How to deal with NULL values in regexp_substr?

Similar Questions

• dealing with null values

  HI people
  
  req1:
  It may be a specific difference between "not" and "does not exist".
  which is the best 'there is no' or 'no', when it nulls.i.e If the subquery contains NULL values.
  
  I hope you people understand my question
  
  
  Thanks in advance
  
  Published by: newbie on February 18, 2011 03:52

  This can be useful
  http://asktom.Oracle.com/pls/asktom/f?p=100:11:3607896341573320:P11_QUESTION_ID:442029737684

• Cumulative total for each day - how to deal with nulls

  I'm on 11.2.0.3.  I want to write a query to calculate a running total of the incidents a day - this request will be used for an APEX line chart.

  
  Sample table and data:

  create table sales (
    id number primary key,
    time_of_sale date,
    item varchar2(20));
  
  
  insert into sales values (1, to_date('02-JAN-2013','DD-MON-YYYY'), 'book');
  insert into sales values (2, to_date('03-JAN-2013','DD-MON-YYYY'), 'candle');
  insert into sales values (3, to_date('05-JAN-2013','DD-MON-YYYY'), 'bicycle');
  insert into sales values (4, to_date('05-JAN-2013','DD-MON-YYYY'), 'football');
  insert into sales values (5, to_date('07-JAN-2013','DD-MON-YYYY'), 'football');
  insert into sales values (6, to_date('10-JAN-2013','DD-MON-YYYY'), 'elephant');
  insert into sales values (7, to_date('10-JAN-2013','DD-MON-YYYY'), 'turtle');
  insert into sales values (8, to_date('10-JAN-2013','DD-MON-YYYY'), 'book');
  insert into sales values (9, to_date('10-JAN-2013','DD-MON-YYYY'), 'candle');
  insert into sales values (10, to_date('10-JAN-2013','DD-MON-YYYY'), 'bicycle');
  commit;
  

  For each of the days between 1 January and 10 Jan I want to get the total functioning of the number of sales.

  So the output l would like is:

  DAY ITEMS_PER_DAY

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

  1ST JANUARY 13 0

  JANUARY 2, 13 1

  JANUARY 3, 13 2

  JANUARY 4, 13 2

  5 JANUARY 13 4

  JANUARY 6, 13 4

  JANUARY 7, 13 5

  JANUARY 8, 13 5

  JANUARY 9, 13 5

  10 JANUARY 13 10

  So even if there is no sale on January 4, we show the total of the previous day.

  If I can create a series of dates using the:

  select trunc((to_date('01-JAN-2013','DD-MON-YYYY')-1) + rownum) day 
  from dual 
  connect by rownum <= 10
  

  And I can't the cumulative aid total:

  select trunc(time_of_sale) as day, sum(count(*)) over (order by trunc(time_of_sale)) as items_per_day
  from sales
  group by trunc(time_of_sale)
  

  Overall, it gives me:

  SQL > select

  2 a.day as the day,

  3 b.items_per_day

  4 of

  5 (select trunc((to_date('01-JAN-2013','DD-MON-YYYY')-1) + rownum) as day

  6 double

  7 plug by rownum < = 10).

  8 (select trunc (time_of_sale) day, sum (count (*)) (trunc (time_of_sale) order) as items_per_day

  9 sales

  10 group by trunc (time_of_sale)) b

  11 where a.day = b.day [+] - replace by parentheses to avoid the formatting of the forum

  12 order by a.day

  13.

  DAY ITEMS_PER_DAY

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

  1st January 13 null

  JANUARY 2, 13 1

  JANUARY 3, 13 2

  Value null 4 January 13

  5 JANUARY 13 4

  Value null January 6, 13

  JANUARY 7, 13 5

  Value null January 8, 13

  Value null January 9, 13

  10 JANUARY 13 10

  It's not exactly what I'm looking for.  I played a bit with a lag, but had no success.

  Any ideas?

  Thank you

  John

  Hi, John,.

