Find the time to intersampling between two channels AI USB 6251

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• find the difference in days between two dates

  Hello world
  
  I'm trying to find out the difference in days between two dates and the execution of the query that I'm passing
  
  SELECT TO_char(sysdate, 'dd/mm/yyyy') - TO_char('15/11/2011', 'dd/mm/yyyy') DAYS FROM DUAL
  
  the error I get is
  
  ORA-01722: invalid number
  01722 00000 - "invalid number."
  * Cause:
  * Action:
  
  Could someone please help.
  
  Thanks in advance

  user10636796 wrote:
  Hello world

  Thanks a lot for all the replies. I am trying to apply it in a statement to my table like this

  SELECT trunc (sysdate) - TO_char (date_last_recommended, ' dd/mm/yyyy') DAYS OF recommendation;

  SELECT trunc (sysdate) - TRUNC (date_last_recommended) DAYS OF recommendation;

• How to find the difference "within weeks" between two date values?

  Hi all

  Jdev version 11.1.1.7.1

  I used two < af:inputDate > & a < af:inputText > < af:panelFormLayout > components. My requirement is, I want to display the difference of weeks between these two day values in the inputText component when the user clicks the shape.

  Any suggestion would be appreciated.

  Kind regards

  
  

  Novel

  You can use this method to get the number of days between the day and date, and then divided by 7 to get the number of weeks

  public static long getDifferenceDaysBetweenTwoDates (d1, d2 oracle.jbo.domain.Date oracle.jbo.domain.Date)

  {

  If (d1! = null & d2! = null)

  {

  return (d1.getValue () .getTime () - d2.getValue () .getTime ()) / (24 * 60 * 60 * 1000);

  }

  return 0;

  }

  Check - http://sameh-nassar.blogspot.in/2014/10/dealing-with-dates-in-java.html

  Ashish

• Find the number of days between the dates...

  Hi friends,

  I am new to this development of blackberry applications so I do not know how to find the number of days between two days. Y at - there any API is avilable otherwise we have to write our own code. In fact, I tried GregorianCalendar but I get error that cannot find the symbol but I already imported net.rim.device.api.util and package java.util also... Please give an idea how to solve this problem.

  with respect,

  s.Kumaran.

  Use DateField class instances to represent dates on the screen.

  And your code will be as follows:

  long date1 = date1DateField.getDate ();

  long date2 = date2DateField.getDate ();

• Count the number of records between two values of keys (BTREE)

  How can I count the number of keys between two values?

  I use python driver and BTREE access method.

  = >

  Ideally, what I want is a set of whole time series data (intervals may change) to a given number of points on average. The keys are timestamps and the values are the data that it takes on average. I need to count the number of records between two timestamps so that I can divide this figure by the number of points I need and data on average. What is the best way to do it?  Or should I keep the timestamp constant intervals and use the RECNO access method?

  Thank you
  (first post btw... and why is there not a lot of people at stackoverflow that answering the questions of Berkeley DB?)

  BDB is an integrated db and there no internal counters or statistics you might grap to use for this.    You will have to do it manually.

  You can create a cursor, grap the records you want, whenever you get the next card that you bump a counter.

  If you are using RECNO, you can use a slider to obtain the number of registration (DB_GET_RECNO), and if all that you data is in

  sequentail records with no missing documents, you can find the total number of take the last rec #-original rec # + 1 to get a count.

  If you pass the SQL API, you can issue a SQL query to give you a count.  Select count (*) where...

  As you enter the data anyway, so better perhaps to count records as you go along.

  Thank you

  Mike

• How to find the time to what zero crossing point occur using zero crossing vi

  Hello 
  Im trying to find the time to the point of inflection of a Bio-Signal with the 2nd derivative, then detecet the point of zero of the derivative signal
  What I need, is to mark the point of inflection on my Bio-Signal at the point where the derivative signal crosses the zero.
  

  The second has two scales on the graph, one on each side.  The first has only a single scale. I do not change the name of VI, so both have the same name.

