Add control characters of the string without full control update

Similar Questions

• REGEXP_REPLACE to add characters in the string

  Hi all
  
  How to add characters between the sctring?
  
  I have a string "1A24B67IO" and you want to convert to "1a 2 - 4B - 67IO.
  
  I tried to use the REGEXP_REPLACE function, but not able to do.

  Hello

  As you have not given the logic behind it... I guess you must insert - so much that even the example you provided.

  With T as
  (
  select '1A24B67IO' col_val from dual
  )
  --
  --End of sample data
  --
  select regexp_replace(col_val,'([[:alnum:]]{3})([[:alnum:]]{2})([[:alnum:]]{4})','\1-\2-\3') from t
  /
  PRAZY@11gR2> /
  
  REGEXP_REPL
  -----------
  1A2-4B-67IO
  
  Elapsed: 00:00:00.20
  

  Kind regards
  Prazy

• replace characters in the string

  Hi all

  Is it possible to index the characters in a string as if it were a table and remove them from the second position and now. I just want with the first letter or number so that I can later concatenate strings to it.

  Thanks for any help!

  -Michael

  Hi Michael,

  Yes. Use the function "Sring subset. At the beginning the value 0 (default value) and the length of 1.

  This will give you the first character in the string.

  Steve

• How to avoid special characters in the string box

  Hello

  is it possible that a special character can be avoided in a box of the string, as authorizing only a - z and 0-9 to enter a box in the chain

  Hi all

  Darin Laurette method is simple.

  AnkitG: You must change the code.

  Good learning for me.

  Kind regards

  Leila

  CLD

• Select tool and exhaust of the characters in the string

  I would like to choose a directory path using the selection functions. The string ends to the position of the backslash,

  Example "C:\Program Files\Microsoft Games" will become "c:".

  How to do this

  Thank you

  JIm

  Thanks for the help, it's the indicator that was truncating the line.

  Sorry for the questions but I'm very new to labview, it is very different from the CVI

  Jim

• REGEXP_REPLACE problem with special characters in the string

  Hi guys,.
  
  I have had problems using the REGEXP_REPLACE function and I'm trying to find a solution.
  
  The closest I can get is this thread:
  
  Re: How to replace the first occurrence of a substring
  
  and that's only because the solution, I'm trying to achieve is the one suggested. I did ppost my problem there, but because it is answered, I don't have an answer.
  
  I'm doing the following:
  

  select 
    regexp_replace('A*|A*|C11|A1|A*||_|OR|H452|C40|A|A*||_|AA|1151|G31|A1|A||_|EX|8HI01|H11|A1|A||_|OCR|H421|B11|A|||_|OR|H434|C11|A|||_|AQA|9990|E10|EP|||_|AA|2151|G31|A|||_|',
    '^(\w|-)*\|{1}',null)
  from dual

  Basically, replace it does not work, because the first part of the string contains an asterisk. If you delete the Asterix, then replace it won't work.
  
  How can I get around this? I can't use a regular REPLACEMENT as I want to remove the first instance of A * and not all instances of the A.
  
  Help, please! I don't know what to do... probably, I mustn't start a loop through the string?
  
  Thanks in advance guys!
  
  Robin

  User_resU wrote:
  I want to remove the first instance of A * and not all instances of the A.

  To decode your regular expression, it seems that you want to remove A * | and not just A *.

  SQL> select   regexp_replace('A*|A*|C11|A1|A*||_|OR|H452|C40|A|A*||_|AA|1151|G31
  |A1|A||_|EX|8HI01|H11|A1|A||_|OCR|H421|B11|A|||_|OR|H434|C11|A|||_|AQA|9990|E10|
  EP|||_|AA|2151|G31|A|||_|',
    2    '^(\w|-)\*\|{1}',null) reg
    3  from dual;
  
  REG
  --------------------------------------------------------------------------------
  
  A*|C11|A1|A*||_|OR|H452|C40|A|A*||_|AA|1151|G31|A1|A||_|EX|8HI01|H11|A1|A||_|OCR
  
  |H421|B11|A|||_|OR|H434|C11|A|||_|AQA|9990|E10|EP|||_|AA|2151|G31|A|||_|
  

• How to divide basedon characters of the string

  var str:String = "M01L01T01S01".

  How to cut based on alphabatics

  I need to change the string above as 'M01_L01_T01_S01 '.