  You want to do the SUM after the outer join, like this:

  WITH days_wanted AS

  (

  SELECT TO_DATE (January 1, 2013 ', 'DD-MON-YYYY') + ROWNUM - 1 day YOU

  OF the double

  CONNECT BY ROWNUM<=>

  )

  SELECT d.day

  SUM (COUNT (s.id)) OVER (ORDER BY d.day) AS items_to_date

  OF days_wanted d

  S sale LEFT OUTER JOIN ON TRUNC (s.time_of_sale) = d.day

  GROUP BY d.day

  ORDER BY d.day

  ;

  Note that we must use COUNT (s.id) instead of COUNT (*) here, because, given that it is evaluated once the outer join, COUNT (*) would be taken into account less than $ 1 a day, because the outer join ensures there will be at least 1 row in the game for each row in days_wanted, whether or not it matches anything in sales.

  It wouldn't work using the old join syntax, too, but I suggest to use the ANSI syntax, especially for outer joins.  It reduces the amount and the complexity of the coding, and which allows to reduce the number of errors.  (Also, it avoids the problem of the display of this site.)

• How to display the line empty as a line with null values

  Hi all
  
  Pls advise me if it is possible to use a single query statement to display
  Empty row (i.e. not a single return line) as a line with null values.
  
  For example,.
  
  Select the names of names_mst whose name = "sgasfgs".
  
  Result:
  Names of
  =====
  < null >

  Hello
  If you desire to join external to double, as shown below, you still get at least a line of production

  SELECT  nm.names
  FROM            dual
  LEFT OUTER JOIN names_mst   nm  ON nm.name='sgasfgs';
  

• Remove data with null value

  Hello

  I'm using Oracle 11 g. I have a table with an id of 3, node, the value column. Combination of the column id and node, that must be taken account for deletion on the registers.

  Here, I need to delete lines with the NULL value in the value column. If for a combination of id and node with non-null values, then I need to delete rows with a null value for this combination.

  If the combination of id, node is not null value then this records should not delete.

  Below table, I need to remove the second row, for which is a value not zero VOICE CAL '10' is there, so I need to delete the row with null values. (VOICE, CAL, NULL)

  Network, FL, there is no value is non-null then I should NOT delete this line.

  This table is to have 100 s of this association, we can delete data in a single delete query?

  Or how I can delete rows with nulls for this combination.

  Tab1

  VALUE OF THE NŒUD ID

  VOICE CAL 10
  VOICE CAL NULL
  NETWORK NULL FL

  Thank you

  Hello

  oradba11 wrote:

  Hello

  I'm using Oracle 11 g. I have a table with an id of 3, node, the value column. Combination of the column id and node, that must be taken account for deletion on the registers.

  Here, I need to delete lines with the NULL value in the value column. If for a combination of id and node with non-null values, then I need to delete rows with a null value for this combination.

  If the combination of id, node is not null value then this records should not delete.

  Below table, I need to remove the second row, for which is a value not zero VOICE CAL '10' is there, so I need to delete the row with null values. (VOICE, CAL, NULL)

  Network, FL, there is no value is non-null then I should NOT delete this line.

  This table is to have 100 s of this association, we can delete data in a single delete query?

  Or how I can delete rows with nulls for this combination.

  Tab1

  VALUE OF THE NŒUD ID

  VOICE CAL 10
  VOICE CAL NULL
  NETWORK NULL FL

  Thank you

  You can do this in a single DELETE statement (it is not a request), using an EXISTS or IN the subquery.  For example:

  REMOVE table_x m

  WHERE the value IS NULL

  AND THERE ARE)

  SELECT 0

  FROM table_x s

  WHERE s.id = m.id

  AND s.node = m.node

  AND s.value IS NOT NULL

  )

  ;

  If you would care to post CREATE TABLE and INSERT instructions for the sample data, and then I could test it.