  Lynn

• Dynamic calculation of the number of days between two dates in a table

  Hello

  I'm working on request where I dynamically calculate the number of days between two dates in a table.

  The calculation must be dynamic, i.e., when I recover the Start_date and End_date and move to the field following (call_duration) in the same row, the difference must be calculated dynamically in this area and make sure the field read-only.

  APEX version: 5.0

  Hi BO123,

  BO123 wrote:

  Hello

  I'm working on request where I dynamically calculate the number of days between two dates in a table.

  The calculation must be dynamic, i.e., when I recover the Start_date and End_date and move to the field following (call_duration) in the same row, the difference must be calculated dynamically in this area and make sure the field read-only.

  APEX version: 5.0

  one of the way to do this by calling ajax on change of end_date.

  See the sample code given below to fetch the resulting duration and making the field read only after calculation

  Step 1: Change your page

  under CSS-> Inline, put the code below

  .row_item_disabled {
    cursor: default;
    opacity: 0.5;
    filter: alpha(opacity=50);
    pointer-events: none;
  }
  

  Step 2: Create on demand Ajax process I say CALC_DURATION

  Please check Procces Ajax, see line 6.7 How to assign a value to the variable sent by ajax call

  Declare
    p_start_date  date;
    p_end_date    date;
    p_duration number;
  Begin
    p_start_date  := to_date(apex_application.g_x01);
    p_end_date    := to_date(apex_application.g_x02);
  
     --do your calculation and assign the output to the variable p_duration
    select p_end_date - p_start_date into p_duration
      from dual;
  
    -- return calculated duration
    sys.htp.p(p_duration);
  End;
  

  Step 3: Create the javascript function

  Change your page-> the function and the declaration of the Global Variable-> put the javascript function

  You must extract the rowid in the first place, for which you want to set the time, see line 2

  assuming f06, f07 and f08 is the id of the start date, and end date columns respectively, and duration

  See no line no 8 how set the value returned by the process of ajax at the duration column

  Replace your column to the respective column identifiers in the code below

  function f_calulate_duration(pThis) {
    var row_id  = pThis.id.substr(4);
    var start_date = $('#f06_'+row_id).val();
    apex.server.process ( "CALC_DURATION", {
    x01: start_date,x02: $(pThis).val()
  }, { success: function( pData ) {
  // set duration to duration column
  $('#f08_'+row_id).val(pData);
  // disable duration column
  $("#f08_" + row_id).attr("readonly", true).addClass('row_item_disabled'); }
  });
  }
  

  Step 4: choose the end date call the javascript function

  Go to report attributes-> edit your Date column end-> column-> Attrbiutes element attributes-> put the code below

  onchange="javascript:f_calulate_duration(this);"
  

  
  

  hope this helps you,

  Kind regards

  Jitendra
  

• JavaScript anomaly on the number of days between two dates

  Use ApEx 4.0, I found an anomaly in a javascript code that calculates the number of days between two dates, the current_date and the past_date. If the past and present is the or before March 10, 2013, and the current_date lies between 10 March 2013 and November 3, 2013, the number of days will be from 1 day to less than the actual number. Between November 3, 2013 and on 4 November 2013, the increments of number by 2, then the count will be accurate from this date forward.
  
  Here are some examples:
  
  March 10, 2013 = 69 days of December 31, 2012
  March 11, 2013 = 69 days of December 31, 2012
  March 12, 2013 = 70 days of December 31, 2012
  
  November 3, 2013 = 306 days in December 31, 2012
  November 4, 2013 = 308 days in December 31, 2012
  
  11 March should be 70 and 12 March should be 71. November 3 is 307 and 4 November corrects the number of fake, which began March 11.
  
