Como get the number of license

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  Hi team,

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  $ServiceInstance = get-view ServiceInstance

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  Can someone help me please.

  Power required:

  If there are 4 cottages with 2 CPU each running in evaluation mode, I need output like '8 '.

  It was my mistake...

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  I bought LR 6 and installed on the desktop and laptop. Now desktop mainboard has failed and hat to be replaced wit consequence to reinstall Windows 7 and 6 LR. Now, I exceed the number of licenses. Should I still buy LR8?

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  I'm unable to install the second copy LR6. I get the windows with the message that I've exceeded the number of licenses.

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  Concerning

  Dieter Kolonko

  Von: John T Smith

  Gesendet: Mittwoch, 19. Juni 2015 21:02

  Year: Kolonko Dieter

  Vacancies: I bought LR 6 and installed on the desktop and laptop. Now desktop mainboard has failed and hat to be replaced wit consequence to reinstall Windows 7 and 6 LR. Now, I exceed the number of licenses. Should I still buy LR8?

  I bought LR 6 and installed on the desktop and laptop. Now desktop mainboard has failed and hat to be replaced wit consequence to reinstall Windows 7 and 6 LR. Now, I exceed the number of licenses. Should I still buy LR8?

  created by John T Smith https://forums.adobe.com/people/JohnTSmith > in download, full installation, commissioning - see the discussion of https://forums.adobe.com/message/7667228#7667228>

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• get the number in the declaration of DBMS

  Hello everyone,

  my friend, I create the following code, but I've always faced a problem to display the result of County all the

  the code is:

  ------------------------------

  I insert a count (*) to get the number of employees

  but he keep get only '1' to the last of the result in the following way:

  DETAILS OF THE EMPLOYEE:

  YJOSEPH5 | YOUSEFJOSEPH | 15 NOVEMBER 15

  HYOUSEF10 | HASSANYOUSEF | 15 NOVEMBER 15

  HYOUSEF17 | HASSANYOUSEF | 15 NOVEMBER 15

  HYOUSEF14 | HASSANYOUSEF | 15 NOVEMBER 15

  HYOUSEF16 | HASSANYOUSEF | 15 NOVEMBER 15

  COUNT: 1

  the code

  ------------------------------------

  DECLARE
  V_USER_ID VARCHAR2 (50);
  NUMBER OF V_EMP_NO;
  V_FIRST_NAME VARCHAR2 (30);
  V_LAST_NAME VARCHAR2 (30);
  DATE OF V_REG_DATE;
  NUMBER OF V_COUNT;
  V_SQL_SATEMENTS VARCHAR2 (1000);
  NCOUNT THE NUMBER;
  CURSOR C IS SELECT USER_ID, EMP_NO, FIRST_NAME, LAST_NAME, REG_DATE, COUNT (*) FROM EMP
  USER_ID, FIRST_NAME, LAST_NAME, EMP_NO, REG_DATE GROUP;
  BEGIN
  OPEN C;
  SELECT COUNT (*) FROM NCOUNT FROM USER_TABLES WHERE TABLE_NAME = 'EMP ';
  IF (NCOUNT < = 0) THEN
  V_SQL_SATEMENTS: =' CREATE TABLE EMP)
  USER_ID VARCHAR2 (40) PRIMARY KEY,.
  NUMBER OF EMP_NO,
  FIRST NAME VARCHAR2 (30),
  VARCHAR2 (30) LAST_NAME,.
  DATE OF REG_DATE)';
  
  RUN IMMEDIATELY "DROP TABLE EMP;
  IMMEDIATELY RUN V_SQL_SATEMENTS;
  END IF;

  SELECT NVL (MAX (EMP_NO), 0) + 1
  IN V_EMP_NO
  YOUSEF. EMP;
  INSERT YOUSEF. EMP (USER_ID, EMP_NO, FIRST_NAME, LAST_NAME, REG_DATE)
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  LOOP
  EXTRACT THE C IN V_USER_ID, V_EMP_NO, V_FIRST_NAME, V_LAST_NAME, V_REG_DATE, V_COUNT;
  WHEN THE EXIT %C NOTFOUND;

  DBMS_OUTPUT. PUT_LINE (V_USER_ID |'|) ' || V_FIRST_NAME | V_LAST_NAME |' | ' || V_REG_DATE);
  END LOOP;
  DBMS_OUTPUT. PUT_LINE ("COUNT:" | "|") V_COUNT);

  CLOSE C;
  END;
  /

  ALL FRIENDS,

  THANKS FOR EVERY BODY,

  I JUST FOUND THE RIGHT WAY TO DO

  FOR THOSE WHO WANT TO SEE:

