regular expression - get the numbers from a string

Similar Questions

• How to get the numbers from a string?

  Hi all
  
  I have a string Let's say
  '0a1b2c3d4e5f '.
  
  I want to retrieve all the numbers out of it.
  
  012345
  
  Please Guide Me.
  
  Thank you

  Try

  SELECT REGEXP_REPLACE ('0a1b2c3d4e5f','(*[^[:digit:]]) ', ") double;

• SQL query to get the numbers from 0 to 99

  How can we write a sql query to get values from 0 to 99... This should not come from any table

  SELECT LEVEL - 1
    FROM  DUAL
    CONNECT BY LEVEL <= 100
  /
  

  SY.

• By adding a '%' to the numbers from a string

  I have an expression that works in the animation of a percentage based on a slider control.

  n = thisComp.Layer ("null 1"). Effect ("the 2 pie slice") ("Slider")

  nRound = Math.round (n * 10) / 10;

  check = 1% nRound

  If (check == 0) {}

  "" + nRound + ".." 0"+ « % »

  } else {}

  nRound

  }

  What I can't understand, why is the '%' appears only when the number is full and not there permanently? After reading Dan Ebbert page, it shows that you can add in fact (which I thought I did it correctly) but only on integers. When he drives through the 'tenth' the % does not appear.

  I'm sure it's something simple, but I can't understand it.

  Thanks for your help.

  In your if/else statement you only add the % in cases where it is an integer. You will need to change to:

  If (check == 0) {}

  "" + nRound + ".." 0"+ « % » ;

  } else {}

  "' + nRound + '% ';

  }

  Also, here is another way to write the expression:

  n = thisComp.Layer ("null 1"). Effect ("the 2 pie slice") ("Slider");

  n.value.toFixed (1) + '% ';

• Extract numbers from a string

  Hello
  i'numbers in the neck of string arrecnote
  
  Please help me in this...
  
  Published by: smile on October 7, 2010 05:33
  
  Published by: smile on October 7, 2010 06:14

  Try this to extract the numbers from a string:

  SQL> ed
  Wrote file afiedt.buf
  
    1  with t as (select 'hello-cycle monthly settlement;  865.15 SDR
    2  
  multilat net 218.15 (USD) USAWW-USACC-2007/10' col1 from dual)
    3  SELECT * FROM
    4  (SELECT REGEXP_SUBSTR(col1,'[0-9]+.[0-9]+',1,lvl) col1 FROM
    5  (select col1,level lvl
    6  from t
    7  connect by level <= LENGTH(col1) - LENGTH(REPLACE(col1,' ')) + 1))
    8* WHERE col1 IS NOT NULL
  SQL> /
  
  COL1
  --------------------------------------------------------------------------------
  865.15
  218.15
  2007/10
  
  SQL> 
  
  -- or if you just want the numbers with its decimal
  
  SQL> ed
  Wrote file afiedt.buf
  
    1  with t as (select 'hello-cycle monthly settlement;  865.15 SDR
    2  
  multilat net 218.15 124 (USD) USAWW-USACC-2007/10' col1 from dual)
    3  SELECT * FROM
    4  (SELECT REGEXP_SUBSTR(col1,'[0-9]+[.][0-9]+',1,lvl) col1 FROM
    5  (select col1,level lvl
    6  from t
    7  connect by level <= LENGTH(col1) - LENGTH(REPLACE(col1,' ')) + 1))
    8* WHERE col1 IS NOT NULL
  SQL> /
  
  COL1
  --------------------------------------------------------------------------------
  865.15
  218.15
  
  SQL> 
  

  Published by: AP on October 7, 2010 05:43

• Regular expression - get longer channel number

  I think it's easy, but I can't get it. Lets say I have a string "A1_1000", I would like to remove the 1000 using regular expressions. When I fed regular expression Match I get '1', which is not the longest number. I know other ways to do it, but I want to clean the solution in one step. Anyone know the regular expression to the right to achieve this? Thank you!

  ceties wrote:
  It is the best solution I've been able to come. I wonder if there are "more smooth way" without the cycle of.

  Since several checks are needed, I would tend to beieve we must travel opportunities. in this example

  

  I start to check at '0' offset in the string to a number. Provided that I have find a number, I check if it is longer that any previous number, I found and if so, store the new number more in the shift register.

  Have fun!

  Ben

• Vista 64-bit now I changed my OS 32-bit now my web cam does not work, tried to get the drivers from dell but did not help

  of the vista 64-bit now, I changed my OS to Vista 32 bit , now my web cam does not work, tried to get the drivers from dell, but they are not available there. My laptop is studio-1537... Remember - this is a public forum so never post private information such as numbers of mail or telephone!

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• get the length of each string in a column

  Hi all

  Is there a way to get the length of each string in a column:

  for example:

  If the table contains lines like below:

  Select 'test test1 test31' col1 of double

  Union of all the

  Select 'sectest1 sectest234' double col1

  The output should be as below:

  col1

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

  4 5 6

  8 10

  That is to say get the count of each separated from the chain of space in a column. Is this possible using regexpr_substr?

  Thanks in advance.

  Roy.

  Try the below

  SELECT LISTAGG(val1,' ') IN ONE GROUP)

  New_val ORDER BY lvl)

  Of

  (SELECT LENGTH (REGEXP_SUBSTR (col1,'[^] +', 1, LEVEL)) val1,)

  RN,

  LEVEL lvl

  Of

  (SELECT col1,

  ROWNUM rn

  Of

  (SELECT 'test test1 test31' col1 OF double)

  UNION ALL

  SELECT 'sectest1 sectest234' double col1

  )

  )

  CONNECT BY LEVEL<= regexp_count(col1,'="">

  AND PRIOR rn = rn

  AND PRIOR DBMS_RANDOM. VALUE IS NOT NULL

  )

  GROUP BY rn;

  OUTPUT:-

  =======

  col1

  ----

  4 5 6

  8 10

  Thank you

  Ann

• Remove the HTML from a string

  I have some data which are recorded by a rich text editor, but I need to display it in a datagrid without HTML tags.
  
  Someone at - it a regular expression for the elimination of all of the HTML in a string?
  
  Y at - it another way to do this in flex.
  
  Thank you

  Nevermind I found it. In case anyone is interested.

  [var myPattern: RegExp = /<[^>] * > / g
  txt = txt.replace(myPattern,"")

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