replace characters in the string

Similar Questions

• Formula to replace characters in a string

  Hi all

  Is it possible to replace characters in a string somewhere if the string has no specific pattern for these characters? I use the replacestr function to replace characters in a string (i.e. ABC), but the chain is in the ABC format * where * are two random characters. The result I'm looking for is to replace ABC * with the NodeID.

  ReplaceStr (propvalue (Core.Abbrev), ABC, ID (), T),) This formula works, but is either two additional characters in the end, after ABC. How can I cut that?

  Thank you

  If)

  Not (Equals (Integer, length (PropValue (Core.ABBREV)), length (ReplaceStr (PropValue (Core.ABBREV), ABC, T))),

  ReplaceStr (PropValue (Core.Abbrev), PropValue (Core.Abbrev), ID (), T),

  PropValue (Core.Abbrev)

  )

• 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

• 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|||_|
  

• 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

• 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

• 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

• by replacing. with \. in the string

  Hello guys

  I'm doing this script to work

  There is no environmental variable called removeip and there multiple IP addresses in it... e. g

  Event Manager environment removeip 172.17.1.0 172.17.100.0 172.31.8.0

  I use these addresses one by one in another script in regexp and so I need

  to replace points by------. So it will match correctly.

  Event Manager applet rd
  event no synchronization No.
  set action 010 string $removeip
  020 action updates $string
  030 string first action "."$string"
  040 set index action $_string_result
  050 action updates $index
  action string 060 replace '$string' $index $index '-.»
  070 action updates $string

  but action 060 let me use $index, it only accepts that the numbers - so it's useless in this form for me.

  Can you please show a way around it in applet?

  Thank you

  Leo

  Some things are better suited for Tcl.  It would be good to see your overall solution / problem such that there could be a better way to solve the system.

  However, you can use regexp to rebuild your IPs as you wish.

  foreach ip $removeip

  RegExp "^ ([0-9] +)------." ([0-9]+)\. ([0 - 9.] (+)"$ip match o1 o2 rest

  RegExp "^ ([0-9] +)------." ([0-9] +) ' $rest match o3 o4

  escip the value "$o1\.$o2\.$o3\.$o4".

  ! Do something with $escip here

• How to replace characters in the Recordset?

  Hello
  
  I need to replace some special characters, the has and o, in a recordset with & aring; & licence; and & ouml;
  
  How is that done?
  
  Thnaks!

  Thank you, Julian, which works very well now.
  Thanks again!

• Add control characters of the string without full control update

  Hello

  I want to have a chain of control where the text has different colors (like DockLight fx).

  What I CAN do now, writes a string with several lines to a control of the chain.

  Then format each line in a different color.

  But it fails so when I want to add a new text, with a new color.

  When I update the chain control, all previous formatting is crushed, and all lines have the same color.

  I want to be able to:

  Add string1 with blue color.

  Add string2 with green color

  Add string3 with red color

  etc.

  Is this possible?

  Thanks in advance!

  Awesome! Glad I could help.

• 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) {}

• How to remove the first 11 characters of a string

  With the help of ' XML Publusher Desktop / generator model for Word / 5.6 Build 45'
  
  I use XML Publisher. With the help of a RTF model, to generate purchase orders by email of the purchase of the Oracle.
  
  The fields are drawn from the XML code generated by Oracle.
  
  One of the fields is the place of delivery Description field.
  
  Will appear in the report as follows:
  
  * & lt;? SHIP_TO_LOCATION_NAME? & gt; *
  
  When I attach a test XML file in Word and view the output, one of my test case returns a description of the site:
  
  DON'T code USE County Hall of Global location
  
  What I have to do is to remove the * DO NOT USE * according to the description of the location.
  
  I could do is a Find / Replace, or a substring to ignore the first 11 characters from the string. But I don't know how to do it.
  
  Any help would be much appreciated, because it is a production problem that is causing a bit of a small problem for us!
  
  Thank you very much.

  Hello
  Maybe you know this syntax :)

  Rahul

• 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.