  -----------------------------------------------------------

  DECLARE
  V_USER_ID VARCHAR2 (50);
  NUMBER OF V_EMP_NO;
  V_FIRST_NAME VARCHAR2 (30);
  V_LAST_NAME VARCHAR2 (30);
  DATE OF V_REG_DATE;
  NUMBER OF V_COUNT;
  V_SQL_SATEMENTS VARCHAR2 (1000);
  NCOUNT THE NUMBER;
  CURSOR C IS SELECT USER_ID, FIRST_NAME, LAST_NAME, EMP_NO FROM EMP REG_DATE;
  BEGIN
  SELECT COUNT (*) IN THE EMP V_COUNT;
  OPEN C;
  SELECT COUNT (*) FROM NCOUNT FROM USER_TABLES WHERE TABLE_NAME = 'EMP ';
  IF (NCOUNT<=0)>
  V_SQL_SATEMENTS: =' CREATE TABLE EMP)
  USER_ID VARCHAR2 (40) PRIMARY KEY,.
  NUMBER OF EMP_NO,
  FIRST NAME VARCHAR2 (30),
  VARCHAR2 (30) LAST_NAME,.
  DATE OF REG_DATE)';
                                           
  RUN IMMEDIATELY "DROP TABLE EMP;
  IMMEDIATELY RUN V_SQL_SATEMENTS;
  END IF;

  SELECT NVL (MAX (EMP_NO), 0) + 1
  IN V_EMP_NO
  YOUSEF. EMP;
  INSERT YOUSEF. EMP (USER_ID, EMP_NO, FIRST_NAME, LAST_NAME, REG_DATE)
  VALUES (SUBSTR (: LAST NAME, 1, 1) |: LAST_NAME |) V_EMP_NO, V_EMP_NO,:FIRST_NAME,:LAST_NAME,TO_DATE(SYSDATE,'DD-MON-YY'));
  DBMS_OUTPUT. PUT_LINE (' DETAILS OF THE EMPLOYEE: "");
  LOOP
  EXTRACT THE C IN V_USER_ID, V_EMP_NO, V_FIRST_NAME, V_LAST_NAME, V_REG_DATE;
  WHEN THE EXIT %C NOTFOUND;

  DBMS_OUTPUT. PUT_LINE (V_USER_ID |'|) ' || V_FIRST_NAME | V_LAST_NAME |' | ' || V_REG_DATE);

  END LOOP;

  CLOSE C;
  DBMS_OUTPUT. PUT_LINE ("COUNT:" | "|") V_COUNT);

  END;
  /

• How to get the number of entries in the navigation menu?

  

  How to get the number of entries in the navigation menu? Check the image. What it shows 8 number of customers and 10 products and 10 orders in the list. How to get it?

  --

  Thank you

  Hello

  Create items of the request saying, "CUSTOMERS", "PRODUCTS" and "ORDERS".

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  I have a column in my form which lists the days in a month. I want to configure a hidden field that calculates the total number of days in a month, based on the month and year of the field inputs. The number of days will determine what appears on the column. For example, if I put 4 months, and 2016 in the field of the year, I get 30 in the hidden field. Thus, on the column 'Day', I'll have numbers 1-30. Or if I put 2 months and 2016 in the field of the year, I get the 29 in the hidden field. If the numbers 1-29 appears in the column 'day '.

  Found this on some forum javascript code:

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function daysInMonth(month,year) {
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}

//July
daysInMonth(7,2009); //31
//February
daysInMonth(2,2009); //28
daysInMonth(2,2008); //29

  I do not know how to convert this code in JavaScript to adobe and don't really know how to use it. All I know how to do is to configure the field values for the field month and year as variables. I am a novice programmer and would appreciate it really all the help I can get. Thank you in advance!

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  I would like to consider making more general code, so if you have a date string that includes at least the month and year we could just call the function and get the number of days for that month.

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  }

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  Event.Value = "";

  If (nMonth! = "" & nYear!) = "") {}
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  var nDaysInMonth = daysInMonth (MyDate); get the number of days;

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  Hi all

  My need is to get the number of the day of the week - any national local is used.

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  I heard that some countries start the week of Sunday.

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  PL/SQL Release 11.2.0.1.0 - Production

  "CORE 11.2.0.1.0 Production."

  AMT for Linux: Version 11.2.0.1.0 - Production

  NLSRTL Version 11.2.0.1.0 - Production

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  2621671 wrote:

  Hi all

  My need is to get the number of the day of the week - any national local is used.

  For example, Saturday's 6th day of the week in my country, because the week starts on Monday.

  ...

  That's what you asked for, whatever your NLS settings:

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  I heard that some countries start the week of Sunday.

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  How can I make it?

• Get the number of CPU of the machine virtual and carrots

  Hello

  I tried the following, but failed to get the number of CPUS the VM and carrots. What I am doing wrong?

  

  Of course, try like this

  $myCol = @)

  foreach ($cluster Get-cluster)

  {

  foreach ($vmhost in ($cluster |)) Get - VMHost))

  {

  foreach ($vm in (Get-VM-location $vmhost)) {}

  $VMView = $vm | Get-View

  $VMSummary = "" | Select NOMCLUSTER, HostName, VMName, VMSockets, VMCores, CPUSockets, CPUCores

  $VMSummary.ClusterName = $cluster. Name

  $VMSummary.HostName = $vmhost. Name

  $VMSummary.VMName = $vm. Name

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