Words in the string?

Similar Questions

• Choose the Second word of the string

  How to get the second word in a string. Each word is separated from the space. If there is only a single word, to return the first word

  For example:

  "abc of anus ijk" I need to go back to

  If the string is

  'abc', I need to return to abc

  Thanks in advance

  Hello

  Try this:

  Select NVL (regexp_substr (your_string, "[^ [: space:]] +' 1, 2"), your_string) double

• Replace the last occurrence of a word in a string

  Hello

  I need to replace the last occurrence of a word in the string. Example of form:

  "I like fruits and vegetables as" need to replace the last occurrence of 'like' that is just before the vegetables and not 'as' before the fruit.

  Something like that!

  SELECT
      REGEXP_REPLACE('I like fruits and also like vegetables','like','hate',1,2) data_col
  FROM dual;
  

• Certain word in a string

  Hi all

  I'm doing a reclassification project e-learning. And I'm supposed to create a series of questions for each topic by using labview. So is it possible that I can match certain word in the string to make sure that is the correct answer? Because I'm sure that respond to all users that the entry will be different. So, I go back to the main point in response. Is there anyway that I can do?

  Really appreciate your help!

  Thank you!!

  Here's another option (based on the code of Jeff).  Turn on the conditional Terminal on the loop FOR and change it to "continue if TRUE."  In this way, the loop closes as soon as a failure is located.  Just pass the result straight out of the loop.  If any do not, then the loop FOR will come out on its own (from automatic indexing) and a pass is passed out.

  

• How to get the string contents of itemByRange?

  Hi all

  Here's what I'm trying to achieve:

  I selected a Word and you want to get the content of the 1^st word in the string (the name of a person).

  

  In fact, the selected text is the source * a hyperlink and my goal is to get the name of the person designated by him.

  [0] is the same as hyperlink.source.sourceText

  Since I can not get the paragraph or line where the selected word is (or simply do not know how to do this), I'm trying to get the full story: from the 1^st character of the selected word and moving backward character by comparing its content. When it hits '\r', I know it's the beginning of the line. This way I can get the part of the line, I'm interested with the itemByRange() method, which returns "Array of Character."

  Main();
  
  function Main() {
      var doc = app.activeDocument;
      var sel = app.selection[0];
  
      GetName(sel);
  }
  
  function GetName(txt) {
      var lastCharacter = txt.characters[0];
      var character = txt.characters[0];
      var story = txt.parentStory;
      var index = lastCharacter.index;
  
      while (character.contents != "\r") {
          character = story.characters[index];
          index--;
      }
  
      // Array of Character itemByRange (from: varies, to: varies)
      // Returns the Characters within the specified range. 
      var arr = story.characters.itemByRange(character, lastCharacter);
  
      var str1 = arr.contents[0];
      var str2 = str1.match(/^\S+/);
  
      return str2;
  }
  

  My problem is that I can't get the content- str1 - by script. However, I can see in the data browser that it is there. If I click on str1, it appears:

  

  If I type str1 in the Console, I also get:

  

  I also tried to use:

  arr.getElements)

  arr.getElements () [0]

  .silence arr.getElements () [0]

  …

  etc.

  but that didn't work either.

  I must be missing something very obvious, but I'm stuck. Can someone help me, please?

  Kind regards

  Kasyan

  Hello

  I did something wrong or

  App.Selection [0]. Lines [0]. Words [0] .silence

  is a 1 Word, you are looking for?

  I hope that...

• Replace the text in the string

  I would like to know how I could continue to write something that would allow a string of variable length to replace in a longer string, subject 'inner chain' is bounded by something.
  
  In other words, assume the string is something like

  Tiger Woods +15/09/2010+

  where the character was more used as a delimiter. I would like to extract this possibility to leave Tiger Woods. I can't do a simple substr, because the string can be

  +15/09/2010+ Tiger Woods

  . It could even be

  Tiger +15/09/2010+ Woods

  . Or it could be

  Tiger Woods +September 2010+

  . In all cases, the correct return is plain Tiger Woods.
  
