Remove 10 characters of this string

Similar Questions

• 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

• String = testname (anothername), how can I remove just (anothername) of this string

  Someone at - it techniques?

  Thank you!

  A behold, even if the regular expression could be better defined to handle more possible scenarios - it assumes a space leading the opening parenthesis...

  var string: String = "Stuff (in parentheses) going." only tested for this channel

  var model: RegExp = / \(.*\) /;

  string = string.replace (model, "");

• 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 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.

• 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

• 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

• remove special characters

  Hello

  How to remove special characters except '-'',' '. " in a string.

  Database: 10g

  Thank you

  Hello

  For example, use the REGEXP_REPLACE

  REGEXP_REPLACE (str

  , '[^-,.]'

  )

  I hope that answers your question.

  If this isn't the case, please post a small example data (CREATE TABLE and only relevant columns, INSERT statements) for all of the tables involved and the results desired from these data.

  In the case of a DML (UPDATE), for example, the sample data should show what looks like the tables before the DML, and the results will be the content of the or the tables changed after the DML.

  Explain, using specific examples, how you get these results from these data.

  Always say what version of Oracle you are using (for example, 11.2.0.2.0).

  See the FAQ forum: https://forums.oracle.com/message/9362002#9362002

• Remove unnecessary characters, space and channel numbers

  Oracle 11.2.0.1

  Windows

  I have data from the table like this:

  David Parker

  Robert Josh

  Hick 1/4

  John ' Fowler

  Bill.Gates

  Herman Patrick «»

  Andrew % Clark

  I want to delete all the numbers, special characters, more than one space between the words and make the correct name.  How can I remove extra characters from string.

  In fact, we get data that has sometimes special characters, including numbers and more spaces in name and it still creates a bad output query.

  Thank you.

  Or...

  SQL > t

  2 ALSO (SELECT "David Parker' DOUBLE str")

  3 UNION ALL

  4. SELECT 'Robert Josh' DOUBLE

  5. ANY TRADE UNION

  6. SELECT ' Hick' 1/4 OF the DOUBLE

  7 UNION ALL

  8. SELECT q ' [John 'Fowler'] ' FROM DUAL

  9 UNION ALL

  10. SELECT 'Bill.Gates' FROM DUAL

  11 UNION ALL

  12. SELECT q'[Herman 'Patrick']' FROM DUAL

  13 UNION ALL

  14 SELECT q'[Andrew % Clark]' FROM DUAL)

  15 SELECT regexp_replace (str, "[^ A-Za-z] + ',' '") final_str

  16 FROM t

  17.

  FINAL_STR

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

  David Parker

  Robert Josh

  Hick

  John Fowler

  Bill Gates

  Herman Patrick

  Andrew Clark

  7 selected lines.

  SQL >

• How use Bridge to rename the batch to remove all characters after the 32nd character?

  I would like to batch rename to remove all characters after the 32nd character. Is this possible?

  Substitution string use with use of regular expressions is selected.

  In the use of the search field

  ^(.{32})(.*)
  

  In the field using replace

  $1
  

• The analysis of the non-numeric characters in a string

  Hello

  I am communicating a robot to LabVIEW (2013) via TCP/IP and sockets. The robot is the server, and LV is the customer.

  I start with the target data in this format (in a string):

  "[0 ~ 100 ~ 0] ~ [0.96593 ~-0.25882 ~ 0 ~ 0] ~ [1 ~ 0 ~ 0 ~ 0] ~ [150 ~ 50]."

  [X, Y, Z], [Q1, Q2, Q3, Q4], [C1, C4, C6, Cx], [TCPVel, OrientVel] except replacement ',' (comma) with a ' ~ ' (tilde) because I load the target data in a spreadsheet which is a comma-delimited file.

  In LV, I take this string, the analysis of the chain and split individual items. I then pulled the singular elements as unique, 32-bit of real numbers. I take each one and convert them individually to a string, concatenate them and send them to the robot. The robot decompresses these raw bytes individually and converts them to a target of robot (position in space).

  I train (BT) check if the user has sent a bad character (non-digital) in the target data. Example of this is:

  [- 50 ~-150 ~ 0] ~ [0.96593 ~-0 ] [B2has5D882 ~ 0 ~ 0] ~ [1 ~-1 ~ 0 ~ 0] ~ [150 ~ 50]

  Now in BT, when I'm scanning the chain, these bad character becomes the entire element to zero and everything else then to zero as well. See attachment for example screen similar.

  My question is, my VI, I can analyze all non-numeric characters (that are inside the parentheses and characters that are NOT a ' ~ ' (tilde))? In this way, I can always get this item numbers and do not have this element or other items turn to zero?

  Thanks in advance for any help!

  Sorry for the question of the test

  SM

  
  

• actively remove unwanted characters

  Hi, I am writing a function to actively remove unwanted characters from an input field. It will run each time a key event occurs on the selected input fields. It will remove the quotes and things like that. That's what I have so far...

  void test(){
      char unwantedCharSet[50]={"abcD"};
      char tempString2[200];
      int matchedCharIndex;
      GetCtrlVal (panelHandle[MAIN], MAIN_STRING,tempString);
      do{
          RemoveSurroundingWhiteSpace (tempString);
          matchedCharIndex = strcspn (tempString, unwantedCharSet);
          if(matchedCharIndex==strlen(tempString)){
              SetCtrlVal (panelHandle[MAIN], MAIN_STRING, tempString);
              return;
          }
          else{
              strncpy (tempString2, tempString, matchedCharIndex);
              CopyString (tempString, matchedCharIndex, tempString,matchedCharIndex+1,(strlen(tempString)-matchedCharIndex));
              strcat (tempString2, tempString);
          }
          sprintf(tempString,"%s",tempString2);
      } while(1);
  return;
  }
  

  It's still glitchy. Does anyone have advice to do this effectively?

