Length of the string, excluding the input character/string

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  They used to work - could be almost ten years (I think version 9 - not tested yet)

  As you can see, you can 'encode' the number of characters, padded in order to set the right length of the decrypted string

  
  

  create or replace FUNCTION ' CRYPTIN "(p_str in varchar2) return varchar2 as". "

  l_data VARCHAR2 (255);

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  -platelets p_string with pad_chr containing the number of characters padded (1 to 8)

  -make the string length a multiple of 8 as required by decipher

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  pad_char char: = chr (0);

  pad_chars varchar2 (8);

  Start

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  end;

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  create or replace FUNCTION ' CRYPTOUT "(p_str in varchar2) return varchar2 as". "

  l_data VARCHAR2 (255);

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  -Removes from p_string upholstered characters (1 to 8)

  pad_length pls_integer;

  Start

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  pad_length: = pad_length - ascii (substr(p_string,pad_length,1));

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  end;

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  end;

  Concerning

  Etbin

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  How cut the length of the string IE channel name here, if it exceeds 20 characters or letters. The channel name is dynamically analyzed the spreadsheet/xml file.

  My Council of Europe in the .js file is

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  Alert ("this is lang" + str);
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  "Str += '.

  ";
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  "+ ""+""+"
  ";
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  }

  nerateHTMLMarkup: function (i, channel) {}
  Alert ("this is lang" + str);
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  "Str += '.

  ";
  str += "" +
  "" +
  "" + channel ['name'] + "" +.
  "" + "
  ";
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  "+ "" + "
  ";
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  return str;

  etc.

  Give mthe rhythm and the exact code where Ihave tio get implemented.

  in the code, replace change

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  channelName var = channel ['name'];

• Lesson1 solutions, length of the string, really a God's solution?

  Hello, I was just looking at the solutions proposed for lesson 1 homework and I came on line 65, ex 2 solution where odd length of the string is calculated this way:

  (s.length () & 1) == 1

  Here, we are a group of users of Java for a long time and we still don't completely understand how it really works.

  Do the bitwise AND operation on the length of the string 1 and audit, seems really strange.

  Either way, I suppose it works somehow but is it really a good solution providing a code which may include only a Ninja of Java for the black belt? If you use it in production which will be able to keep it?

  the pit of my ignorance, I wrote this:

  ! (p.length (% 2) == 0)

  Thank you and have a nice day!

  It's probably again my fault.

  I remember presenting examples of lambda at the beginning I used (I % 2 == 0) to test whether a number has been the same. One of our graphics guys chastises me for using rest (%) instead of bits - and ((i & 1) == 0). In C, the remainder operator is usually compiled to some form of a division operation, which is much slower than a - bit operation and exploitation. (An optimizer may be able to replace i with i 2% & 1 but optimizations in c may be quite subtle). In any case, I started to use at the level of the ILO - and in my examples.

  Then I started getting complaints about the use of the bitwise operators because nobody knew how they worked.

  In any case, in Java, the result is well-defined, as a type int is specified as a 32 bit two complement binary number, so I & 1 will tell you if it's odd or even. The force JIT compiler - reduce i % 2 to i & 1, but when the interpreter is running it will go through the slow lane. As someone said, it's a micro-optimisation anyway.

  A ride is what % is the operator remains , not a modulo operator, while for positive numbers, the result is the same. For negative numbers, the result can be negative! So, be sure to not try for an odd number using i % 2 == 1 because it will fail for an odd, negative number, the result will be-1. Test! = 0, or better still simply use at the level of the ILO - and. Of course, the string length is not a problem because it is still not negative, but the problem may occur in other situations, in particular array of indexation.

• get the length of each string in a column

  Hi all

  Is there a way to get the length of each string in a column:

  for example:

  If the table contains lines like below:

  Select 'test test1 test31' col1 of double

  Union of all the

  Select 'sectest1 sectest234' double col1

  The output should be as below:

  col1

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

  4 5 6

  8 10

  That is to say get the count of each separated from the chain of space in a column. Is this possible using regexpr_substr?

  Thanks in advance.

  Roy.