  Im not an expert in manipulating strings, so it is perhaps not the best way but this one solution

  var str:String = "M01L01T01S01"var pattern:RegExp = /[a-zA-Z]\d*/ig;var results:Array = str.match(pattern);var output:String = results.join("_");
  

• How to get individual characters in the string

  Hello

  I have a question how to get the numbers on the back the strings.

  mc_invis_table_1

  mc_invis_table_2

  mc_invis_table_30

  mc_invis_table_100

  mc_invis_table_200

  mc_invis_table_500

  mc_invis_table_264

  My strings have a standard naming mc_invis_table_ followed by numbers from 1 to 1000.

  Can I get the numbers on the back of each channels above? Like 1,2,30,100,200,500,264 and so on...

  Thank you

  Zainuu

  var str:String = "mc_invis_table_200";
  var res: String = str.substr (15, str.length);
  trace (res);

  hope it solves

• How to remove special characters from the string using translate() without typing all special characters?

  Hi all

  I am trying to remove special characters without the help of regular expressions.

  translate (the column name or string,'!@#$ & * (* () _) * "" :} {?}) >? /, «, » ')

  I want to eliminate this manual process to give all special characters using a chr() or ascii() function.

  Please show me the way.

  Thanks in advance

  Similar to the solution of Michael...

  SQL > ed
  A written file afiedt.buf

  1 with t as (select "[it comes of the #] [more amazing!") Test @# "$* & $%) assuming chain cost $ 5 000' double Str)
  2, i like (select level 1 c from dual connect by level<=>
  3 less
  4 Select + 32 (level-1) double connect by level<=>
  5 less
  6 select + 58 (level-1) double connect by level<=>
  7 less
  8 select + 91 (level-1) double connect by level<=>
  9 less
  10. Select 123 + (level-1) from dual connect by level<=>
  11 less
  12. Select 255 double
  13            )
  14, ts as (select level r, substr (str, level 1) c
  15 t
  16 connect by level<=>
  17             )
  18, tf as (select row_number() (order for r) r
  19                    ,ts.c
  20 TS
  21 I join on (i.c = ascii (ts.c))
  22             )
  23 select replace (sys_connect_by_path(c,'!'),'! ') Str
  24 TF
  25 where connect_by_isleaf = 1
  26 connect r = prior r + 1
  27 * start with r = 1
  SQL > /.

  STR
  -----------------------------------------------------------------------------------------------------------------------
  Thisisthemostamazingtest¸astringcosting5000

  Or something as horrible as this...

  SQL > ed
  A written file afiedt.buf

  1 with t as (select "[it comes of the #] [more amazing!") Test @# "$* & $%) assuming chain cost $ 5 000' double Str)
  2, I like (select replace (sys_connect_by_path (chr (c), 'A'), 'A') as tr)
  3 of)
  4 select c, rownum r
  5 (select 32 + (level-1) as the double connection by level c<=>
  6                         union
  7 select + 58 (level-1) double connect by level<=>
  8                         union
  9 select + 91 (level-1) double connect by level<=>
  10 the union
  11. Select 123 + (level-1) from dual connect by level<=>
  12 union
  13. Select 255 double
  14 tri 1
  15                        )
  16                  )
  17 where connect_by_isleaf = 1
  18 log r = prior r + 1
  19 start with r = 1
  20            )
  21 select translate (str, 'A' |) TR, 'A') as str
  22 * t, I
  SQL > /.

  STR
  --------------------------------------------------------------------
  Thisisthemostamazingtest¸astringcosting5000

• ADF MOBILE: how to add an item in the list without refresh the entire list

  Hello

  I am using this method to add an item to a list (I've already added the item to the data model):

  
  

  providerChangeSupport.fireProviderCreate (String providerKey, int newRowKey, object newValue);

  
  

  I don't want refresh the entire list, rather, I want to just add a new item to the list, that's why I want to use this method.

  
  

  I want to know, what value should I use for the second argument? i.e. 'int newRowKey '.

  
  

  Thank you
  

  
  

  Hello

  the event of change of provider in your case seems to actually create a new line. So the int is the rowKey (index PK or line) of the new line. There is no other choice to simply add an element to a list (ADF Mobile provides no component handles)

  See: Embedded Java/ADF for beginners: ADF Mobile: refresh your valuable data

  Frank

• Add special characters in the legends

  Is it possible to add a special character to a legend, other than the VERY limited selection of the Insert symbol in the properties option? I need to put the arrow as part of a statement? I'm new to Captivate and use a trial version to see if the company should buy it. Could be my problem?