  Thanks in advance!

  It was my solution, though, it takes a long time to perform. Someone has any advice on a better approach (more fast/cleaner)?

  /*=====================================================================*/
  //  TEST
  /*=====================================================================*/
  void test(){
      GetCtrlVal(panelHandle[MAIN],MAIN_STRING,tempString);
      SetCtrlVal(panelHandle[MAIN],MAIN_STRING,filter(tempString,unwantedCharSet));
  return;
  }
  /*=====================================================================*/
  //  FILTER
  /*=====================================================================*/
  char *filter(char *inputString,char *filterString){
      int ptr=0,matchedCharIndex=0;
      char outputString[100];
      do{
          matchedCharIndex = strcspn (inputString,filterString);
          if(matchedCharIndex!=strlen(tempString)){
              strcpy (outputString, inputString);
              for(ptr=matchedCharIndex;ptr<=strlen(inputString);ptr++){
                  outputString[ptr]=inputString[ptr+1];
                  DebugPrintf("%s\n",outputString);
                  if(outputString[ptr]=='\n'||outputString[ptr]=='\0')
                      break;
              }
              strcpy(inputString,outputString);
          }
          else
              return inputString;
      } while(1);
  return inputString;
  }
  

• In the VI library OR: extract, control of the channel numbers are loaded automatically every time that the VI is open with "count to five: a 2 three 4.0 five." Whence this string of data?

  Even after the removal of the string "count to five: a 2 3 4.» 5."since the control of the chain and its replacement by another string, the original string returns after that the VI has been saved then reopened. Whence this string of data? I have attached a copy of the library feature. In my application, I was able to work around the problem by replacing the chain with a constant string control. But I'm still curious to know what is happening.

  Thank you
  Chuck

  Chuck,

  Chain drive was scheduled by default with the string you see.  To change it, enter the new string, right-click on the control and select

  Operations on the data > default font of the current value

  Now, save your vi.

• Someone met with this string? HKEY_USERSYS-1-5-21-1201331196-1528398142-3083030346-1000

  One of my friends noticed this string posing as a user in his notebook. She had not put in place, and no one else has access to it. We tried everything that we can think of to get rid of him, as it appears as a user with special privileges just about every program properties > Security tab. A regedit found as a user in the registry of the user software account and regular/administrative of all system folders under. Impossible to remove even when taking possession. An Internet search gave no results of all search engines.

  It is important to note that its system has been completely infected with a variety of bugs (Trojans, viruses, worms, etc.). We have been able to get rid of all but a string of unknown origin. I have a feeling that it is a bad worm, bad. The anti-malware/spyware programs or antivirus detects it. Who has been able to detect it is Spybot Search and Destroy. All other popular are unable to detect it, even the paid versions.

  I would appreciate your comments and advice that you might be able to share. Thank you.

  Hello

  This should be normal to not try to remove or modify.

  It should be under: HKEY_USERS

  The names may vary:

  HKEY_USERS\. DEFAULT

  HKEY_USERS\S-1-5-18

  HKEY_USERS\S-1-5-19

  HKEY_USERS\S-1-5-20

  (These two will vary with the reality of the user who is logged)

  HKEY_USERS\S-1-5-21-1758106260-3148989398-2386694644-1000

  HKEY_USERS\S-1-5-21-1758106260-3148989398-2386694644-1000_Classes

  HKEY_USERS
  http://pcsupport.about.com/od/termshm/g/HKEY_USERS.htm

  I hope this helps.

  Rob Brown - Microsoft MVP<- profile="" -="" windows="" expert="" -="" consumer="" :="" bicycle=""><- mark="" twain="" said="" it="">

• Removes all characters except numbers

  Hi all

  I want to reformat to following phone numbers. As you can see it in what follows, I want to delete all characters except numbers and it does not start wirh 0 0 at the beginning of the f-number. And the length is 11. What do you recommend?

  (543) 456-35-50 = > 05434563550

  (555) 555-55-55 = > 05555555555

  (210) 111-11 / 11 = > 02101111111

  Thank you

  Hello

  3018678 wrote:

  Hi all

  I want to reformat to following phone numbers. As you can see it in what follows, I want to delete all characters except numbers and it does not start wirh 0 0 at the beginning of the f-number. And the length is 11. What do you recommend?

  (543) 456-35-50-online 05434563550

  (555) 555-55-55-online 05555555555

  111-11 / 11 (210)-online 02101111111

  Thank you

  To remove all characters except the 10 digits, you can use REGEXP_REPLACE

  REGEXP_REPLACE (str, '\D')

  To put a "0" at the beginning of the string (if it isn't already begin with a '0'), you can use REGEXP_REPLACE even once:

  REGEXP_REPLACE (REGEXP_REPLACE (str, '\D')

  , '^([^0])'

  , "0\1".

  )

  I don't think that there is a way to do both things wih a single integrated function call.

  There is a requirement on the length being 11?  What happens if that comes into conflict with one of the other requirements?  For example, if the string is already long of 11 characters, but does not begin with a '0', do you want to add a "0", which makes 12 characters?  What happens if the length, after removing the numbers no, is very small, say 1?  (You can use LPAD adds however many ' 0 s you need.)  If the length (after removing the numbers) is 12 or more?  What happens if the string does not contain any what numbers at all?

  If these things are questions, then post CREATE TABLE and INSERT statements for some examples of data, accurate results, you want that data, and explanations.

  Check out the Forum FAQ: Re: 2. How can I ask a question on the forums?