  Try the below

  SELECT LISTAGG(val1,' ') IN ONE GROUP)

  New_val ORDER BY lvl)

  Of

  (SELECT LENGTH (REGEXP_SUBSTR (col1,'[^] +', 1, LEVEL)) val1,)

  RN,

  LEVEL lvl

  Of

  (SELECT col1,

  ROWNUM rn

  Of

  (SELECT 'test test1 test31' col1 OF double)

  UNION ALL

  SELECT 'sectest1 sectest234' double col1

  )

  )

  CONNECT BY LEVEL<= regexp_count(col1,'="">

  AND PRIOR rn = rn

  AND PRIOR DBMS_RANDOM. VALUE IS NOT NULL

  )

  GROUP BY rn;

  OUTPUT:-

  =======

  col1

  ----

  4 5 6

  8 10

  Thank you

  Ann

• How to use the input as VC Atttibutes string?

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  The workflow that I am putting in place is:

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  2-search 'TOTO' virtual machine in the vCenter, if found then

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  For stage 1 and 2, it works, I can search based on the input of the chain and get a result.

  But for step 3, I use another workflow 'Create virtual machine' and we must put a VC:VirtualMachine.

  And it does not work because the type I get to step 2 is not compatible with the entry for step 3.

  Progressive input signal 3 waiting for VC:VirtualMachine

  And the output of step 2 is the table / VC:VirtualMachine.

  I tried to change the input type for step 3 as table / VC:VirtualMachine.

  Whenever I got the error message:

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  I don't want the workflow using the VC search plugin, but I need to set the input parameter.

  If someone has an idea.

  Thank you all.

  Keep the entry for step 3 as VcVirtualmachine (only one), but add an extra step:

  2.5: get the first (or last, should not question: I hope that you will find only one :-)) table.)

  You can do it in Javascript, using the method. pop() from an array.

  var sourceVM = foundVMs.pop () / / sourceVM is of type VcVirtualMachine, foundVMs table/VcVirtualMachine

  See you soon,.

  Joerg

• How to get the length of the string a vm?

  Hello

  I'm trying to automate some consolidation of virtual machines and would like to add script. I have over 800 virtual machines manually would not work.

  I would like to

  $allVApps = get-CIVApp

  foreach ($CIVm to ($allVApps |)) Get - CIVM)) {}

  find the number of channels and if greater than 8 consolidate them.

  }

  Someone has this practice? or something else?

  I'm in a cloud environment vSphere 5.5, installed latest powercli.

  Thank you

  Don't know why this does not work for others, but what I posted before work for me.  Here is a more complete version of how get the those of VAPP VM.  There are no doubt more direct ways, but that how I made it so far:

  1. get the VAPP

  $vApp = get-CIvApp-name 'Test-VAPP.

  2. get the VMS of vApp

  $VMs = $vApps.ExtensionData.Children.vm

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  I hope this helps.

• Delete the last character in the input field

  How to remove the last character of an input field?

  It deletes all

  When you click on:

  SYM.$("textfield"). Empty();

  Thank you!

  Try this:

  var inputTxt = sym.$("textfield").text ();

  var truncatedTxt = inputTxt.slice (0, -1);

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• find the length of the string

  Hi all
  
  I have a srting that contains two or more channels as for instance, it's my channel (think of this total string as one string) "Bronson, Ms. Jamie", "elder, Ms. Francis '.
  Now I want to calculate the length of the string of replacement of commas and apostrophes with null.
  
  Can someone pls help me with this?
  
  Thank you

  SQL>
    1  with t as (
    2   select '''Bronson, Ms. Jamie'',''Elder, Mrs. Francis''' as mystring
    3  from dual)
    4  select * from t;
  
  MYSTRING
  ------------------------------------------
  'Bronson, Ms. Jamie','Elder, Mrs. Francis'  
  
  SQL>
    1  with t as (
    2   select '''Bronson, Ms. Jamie'',''Elder, Mrs. Francis''' as mystring
    3  from dual)
    4  select replace(replace(mystring, ',', null), '''', null) from t;
  
  REPLACE(REPLACE(MYSTRING,',',NULL),
  -----------------------------------
  Bronson Ms. JamieElder Mrs. Francis
  
  SQL> l
    1  with t as (
    2   select '''Bronson, Ms. Jamie'',''Elder, Mrs. Francis''' as mystring
    3  from dual)
    4  select length(replace(replace(mystring, ',', null), '''', null)) as string_len from t;
  
  STRING_LEN
  ----------
          35