  Point is, I want to remove everything between the two more brands wherever they appear and return what is left.
  
  I'm guessing that maybe that's a job for regular expressions, but I dread, so yes, because I'm not very good for them!
  
  Any help appreciated, so. :-)

  and to be complete - as says hoek - regular expressions can do it well.

  SQL> with t as (
    2      select 'Tiger Woods +15/09/2010+' str from dual union
    3      select '+15/09/2010+ Tiger Woods'  from dual union
    4      select 'Tiger +15/09/2010+ Woods' from dual union
    5      select 'Tiger Woods +September 2010+' from dual
    6      )
    7  select regexp_replace(str,'\+.+\+')
    8    from t;
  
  REGEXP_REPLACE(STR,'\+.+\+')
  --------------------------------
   Tiger Woods
  Tiger  Woods
  Tiger Woods
  Tiger Woods
  
  4 rows selected.
  

  Although you will notice let outdoor spaces, you can do another to delete extra spaces.

• Oracle regular expressions - splits the string into words for

  Hello

  Nice day!

  My requirement is to split the string into words.

  So I need to identify the new line character and the semicolon (;), comma and space like terminator for string entry.

  Please note that I am currently embedded blank and the comma as separator, as shown below.

  Select regexp_substr('test)
  TO
  string in words, "([^, [: blanc:]] +) (', 1, 1) double;"

  How to integrate the semicolons and line break characters in regular expression Oracle?

  Please notify.

  Thanks and greetings

  Sree

  This has nothing to do with REGEXP. Is SQL * more parser does not not a semicolon at the end of the line:

  SQL > select ' testto, mm\;
  ERROR:
  ORA-01756: city not properly finished chain

  SQL >

  Just break the chain:

  SQL > select regexp_substr ('testto, mm\;' |) '
  2 string into words
  3 \w+',1,level ',') of double
  4. connect by level<= regexp_count('testto,mm\;'="" ||="">
  5 string in words
  6      ','\w+')
  7.

  REGEXP_SUBSTR ('TESTTO, MM\;' |') STRINGIN
  --------------------------------------
  Testto
  mm
  string
  in
  Words

  SQL >

  Or modify SQL * more the character of endpoints:

  SQL > set sqlterm.
  SQL > select regexp_substr ('testto, mm\;)
  2 string into words
  3 \w+',1,level ',') of double
  4. connect by level<=>
  5 string in words
  6      ','\w+')
  7.

  REGEXP_SUBSTR ('TESTTO, MM\;) STRINGINTOWO
  --------------------------------------
  Testto
  mm
  string
  in
  Words

  SQL >

  SY.

• Print every word in the new line for a space in a string

  Hi all

  I would print every word in the new line, if there is a space in the string.

  Is there a better way using regexp?

  declare

  number of v_count;

  v_text varchar2 (1000): = 'Oracle Database 11 g Enterprise Edition Release 11.1.0.7.0 - Production | " ';

  v_single_text varchar2 (1000);

  number of v_space_pos;

  Start

  Select regexp_count (v_text,' ') in double v_count;

  While v_count > 0

  loop

  v_space_pos: = instr (v_text,' ');

  v_single_text: = substr(v_text,1,v_space_pos-1);

  dbms_output.put_line (v_single_text);

  v_text: = substr (v_text, v_space_pos + 1);

  v_count: = v_count-1;

  end loop;

  end;

  Thank you

  Rambeau

  HI - try as

  with t as (select 'Oracle Database 11 g Enterprise Edition Release 11.1.0.7.0 - Production' |) "" double str)

  Select regexp_substr (str,'[^] +', 1, level)

  t

  connect by level<=regexp_count(str,'[^>

• regexp_substr - how to search for two different words and result based on returning a different part of the string