  Hi LyonLover and welcome to our community

  No, the trial version has nothing to do with anything. I know that many people use tests that are limited in capabilities. But with Captivate, the trial is the 'real deal' and your only limit is the time and the fact that you can not disable the expiration of any project that you create until you register.

  If you're not finding the characters you want in the settings Insert symbol (and I assume that you have already clicked on the link "other...) ("to bottom of the first dialog box see you) you could try clicking Start > run... (On Windows) and type charmap. And then press ENTER. This should display the applet of the Windows character map. Once there, you can locate and select any character you want, then copy to the Windows Clipboard and paste it in the Caption dialog box. You can find that you need to change the font scheme, you watch when you have selected the character.

  I hope this will help... Rick

• Divide the string without any delimeters

  A little banal theme, but...

  I have a string saying 1010002. I need to separate groups of two digits itno - 1 01 00 02, if it is uneven - the most significant byte is the only one.

  Thanks in advance.

  If you check his comment he doesn't want it reversed, does?

  

  

• Add color themes in the library without mounting

  If I search the site of the color themes and add to a collection, I am sent edition theme. If I want to return to my search, I have to enter the terms again and start all over again. If I just try to go back, it's going to start me on explore.

  How can I avoid this? I prefer not to go to the editor at all, but to all least, how can I return to my search results?

  I am happy to get in office applications, if this is the solution, but I don't think it works more.

  Thank you!

  Hi Stanford,

  Sorry it took me so long to respond.

  Seems that the web application "guess" that record in your personal library means that you will probably want to make changes to the theme that you make when you first clicked on check-in at the library. I couldn't find an alternative for editing this page. And you are right. Research themes within the office simply launches the web application.

  In other words... I have no solution to keep you except to just copy and paste your search terms. Wish I did.

  Sue.

• Wildcard characters in the string for NthWord

  I have a script that looks through the pages to see if the 70th word on each page is equal to a code number.

  This number is unique for each instance: for example 1090089, 10704591 but always will start with number one.

  I am not coming with an expression javascript for a wild-card search call, the rest of the script is allowed.

  Can someone please offer assistance with the Joker of the script search component works correctly?

  var this.numPages = NUMPAGES;

  for (var k = 0; k < numpages; k ++)
  {
  {
  numWords var = this.getPageNumWords (k);
  for (var i = 0; i < numWords; i ++)
  {
  var ckWord = this.getPageNthWord (k, i, true);

  var q = this.getPageNthWordQuads (k, i);
  Convert quads in the rotation by default user space
  Userspace used by links.
  m = (new Matrix2D).fromRotated(this,0);
  mInv = m.invert)
  r = mInv.transform (q)
  r = r.ToString)
  r = r.split(",");

  If (ckWord == '1' + / d\d\d\d\d\d\ / & & I == 70))
  {
  Console.println ("Code number is" + "Page" + (k + 1));
  }
  }
  }
  }

  It is not clear what restrictions there are on the code you are looking for, but the following will match a string that starts with "1" followed by six digits or more:

  If (ckWord.match(/^1\d{6,}$/) & I == 70) {}

• VISA read lose characters beyond the end of the output string

  Hello

  I wrote a VI to take a string of output data of an ardunio Uno and analysand. I use the vi read Visa to enter the output channel of the unit. In the end I will connect a device that actually gives the value in this type of format string: (#80212164,2289,2292,2296,2300,2328,2289,2297,2290,2300,2308,2292,2295,2298,2289,22,24,0 *).

  So after a large number of loops, the program starts to drop the last characters of the string that it generates. If the string of Visa Read output reads something like (#80212164,2289,2292,2296,2300,2328,2289,2297,2290,2300,2308,2292,2295,2298,2289,22,24,). The only way to solve this problem, once it has occurred must completely close labview (completely). Once I open again and start the program running, all is well in the world.

  Has anyone had this problem? I tried to debug it in different ways and the only weird symptom I have other Visa Read function lose a few characters of the string is the fact that by looking at the bytes to the Port after I read visa, is that she starts showing five bytes instead of zero.

  my last attempt at resolving this issue is attached.

  Note: The Ardunio outputs a string of this format every 2 seconds with the values being incrimental on a specified range. (80212164,2289,2292,2296,2300,2328,2289,2297,2290,2300,2308,2292,2295,2298,2289,22,24,0 # *)

  You can configure the * your character of termination if your Arduino does not send the carriage return or line feed.  Use the configure a Serial Port to ensure that your settings are correct.  You can also use this VI to increase your timeout so that you no longer need your waiting.