Remove characters from a string

Similar Questions

• extract characters from a string

  Hi gurus,
  I want to extract the first, third, fifth, seventh... characters in a string.
  not to extract characters from a subset of characters in a select query in oracle 9i database.
  
  for example (1):
  Assume that a string like this "SUE" is that I am 'ACEGI"of the source string
  and the source can be any valid string, I just give an example here.
  
  Example (2) of "kalpataru' in klatau'
  
  Please guide me for this

  for 9i:

  SQL> with t as (
   select 'ABCDEFGHIJ' str from dual union
   select 'kalpataru' from dual
  )
  --
  --
   select str,
          trim(extract(xmlagg(xmlelement(e, substr(str,level,1)) order by level),'//text()')) str2   from t
    where mod(level,2) = 1
  connect by level <= length(str) and prior str = str and prior sys_guid() is not null
    group by str
  /
  STR        STR2
  ---------- --------------
  ABCDEFGHIJ ACEGI
  kalpataru  klaau         
  

• 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

• How to remove characters from a file name using applescript

  Hello

  I use an AppleScript to write files from an application.

  The files are written as JPEG and PDF files

  I want to omit parts of the name of the JPEG file

  Name of the file looks like this, 70618_MEN_327926_2200x3480_1.jpg

  I need to be renamed on output, 327926_1

  That is to say, delete all characters to the end of the second line of underscores, so all characters after the third underscore and leave all the characters after the fourth underscore character.

  Hello

  You can use a Manager, like this (Manager contains explanations):

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

  tNom set to "70618_MEN_327926_2200x3480_1.jpg".

  Set newName to my getChars (tName)

  -the result is "327926_1.jpg".

  on getChars (t)

  value {oTid, point text delimiters} {text item delimiters, '_'}

  the entire at t text elements - divide the text using the line underscore as separator, the result is a list of items

  If (count l) > 4 then - at least 4 characters to underscores

  the charsAfter4_ value to (points 5 thru - 1 l) in the form of text - the text after the fourth underscore

  the value charsBetween2_3_ in section l 3 - the text between the second and the third underscores

  the value t charsBetween2_3_ & "_" & charsAfter4_ - concatenate two strings

  end if

  set the text of the delimiters to oTid - reset point of text delimiters

  return t

  getChars end

• Remove characters from a field

  Hello
  
  What is the syntax to remove "" go around a field value?
  
  Will that eliminate the 3 series of ' ' value? (as in "'' value' '")
  
  Thank you

  >
  A select statement, the "regexp_replace" works perfectly.

  When I try to do an update, I get an 'incorrect table name' error - with or without the apostrophes...

  I couldn't get the first to work at all.

  the script;

  UPDATE

  SET "FIELDHEADING' = regexp_replace ('FIELDHEADING', ' ^ (" "" ")(.*) (" "" ") $', '\2')

  OF TB_REV_1
  >
  It is not at all what you're trying to reach. What is FIELDHEADING? What is a column? If so, it doesn't take much quotes in DML. See doc UPDATE: http://download.oracle.com/docs/cd/E11882_01/server.112/e10592/statements_10008.htm#i2067715

  Display all such information as described in the article "How to ask questions" is the best way to get issues resolved.

  The regexp_replace works as follows:

  The first argument is the string to match.

  The second is the regular expression, the string is compared to.

  The third argument has the value will replace the first string substrings that match the regular expression.

  ^ - indicates the beginning of the string
  $ - indicates the end of the string
  . -matches any character
  * - means zero or more instances of the previous item
  () - items in groups.

  If ' ^ ("" "")(.*) ("" "") $"matches all of the first chain - all characters - if it starts and ends with groups of 3 quotes.

  \2 - is a reference back, referring to the groups in the regular expression in the order, so \2 means in this case (. *): all the characters between the quotes groups.

  All correspondence to the rational expression is replaced with the 2nd element in the expression.

  Takes more time to explain what to write... :-)

  http://en.Wikipedia.org/wiki/Regular_expression

• How to distinguish the figures and characters from a string

  I am looking for a simple way to do this or any built-in subVIs that can do this.

  I have 3 strings "1234c", "ab3dc". 'XYA', I want to identify those who have numbers and to reclaim all of the numbers. So an important step is to distinguish the character numeric and non-numeric characters. How can I do it in Labview with subVIs integrated?

  Thank you

  [0-9] is looking for a number, "+" makes the search for 1 or more of them (it is "greedy", so it will get as many as there are has).  I added a rope in addition to your list to show that the Pattern Match will match only the first instance of a number.  If there's a chance you have data as the last element that you need to run it in a loop.  Decimal string number can also work if a figure is the first character that it finds.