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• How to deal with operating system windows 8.1 and apps on usb device

  Hi, I have a laptop Lenovo G505. Windows 8 64-bit was preinstalled in it. Later, I upgraded to 8.1. Laptop is almost 9 months and still under warranty. Since a few days, built Lenovo solution Center alert this analysis of hardware detected failures in one or more devices. On suggestion of Lenovo, I ran hdtune pro app in my machine and it also detected 3 hard drive errors. As my machine is under warranty, Lenovo is ready to replace my hard drive with a new one. But they say that it is the responsibility of the customer to return the windows 8.1 operating system and install in the new drive and their liability is limited to the replacement of the hard disk only. I am not computer savvy and I request concerned to guide me how to deal with operating system windows 8.1 on external media like USB, DVD, and install in the new drive. I also installed a number of apps and paid games. Therefore, I would like to resume these games also files so that I need not buy again. Assistance in this regard is required.

  Hello Mohan,

  Thanks for posting your query in Microsoft Community Forum.

  I can of course understand the situation and will be happy to help with your query.

  Let me ask you;

  • You have the key product for Windows 8 or 8.1?

  You can create an installation media (DVD/USB) for Windows 8.1 and re-install when you want. However, in order to create an installation media for Windows 8.1, you need the product key. If you don't have the product key for Windows 8.1, you can not create installation media.

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  As far as your paid apps (purchased by store) are concerned, you should be able to install them as soon as you sign in with the same Microsoft Account where apps were purchased.

  You can see the steps indicated by Ronnie Vernon replied on 24 December2013, re - install paid apps.

  You can follow the steps below to create installation media.

  1. Run the program Windows 8.1 configuration or installation of Windows 8 (depending on the availability of the product key for the specific operating system.

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  2. Plug the USB key into a USB port on your PC.

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  To create a DVD

  1. Click on the ISO file , and then select the location where to save the ISO file.

  2. Insert a blank recordable DVD into your DVD drive.

  3. Burn the ISO of the DVD file using Windows disc Image burner, or another program DVD burning.

  Note: Ensure that the USB or DVD key that you use has nothing else stored on it and has a sufficient free space to download the Windows 8.1 Installer: about 2.3 GB for the 64-bit version.

  You can also read the article below to create a USB stick recovery.

  Create a recovery USB disk

  Hope this information is useful. Please feel free to answer in the case where you are facing in the future other problems with Windows.

  Thank you.

• After the recent update on 06/01/2016, pdf files created on the Tamil language there are blank pages. How to deal with all this?

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• How to deal with the change in scope

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  September, a company (X) is thrown into a perimeter (a-frame) and granted in a different consolidation perimeter (part B).

  I need perform the following acts:

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  (B) BAT profits/losses of X in September, October and December, must contribute to the results of scope B until December.

  (Operations under A) and B) can be automated? No idea how to do?

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  Thank you very much!

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• full outer join: how to get without null values?

  full outer join: how to get without null values?

  Hello

  Please go well this url

  http://asktom.Oracle.com/pls/asktom/f?p=100:11:999478429860455:P11_QUESTION_ID:6585774577187

  Thank you
  Prakash P

• How to deal with the table still growing?

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• How to deal with Menu select when there are NULL or a value with SQL

  I have a Menu(:P_55X) SELECT with:

  "NULL VALUE DISPLAY" = NULL

  'NULL RETURN VALUE' = NULL

  The graphical query is as follows.

  
  

  SELECT NULL LINK,
  START_DATE "FOR_MONTH",
  ROUND(AVG("SLA"),3) "SLA_UPTIME"
  from SLA_TIMES_ZABBIX
  LEFT JOIN (SELECT DISTINCT NODE_NAME,OS_TYPE from CMS.CMS_NODE_OS where remove_dt is NULL) OSNODE on LOWER(SLA_TIMES_ZABBIX.NODE_NAME)=LOWER(OSNODE.NODE_NAME)
  where OS_TYPE=:P55_X group by START_DATE order by START_DATE ASC;
  

  How can I accout for the (: P55_X) variable is sometimes NULL and not null sometimes. which means I can't use the '=' when values is null because I need an IS NULL it.

  The column in the database has a value or is set to NULL.

