. It is on the lower part of the airport, probably integrated in the base of carpet of MOSS. The model number starts with an 'A' follow-up of four digits. Please post back this information.

  (3) what version of AirPort Express you have?  Please look for the model number on the side of the Express. It is difficult to see, with grey lettering in a certain context, so you need to perhaps good lighting and glasses to discern the number of reading.  As extreme, the Express, model number starts with an 'A' followed by four digits. Please post back this information.

  Once we have the information we need, we know more about how to proceed from this point.

• Not a GROUP BY clause error

  Hi all
  
  Can someone tell me what is wrong with the query below? I get an exception is NOT a GROUP OF CLAUSE:
  

  SELECT
          LAST_DAY((xxx.timevalue)) as  SUMMARYDATE
        , xxx.acct AS ACCTNUM
        , xxx.referenceid AS SEARCHTYPE
         , CASE WHEN NVL(xxx.PRIMARYSRCH_IND,'N') = 'Y' THEN COUNT(xxx.searchid) ELSE 0 END  Cnt_PrimarySearches
                 , CASE WHEN NVL(xxx.primarysrch_ind,'N') = 'N' THEN COUNT(xxx.Searchid) ELSE 0 END  Cnt_SecondarySearches
       
        FROM XXX_Tablename xxx
        WHERE
         xxx.referenceid NOT IN ('Alert','Search','Collection')
        GROUP BY
    LAST_DAY((xxx.timevalue))  
    , xxx.acctnum
    , xxx.referenceid 

  Hello

  Try this

  SELECT
          LAST_DAY((xxx.timevalue)) as  SUMMARYDATE
        , xxx.acct AS ACCTNUM
        , xxx.referenceid AS SEARCHTYPE
         ,COUNT( CASE WHEN NVL(xxx.PRIMARYSRCH_IND,'N') = 'Y' THEN (xxx.searchid) ELSE 0 END  Cnt_PrimarySearches)
                 , COUNT(CASE WHEN NVL(xxx.primarysrch_ind,'N') = 'N' THEN (xxx.Searchid) ELSE 0 END) Cnt_SecondarySearches
  
        FROM XXX_Tablename xxx
        WHERE
         xxx.referenceid NOT IN ('Alert','Search','Collection')
        GROUP BY
    LAST_DAY((xxx.timevalue))
    , xxx.acctnum
    , xxx.referenceid
  

  Not tested

• Thunderbird does not recognize groups of outlook express news. What can I do to fix?

  Years, I had no problem emailing messages to my Outlook Express news groups. Now, when I try to do, I get a message saying that the news group "is not a valid email address because it is not of the form user@host.» Thanks for your help.

  See the bug:

  • https://Bugzilla.Mozilla.org/show_bug.cgi?id=1060901

  Please register and add comments stating what OS you are using and what version of thunderbird. then Vote for the bug - this at the top:

  • Importance:-normal (vote)

  (Vote) is a link: click on it. Select the check box for the bug and click the button "change my voice."

  Workaround solution:

  Comment of Fabian Rodriguez 2014-09-04 06:19:15 PDT

  Confirmed workaround solutions:

  • All by composing an email open the address book and select the list you want to send, to highlight all the names in the list and drag them to the To: box. This example uses your existing data without changing it.
  • Replace white "" between the words in the descriptions of these lists with an underscore "_". This requires modification of your descriptions of mailing list (s).

• ORA-00979: not a GROUP-BY-error expression

  Hello
  We moved just our project mysql database to oracle 10 database. The migration was made manually, without the use of tools. We had to change many queries to oracle suit. The following query when it is run on mysql works perfectly well. But when I run it in oracle, it gives the error mentioned in the subject.
  
  -----
  Select distinct d.managedserverid, s.hostname, m.port, a.applicationname, max (d.versionid) of table1, table2 m, s table3, table 4 d, where d.applicationid = '1' and d.Active = 'True' and d.managedserverid = m.managedserverid and m.serverid = s.serverid and d.applicationid = a.applicationid group by d.managedserverid;
  -----
  
  Any help would be appreciated.
  
  Thank you
  SAPP

  Hello

  Welcome to the forum.

  When you use the aggregate functions such as min, max, sum, etc. you have to group the columns not aggregated in the select query.

  Try this.

  select d.managedserverid, s.hostname, m.port, a.applicationname, max(d.versionid)
  from table1 d, table2 m, table3 s, table4 a
  where d.applicationid = '1'
  and d.Active = 'True'
  and d.managedserverid = m.managedserverid
  and m.serverid = s.serverid
  and d.applicationid = a.applicationid
  group by d.managedserverid, s.hostname, m.port, a.applicationname
  

  HTH,
  Prazy