  

• Removing spaces from a string

  I have a string where there is a particular sequence, I want to replace.

  The sequence is ' / p e r.

  I tried replaceAll ("-p e r," ""); Tried to replace the charsequence, but doesn't have too much work.

  I am able to replace / by replaceAll("\\\","")

  Somehow if I put string a = "/ p e r ' and

  a.Replace ("" p e r ", pr) then came out / pr.

  But I read this to a BLOB field and there spaces between the two string operations fail somehow.

  Can someone help me on this please. I'm stuck for some time

  Please also let me know if this isn't the right group, because I don't have an answer during my last post here

  See Delete white spaces from strings in Java - Stack Overflow

• How to remove characters from desried place?

  Hi all;
  
  I need to remove the 2nd character desired to delete.
  ex-"123 - 2335-456", I need to delete 2nd onve '-' IE South result be-'123-2335456 ".
  
  How, without use of substr/instr to get 2nd place of "-"...
  
  
  Rgds,
  PC

  What is your version of db. If you can use regexp, you can do

  select regexp_replace('123-2335-456','-','',1,2) from dual
  

• Remove the unused characters from string

  Hi, I try to remove unused characters from a string, but I can't find a function for this.

  The string like: 0000 0000 0000 0000 0000 0000 0130 3130 3030 3030 3004 0000 0000 0000 0000 0000 0000 0000, I need the characters from 01 to 04.

  I need some advice.

  If the characters between the SOH and EOT had a NULL value, then Match Regular Expression can be a choice:

  Find \x01, find something else THAN \x04, then find \x04

  

• 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

• Remove 10 characters of this string

  How to remove the last 10 characters of any string?

  

  \

  Keep throwing me these easier!

• How to remove the first few characters of a string of liquid

  I would like to delete the first three characters of a string of liquid. The question is how.

  If I use the markup according to liquid when I know what are the first characters

  {{it. {{["Estate status'] | strip_html | remove_first:"11."}}

  It will remove the 11 characters.

  But if I don't know what are the first characters, this markup can I use to get "RID" of the first three characters?

  I tried 'split' and 'slice', but it must always know what characters to split or slice.

  Thanks for any help.

  Kind regards

  Carla

  Assign a value into slices and you can have table data that you can get out.

• 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

• Bunch of Oracle on the special characters in a string - translate?

  How to remove special characters in a string, it can contain. , (space), etc.

  SELECT TRANSLATE('123.456 (90-90)', '', '') FROM DUAL
  

  Basically I want to undress and to compare with the db value.

  WHERE TRANSLATE(TABLE.NUMCODE, '', '')  = TRANSLATE('123.456 (90-90)', '', '') 
  

  Use TRANSLATE, since you're on 9i. For example:

  TRANSLATE (your_string,'X,. #! % @$ ^ & * () _-+=', 'X')

  Removes all the s characters in ', #! % @$ ^ & * () _-+=' of your_string.

  SY.

• How to uninstall and remove programs from the list of programs.

  I'm trying to remove a program, but I get a message that I don't have the privileges to remove the program.  is there another way to delete other than using the uninstall feature in the control panel?

  Hello

  Remember logon as administrator.

  Other good choices is Revo Uninstaller, which will eliminate all traces of a program of MyUninstaller which provides additional information, or CCleaner.

  Revo Uninstaller - free
  http://www.revouninstaller.com/revo_uninstaller_free_download.html

  MyUninstaller is a standard utility alternative to the Add/Remove applet of windows operating
  System. It displays a list of all installed applications and allows you to uninstall an application,
  delete an entry to uninstall, and then save the list of all applications installed in a text or HTML file.
  MyUninstaller also provides additional information for installed applications more than the standard
  Add/Remove the applet does not appear: the product name, company, version, uninstall string, installation
  folder and much more.

  MyUninstaller - free
  http://www.NirSoft.NET/utils/myuninst.html

  CCleaner - Tools - uninstall section

  CCleaner - free
  http://www.CCleaner.com/

  To install one of these download - SAVE - go to where you put them - click on - RUN AS ADMIN

  How to remove an uninstall entry in programs and features for Vista and Windows 7
  http://www.Vistax64.com/tutorials/86721-programs-features-remove-uninstall-entry.html

  How to manually remove programs from the add tool / remove programs
  http://support.Microsoft.com/default.aspx?scid=kb;en-us;314481
  I hope this helps.

  Rob Brown - MS MVP - Windows Desktop Experience: Bike - Mark Twain said it right.