  Hello everyone

  I fight with this regexp_substr and I wonder if there is a way to deal with it. I have two string type in a column with the following structure:

  structure 1 always starts with "app.catalog.school.", as the following three examples

  app.catalog.school.DB.BT

  app.catalog.school.B.FTA

  app.catalog.school.ACA.CD

  structure 2 always begins with "app.catalog.admin.", as the following three examples

  app.catalog.admin.C.ABC

  app.catalog.admin.BT.AP

  app.catalog.admin.MI.RT

  If the value of the field begins with "app.catalog.school." it should return everything after the last period (.) as follows

  BT

  FREE TRADE AGREEMENT

  CD

  If the value of the field begins with "app.catalog.admin." it should return the string between the third and fourth period (.) as follows

  C

  BT

  MI

  I actually have a solution to this:

  decode)

  regexp_substr (myfield, ' [^.] +' 1, 3), 'school', regexp_substr (myfield, ' [^.] +' 1, 5), "admin", regexp_substr (myfield, ' [^.] +' 1, 4)

  )

  I am not happy with the solution above using decode(). I think I should be able to solve this problem purely with reg_exp but after hours to try again without success. You have better solution/suggestion?

  I use Oracle version 10.2. Thank you very much.

  Oops: Oracle version 10.2, here to use regexp_replace

  WITH testdata UNTIL
  (SELECT 'app.catalog.school.DB.BT' FROM DUAL str

  UNION ALL

  SELECT 'app.catalog.admin.C.ABC' FROM DUAL
  )

  Select
  Str
  , regexp_replace (str, ' ^. * (admin | school\.)) [ ^.] +)\. ([^.] +). (* $', "\2") res
  of testdata

  RES STR

  "app.catalog.school.DB.BT" 'BT '.

  'app.catalog.admin.C.ABC' 'C '.

  to 11.x, we can use regexp_substr because he has this subexpression setting more pure accident.

  WITH testdata UNTIL
  (SELECT 'app.catalog.school.DB.BT' FROM DUAL str

  UNION ALL

  SELECT 'app.catalog.admin.C.ABC' FROM DUAL
  )

  Select
  Str
  , regexp_substr (str, ' (admin | school\.)) [ ^.] +)\. ([^.] +)', 1, 1, null, res) 2.
  of testdata

  STR	RES
  app.catalog.school.DB.BT	BT
  app.catalog.admin.C.ABC	C

  Post edited by: extended chris227

• How to count the number of words in a string?

  It is only possible by counting the number of white spaces in the string?

  You can use the String.Split method to divide the string into an array using the character space as delimiter, and then check the length property of the resulting table.

• Extraction of text from the strings in a single cell of a row of cells

  Hello

  Is there an easier way to search a text / word of a string in the row of cells, and where a match is found back a header text - see the example below.

  

  The left table contains the initial positions of conformity in groups of companies. I would like to complete the table on the right with compliance positions indicated in the table on the left, using the header text (red zone). Example 1 shows that "C company" is not in line with the left table and returns a "Non-compliant" in the relevant cell for company C. Example 2 A company, C and Z are in line with the left table and returns "comply" in the relevant cell in company A, C and Z respectively. It goes the same for the partial.

  The lines must also be aligned as shown in the left column will contain a single reference (not shown in the example).

  My current approach was to simply build a small matrix and use IF (COUNTIF ($E5, "* company A *" "'), 'In line'," "") - this works fine, but is a bit messy - obtained above table, it's really what I like o would reach.

  I also tried to use nested 'FI', 'find', 'Index' and 'Match' combinations, based on a few positions at the beginning, but not could not get the correct syntax.

  Would appreciate other solutions for the construction of the table to the right.

  Thank you in advance...