  Thank you

  Robert

  where (OS_TYPE = :P55_X OR (:P55_X IS NULL and OS_TYPE IS NULL))
  

• REGEXP_SUBSTR for the list delimited by commas with null values

  Hello
  
  I have a column that stores a list of values comma-delimited. Some of these values in the list may be null. I'm having some trouble trying to extract the values using the REGEXP_SUBSTR function when null values are present. Here are two things I've tried:

  SELECT
     REGEXP_SUBSTR (val, '[^,]*', 1, 1) pos1
    ,REGEXP_SUBSTR (val, '[^,]*', 1, 2) pos2
    ,REGEXP_SUBSTR (val, '[^,]*', 1, 3) pos3
    ,REGEXP_SUBSTR (val, '[^,]*', 1, 4) pos4
    ,REGEXP_SUBSTR (val, '[^,]*', 1, 5) pos5
  FROM (SELECT 'AAA,BBB,,DDD,,FFF' val FROM dual);
  
  POS P POS P P
  --- - --- - -
  AAA   BBB

  SELECT
     REGEXP_SUBSTR (val, '[^,]+', 1, 1) pos1
    ,REGEXP_SUBSTR (val, '[^,]+', 1, 2) pos2
    ,REGEXP_SUBSTR (val, '[^,]+', 1, 3) pos3
    ,REGEXP_SUBSTR (val, '[^,]+', 1, 4) pos4
    ,REGEXP_SUBSTR (val, '[^,]+', 1, 5) pos5
  FROM (SELECT 'AAA,BBB,,DDD,,FFF' val FROM dual);
  
  POS POS POS POS P
  --- --- --- --- -
  AAA BBB DDD FFF

  As you can see that neither calls work correctly. Anyone know how to change the regular expression pattern to handle null values? I tried various other models but could not get anyone to work in all cases.
  
  Thank you
  
  Martin
  -----
  http://www.ClariFit.com
  http://www.TalkApex.com

  Hi, Martin,.

  That's what you want:

  SELECT
     RTRIM (REGEXP_SUBSTR (val, '[^,]*,', 1, 1), ',') pos1
    ,RTRIM (REGEXP_SUBSTR (val, '[^,]*,', 1, 2), ',') pos2
    ,RTRIM (REGEXP_SUBSTR (val, '[^,]*,', 1, 3), ',') pos3
    ,RTRIM (REGEXP_SUBSTR (val, '[^,]*,', 1, 4), ',') pos4
    ,RTRIM (REGEXP_SUBSTR (val || ','
                        , '[^,]*,', 1, 5), ',') pos5
  FROM (SELECT 'AAA,BBB,,DDD,,FFF' val FROM dual);
  

  The query above works in Oracle 10 or 11, but in Oracle 11, you can also do it with only REGEXP_SUBSTR, without using RTRIM:

  SELECT
     REGEXP_SUBSTR (val, '([^,]*),|$', 1, 1, NULL, 1) pos1
    ,REGEXP_SUBSTR (val, '([^,]*),|$', 1, 2, NULL, 1) pos2
    ,REGEXP_SUBSTR (val, '([^,]*),|$', 1, 3, NULL, 1) pos3
    ,REGEXP_SUBSTR (val, '([^,]*),|$', 1, 4, NULL, 1) pos4
    ,REGEXP_SUBSTR (val, '([^,]*),|$', 1, 5, NULL, 1) pos5
  FROM (SELECT 'AAA,BBB,,DDD,,FFF' val FROM dual);
  

  The problem with your first request was that he was looking for sub channels of 0 or more non-virgules. There was as a substring. consisting of 3 characters starting at position 1, he returned "AAA", as expected. Then there was an another substring, the 0 characters, starting at position 4, so it returned NULL. Then, there was a substring of 3 characters starting at position 5, so he returned 'BBB '.

  The problem with your 2nd request was that he was looking for 1 or more non-virgules. 'DDD' is the 3rd this substring.

  Published by: Frank Kulash, on February 16, 2012 11:36
  Added Oracle 11 example

• How to replace a. null value in obiee report

  Hi gurus,

  In a single column, we get. as the data, how to replace. with the null value

  Please provide your inputs

  Thank you for your time in advance

  Try it out below mentioned formula.

  REPLACE ('service request'. "RS #"(, '.', '') "