WLC 8510 - v8.0.100.0 - APGROUP count how many APs

Similar Questions

• Count how many an array of letters in a set of words

  Is it possible to count for each word to say, 25 words, how many letters of a range of letters in each word? So if the table has been BCEGHIRLWYDO and some of the words were IDOL, SIDEWALK and WAXY, you say IDOL 4, CURB 3 and 2 WAXY?

  (The table will always be 12 letters, I think.) I can count how many of each letter is used for the whole word game, which gives me the table, but do not know how to get a census for each individual word.

  I'm still with contentment using OS X 10.5.8, mini mac, Numbers ' 09, 2.0.5.

  Thanks for all the help I got from you folks.

  A script is effective to do this.

  

  No necessary knowledge of AppleScript.

  1. Copy and paste the script below in the AppleScript editor or Script Editor as it can be called on your machine (in Applications > utilities).
  2. Add the words as needed to the first line of the script (between the {} and each surrounded by "") and edit the letters if necessary.
  3. Click the 'run '.
  4. Click in a cell of numbers (in my example, B2).
  5. Command-v to paste.

  SG

  game gathered to {'IDOL', 'STOP', 'WAXY'}

  theLetters value to "BCEGHIRLWYDO".

  game of pasteStr to «»

  Repeat with w in collected

  the value pasteStr to the pasteStr & w tab & countMatches (w, theLetters) & return

  end Repeat

  Set the Clipboard for pasteStr

  return pasteStr

  at countMatches (theWrd, theLtrs)

  the value n to 0

  Repeat with l in theWrd characters

  If theLtrs l to contains then the value n to n + 1

  end Repeat

  n return

  end countMatches

• How to count how many times a Boolean control is true

  As part of my project using LabVIEW 2012 I'm counting how many times one controls the boolean true value. Please help me...

  

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  And yet anothr approach...

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  The fill color of the same color as the background color, then make the background color transparent.

  Drop a box of color BEHIND the bar and the size correctly.

  He has this strange background 3D with flat bar, but what the Hey, his relatives. If its important a custom color box.

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• I get Windows Live Mail error ID: 0x800CCC6E when I try and send an email to a group of 100 or more. How can I fix it?

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• Separated from 100 GB in 2x50GB - how to make a 100 GB HARD drive?

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  My hard drive can be partitioned? How can I make a hard drive 100 GB of it? East - advised to do? I hope someone could help me more.

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  Expert mode on recovery DVDs installation allows you to install Recovery on defined partition image. If you have only one partition and that something is wrong that all content will be lost.

  The 100 GB is enough space and it can also be split on three partitions where the first (C) will be used for the OS and installing different applications and two others just for the data backup.

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• SUDDEN question MS Windows Mail Security. I don't know how many years counts with Bellsouth / AT & T.

  Cannot Log on to ATT email account. Keeks pop-up Windows security request my username and password.  Using my correct email address and the right password in the pop-up window will allow only connect to ATT once in every 30 to 50 attempts.  The error message in the task of Windows Mail window is the following:

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• WLC 4404 and high availability

  If I have 2 WLC 4404 are supported 100 Access Points for one and I installed 110 Access Point, I can you decide when a WLC down who are the 10 Access Point which must lie down?

  Thanks for your help

  fcostalunga

  Let me understand your question.

  You have two WLC and can support 100 AP. You have 110 APs. If one of the WLC must go down, how you can configure or say which 10 AP will not be able to join?

  If this is your question, here are some solutions I can think.

  1. H-HARVEST. If you enable REAP H to your AP, when one or ALL of your WLCs must go down, the AP will be able to operate as if the WLC has increased. The exception-to-the-rule is to NOT reload the AP and it should be good.

  2. you can configure, CLI only, 100 AP s primary and secondary controller. 10 AP will no config, so they will go into "limbo".

  I hope this helps.

• Count of NOT NULL or NOT

  Hi all

  I am trying to run a query to count how many reports in my system have a valid "RunNumber" in a specified amount of time.  To do this, I need to eliminate some bad entries and null values that are entered in the column RunNumber

  My query looks like this:

  Select round (count (a.ReportNumber) / tot_reports * 100, 2) PERCENTAGE

  report a, b of EMS,

  (select count (ReportNumber) report tot_reports, ems WHERE report.id = emss.id and insertdate BETWEEN 1st January 13 'AND January 31, 13')

  where a.id = b.aid and a.insertdate BETWEEN January 1, 13 'AND 31 January 13'

  and (b.RunNumber not in ('0000000000', 'N/a', 'NA', 'NONE', 'NONE GEN"," NOT Available. ","NOT AVAILABLE","NOT PROVID","REFUSAL","DENIED", 'UNK', 'UNK',"UNK.", 'UNK000',

  ("UNKKNOWN", "UNKNOWN", "UNKNOWN", "UNKNOWN", "UU", "XX", "XXX", "XXXXX", "XXXXXX") or b.Runnumber is not null)

  Tot_reports group;

  When I run this query, I get 97.73%

  There are 44 total records and 1 card has a null value, and 3 contain a 'bad data' I'm trying to weed out.  So techincally the % should be 40 files with a 'good' runnumber.  And this percentage 40/44 * 100 = 90.9%

  I know that the null value is shaking the calculation but I don't how to work around it.  Any help will be greatly appreciated.  Thanks in advance.

  Thank you very much.  This helped me wrap my head around what I was trying to accomplish and it worked perfectly.

• Count (*)

  create table test(tno number(5), tval varchar2(20));
  
  
  begin for i in 1..100 loop
  insert into test values(i,i+10);
  end loop;
  end;
  
  select count(*) from test;
  select count(distinct 1) from test;
  select count(distinct 2) from test;
  select count(distinct 3) from test;
  
  
  COUNT(*)               
  ---------------------- 
  100                    
  
  1 rows selected
  
  COUNT(DISTINCT1)       
  ---------------------- 
  1                      
  
  1 rows selected
  
  COUNT(DISTINCT2)       
  ---------------------- 
  1                      
  
  1 rows selected
  
  COUNT(DISTINCT3)       
  ---------------------- 
  1                      
  
  1 rows selected
  
  
  why the second and third query getting count 1. ??

  Select count (*) of the test;

  -Here, you count how many lines in the test table, s back 100...

  SELECT count (distinct 1) test;
  -Here, you count the number of distinct values in select 1 of the test. (this is always 1) even in other cases

  SELECT count (distinct 2) test;
  SELECT count (distinct 3) test;

  If you need with the number of distinct values in your table

  SELECT count (distinct tno) test.