  Pasel

  This may work for you:

  

  A2 = if (COUNTA ('Table 1 - company compliance Positions': $A2) > 0, IF (LEN ('Table 1 - company compliance Positions': $A2) −LEN (SUBSTITUTE ('Table 1 - company compliance Positions': $A2, A$ 1, "")) > 0, 'table 1 - company compliance Positions': $A$ 1, ' '), "") & IF (COUNTA ('Table 1 - company compliance Positions': $B2) > 0, IF (LEN ('Table 1 - company compliance Positions': $B2) −LEN (SUBSTITUTE ("table 1 - company compliance Positions")) ((((: $B2, a$ 1, ' ')) > 0, "table 1 - company compliance Positions": $B$ 1, ' '), "") & IF (COUNTA ('Table 1 - company compliance Positions': $C2) > 0, IF (LEN ('Table 1 - company compliance Positions': $C2) −LEN (SUBSTITUTE ('Table 1 - company compliance Positions': $C2, A$ 1, "")) > 0, "table 1 - company compliance Positions": $C$ 1, ""), "")

  It's shorthand dethrone select cell A2, and then type (or copy and paste it here) the formula:

  = IF (COUNTA ('Table 1 - company compliance Positions': $A2) > 0, IF (LEN ('Table 1 - company compliance Positions': $A2) −LEN (SUBSTITUTE ('Table 1 - company compliance Positions': $A2, A$ 1, "")) > 0, 'table 1 - company compliance Positions': $A$ 1, ' '), "") & IF (COUNTA ('Table 1 - company compliance Positions': $B2) > 0, IF (LEN ('Table 1 - company compliance Positions': $B2) −LEN (SUBSTITUTE ("table 1 - company compliance Positions")) ((((: $B2, a$ 1, ' ')) > 0, "table 1 - company compliance Positions": $B$ 1, ' '), "") & IF (COUNTA ('Table 1 - company compliance Positions': $C2) > 0, IF (LEN ('Table 1 - company compliance Positions': $C2) −LEN (SUBSTITUTE ('Table 1 - company compliance Positions': $C2, A$ 1, "")) > 0, "table 1 - company compliance Positions": $C$ 1, ""), "")

  Select cell A2, copy

  Select cells A2 at the end of column F, dough

• Need a bit of help with the string read/write DLL

  I have a DLL that communicates via RS232 with some instruments.

  From 2 numbers (parameter number and a number of processes) the DLL will communicate and extract the values.

  I managed to put in place the DLL to get words, characters, long and floats (single precision).

  Now, it is the documentation that accompanies it for strings:

  

  ReadString asks the DLL to get the data of the instrument, if succesfull I have others on call GetString to get the string memory.

  I have a few questions:

  -The values of her star (*) are pointers; As I understand it that it points to a place of memory where the value is stored, is - that correct?

  -How do I know if I pass the value to the DLL, or read? For example with GetString Im not sure if

  I should pass the length of the string to the DLL or if it is read

  -How to get the character returned in a string; IM assuming I need to find somewhere a byte array.

  Now I can also write strings:

  

  As a matter of f:

  -Should I add NULL at the end of the string, or labview does by default?

  My procedure is now: read the input chain, reduce it to the size of the specified string (this is constant according to the setting of the instrument), converted to a byte array and replace the last element of the array with 0 (Null).

  However, the DLL entry expected a character, that I should get in the DLL?

  _Faust wrote:

  I have a few questions:

  -The values of her star (*) are pointers; As I understand it that it points to a place of memory where the value is stored, is - that correct?

  They can be pointers or buffers. Which of the two is ambiguous in the syntax of C. However, in your case I suppose that piMsgId and pcLen be a pointer to a unique value and pszDest a string buffer.

  -How do I know if I pass the value to the DLL, or read? For example with GetString Im not sure if

  I should pass the length of the string to the DLL or if it is read

  Description is not really clear however most of the C API require today the length of the allocated buffer at the entrance to make sure that they don't overrun the buffer length. They may or may not change this value before returning to indicate the amount of data they actually informed. The pcLen parameter passed by reference probably indicates that this function does this way.