  Change the past_date to March 10, 2013 produces the following:
  
  March 10, 2013 = 0 days of March 10, 2013
  March 11, 2013 = 0 days of March 10, 2013
  March 12, 2013 = 1 days of March 10, 2013
  
  But change the past_date to 11 March 2013, product of the correct numbers:
  
  March 11, 2013 = 0 days of March 11, 2013
  March 12, 2013 = 1 days of March 11, 2013
  March 13, 2013 = 2 days of March 11, 2013
  
  I would certainly all help to determine the cause of this anomaly. Here's the javascript code:
  
  var w1 = ($v ("P48_PAST_DATE"));
  W1 = (w1.toString ());
  vmon var = (w1.substr (3.3));
  vyr var = (w1.substr (7));
  var r = (vyr.length);
  If (r == 2)
  vyr. = (parseFloat (vyr) + 2000);
  vday var = (w1.substr (0.2));
  var y = (vmon.concat ("", vday, ",", vyr));
  y = Date.parse (y);
  
  var w2 = ($v ("P48_CURRENT_DATE"));
  var vmon2 = (w2.substr (3.3));
  var vyr2 = (w2.substr (7));
  var vday2 = (w2.substr (0.2));
  var x = (vmon2.concat ("", vday2, ",", vyr2));
  x = Date.parse (x);
  
  var numdays = (x - y);
  numdays = (Math.floor(numdays / 86400000));
  $s ("P48_NUMBEROFDAYS", numdays);

  Did you google for something like "javascript number of days between two dates. I think you will find the explanation to this observation:

  This method does not work correctly if there is an advanced economies jump between the two dates.

  There are examples available to calculate the difference between two dates.

• Calculate the difference in days between two Dates

  Hello

  I'm trying to understand how to calculate the difference in days between two dates using JavaScript in LiveCycle. (Minimum = JavaScript knowledge)

  Where 'Start_Date' and 'Current_Date' are the names of the two dates in the palette of the hierarchy. (the two Date/time field)

  * Current date is using the object > value > execution property > current Date/time

  I need a text or number field showing the difference in days. (Difference_in_Days)

  I noticed the following code is pretty standard among other responses:

  var

  Start_date = new Date (Start_Date);

  var

  Current_Date = new Date (Current_Date);

  var

  nAgeMilliseconds = Current_Date.getTime) - Start_date.getTime ();

  var

  nMilliSecondsPerYear = 365 * 24 * 60 * 60 * 1000 ;

  I know there is lack of code and code above are may not be not correct.

  Please notify.

  OK, that's because of the way that javascript and works of the calculate event.  The field will be filled with whatever the script resolves at the end of execution. Technically, your script does not have a value because the last thing you do is an assignment to a variable.  Change the last line as follows:

  Math.ABS ((firstDate.getTime)

  ((- secondDate.getTime (()) / (oneDay));

  (eliminate the variable assignment) and get rid of the app.alert.  Your script will "return" (have) regardless of the value of calculation from the East and which will be stored in the field.

• Can I synchronize the activity of health between two watches from Apple?

  I have two Apple - one for work, one for out watches. If I change watches in the middle of the day, the follow-up of the first watch of activity is not synchronized with the second watch. Is it possible to enable this so that the follow-up activity is specific for the whole day?

  Thank you!

  N ° it is not possible to synchronize the activity of health between two watches from Apple.

• Could not find the Time Capsule for Time Machine backup drive

  My iMac is running El Capitan and Time Machine is unable to find the Time Capsule backup drive.  I blew the backup several times in hopes that it will fix the problem.   The only thing I found which will allow a return to the top is to unplug the Time Capsule and plug it in again.  The work around works only once, then it's back to "can't find the backup disk.

  For anyone a cela or similar questions are you go in your Utilities folder (in your applications folder) and run the utility airport to see what he says about the disks attached. This cannot work for the readers of time capsule Apple rather than third party readers who have been designated as the backup time drives machine.

• How to make the difference in percentage between two matrices?

  Hello everyone, I just want to ask how to make the difference in percentage between two binary matrices. This is to compare two images converted to matrices in LabView. Thank you.

  Well, in the current image interesting pixels are zero in the reference image. Here is a small example that would take all the zero 100% pixel, then look at the pixels of same in the other image to see how are always 0.

  Modify if needed.