  -How to get the character returned in a string; IM assuming I need to find somewhere a byte array.

  It is not necessary. You can simply set the parameter to a string and must be passed as a C string pointer. In LabVIEW 8.5 or more you can also configure to resize this string for the number of characters as indicates it a different setting of the library to compose the node (CLN), in this case your pcLen. 8.5 you can use a table to explicitedly initialization function create a table Bbyte of the length needed, convert the byte array to string and pass to the CLN parameter. LabVIEW will return automatically only the part of the string up to the NULL character if a CLN is configured as C string pointer

  Now I can also write strings:

  As a matter of f:

  -Should I add NULL at the end of the string, or labview does by default?

  No! If you configure the parameter is a string, passed as a string pointer C, LabVIEW will do it automatically for you.

  My procedure is now: read the input chain, reduce it to the size of the specified string (this is constant according to the setting of the instrument), converted to a byte array and replace the last element of the array with 0 (Null).

  However, the DLL entry expected a character, that I should get in the DLL?

• How to get the string (specified by row and column) of txt file with labview

  Hello world

  How to get the string (specified by row and column) of txt file with labview

  THX

  As far as I know, a text file has no column.  Be more specific.  Do you mean something like the 5th word on line 4, where the words are separated by a space, and lines are separated by a newline character?  You can read from the spreadsheet String function and set the delimiter to a space.  This will produce a 2D channels table.  Then use the table to index and give the line number and column number.

• How to save the string in a file with some special characters

  Hello

  I m using MD5 - llb to create a password.

  So, I want to save this MD5-string to a file and read it later abd compared again.

  Works perfectly with a proplem:

  Some Word create a '\r' in the string of MD5 and when I save this string in a file and read it at the time its read as a '\n' for the comparison does not work.

  My question is: How can I record a string exactly as if it were somewhere and read back (also the invisible characters)

  For example, the word: heinz

  is in MD5: \r\FB^\07\A6\07T\C7\D9\C2\94\AB\C9\1DS\95 (string as display codes indicator)

  When I save the string in a file, then I rerad back: \n\FB^\07\A6\07T\C7\D9\C2\94\AB\C9\1DS\95

  I used the "write text file" and "read text file".

  What should I use to record and play back the same thing?

  THX

  Have you tried right click and deselection "convert EOL?

• Change the color of specific sets of words in the text indicator

  Hi all

  I try to write my first semi-vaste program. He successfully install drivers and serial for my test steps, which are going well. However, I have problems with creating a nice log window. I am currently using a standard text indicator and messages of PASS/FAIL indicator formatting output.

  Basically my Subvi is trying to determine if the string is SUCCESS or FAILURE and assigns a color on this basis. index and index + 4 (success/failure being 4 letter words) are indicators, that within the MAIN program loop to increase or find the next instance of SUCCESS or FAILURE.

  I took elements of the following resources:

  http://forums.NI.com/T5/LabVIEW/is-there-an-easy-method-for-appending-text-to-a-string-control/m-p/8...

  http://digital.NI.com/public.nsf/allkb/6BD344ACA4DEE20A8625692700737E16

  However, it seems that only a few words (which are not same SUCCESS or FAILURE) are awarded randomly red or green. The second passage, the whole journal is red for some reason any.

  I also tried this in the same line as the indicator of the log window, and in this case, the Word changes color, but whenever a new row is written, everything loses its color and is all black again. I can't seem to find a thread with someone using this exact application. Any help at all would be really appreciated. Thanks in advance!

  Getrich wrote:

  Thanks Steve responding. I'm trying the listbox. I am able to change listItems that you have demonstrated, to have colours. I think that your suggestion to make the color of an entire line is acceptable for my application. Is it possible to implement this in the text box? I built a lot around the use of the text box, and I have to change a little subVIs (or at least add some loops) to implement the text correctly area.

  If you want to use a textbox control, you will probably need to use the approach of the link in your message of moose.