  

• Find the time to restore a data file

  RDBMS version: 11.2.0.4

  Platform: Oracle Linux 6.4

  To test our backup RMAN Tape (Netbackup), we have created a table called BACKUPTEST with a single space data file.

  Then, we have removed the data file. Then was restored and recovered with RMAN's data file and tablespace was again online.

  I wanted to show the evidence to my manager that the data file has been restored. But v$ datafile. CREATION_TIME will show the time when the tablespace is imagined. Is there another way, I could find the time when the data file has been restored other than log RMAN.

  Below is an excerpt after the data file has been restored.

  $ sqlplus / as sysdba

  SQL * more: Production of the version 11.2.0.4.0 Fri dec 19 15:08:03 2013

  Copyright (c) 1982, 2013, Oracle.  All rights reserved.

  Connected to:

  Oracle Database 11 g Enterprise Edition Release 11.2.0.4.0 - 64 bit Production

  With partitioning, Real Application Clusters, Automatic Storage Management, OLAP,.

  Options of Data Mining and Real Application Testing

  SQL > select TABLESPACE_NAME, status of dba_tablespaces where nom_tablespace = 'BACKUPTEST ';

  STATUS TABLESPACE_NAME

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

  BACKUPTEST ONLINE

  SQL > alter session set nls_date_format = "DD-MM-YYYY HH24."

  Modified session.

  SQL > select CREATION_TIME of v$ datafile where FILE # = 78;

  CREATION_TIME

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

  19/12/2013-12:16

  SQL > select CREATION_TIME, last_time v$ datafile where FILE # = 78;

  CREATION_TIME LAST_TIME

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

  19/12/2013-12:16

  SQL >

  19/12/2013 12:16 is the hour when the tablespace has been created with this data file no time, it has been restored from a backup RMAN.

  Check the alert.log.  There will be messages about the RESTORATION and RECOVERY actions.

  Hemant K Collette

• To find the months and days between 2 dates

  Hello

  I want to find the months and days between 2 dates.

  For example.

  1 - Date: August 25, 2013

  2 - Date: October 23, 2013

  If we consider each month 30 days, it should give

  August 25, 2013 to August 30, 2013 = 6 days

  01-Sep-2013-30-Sep-2013 = 1 month

  October 23, 2013 to October 30, 2013 = 8 days

  Total = 1 month and 14 days.

  Kindly help as soon as possible.

  Thanks and greetings

  Suresh

  Assuming that d2 > d1,.

  where d)

  Select sysdate d1, sysdate + 56 double d2

  Union all select to_date (March 1, 2013 ',' dd-mon-yyyy "") d1, to_date (March 31, 2013 ',' dd-mon-yyyy ') d2 double

  Union all select to_date (5 February 2013 ',' dd-mon-yyyy ') d1, to_date (March 31, 2013 ',' dd-mon-yyyy "") double d2

  Union all select to_date (February 25, 2013 ',' dd-mon-yyyy "") d1, to_date (March 23, 2013 ',' dd-mon-yyyy ') d2 double

  Union all select to_date (February 25, 2013 ',' dd-mon-yyyy ') d1, to_date (March 31, 2013 ',' dd-mon-yyyy "") double d2

  Union all select to_date (August 2, 2013 ',' dd-mon-yyyy "") d1, to_date (29 October 2013 ',' dd-mon-yyyy ') d2 double

  Union all select to_date (February 1, 2013 ',' dd-mon-yyyy "") d1, to_date (May 31, 2013 ',' dd-mon-yyyy ') d2 double

  Union all select to_date (25 August 2013 ',' dd-mon-yyyy "") d1, to_date ('03-Sep-2013', 'Mon-dd-yyyy') d2 double

  Union all select to_date (July 30, 2013 ',' dd-mon-yyyy "") d1, to_date (August 31, 2013 ',' dd-mon-yyyy ') d2 double

  Union all select to_date (July 31, 2013 ',' dd-mon-yyyy ') d1, to_date (August 30, 2013 ',' dd-mon-yyyy "") double d2

  Union all select to_date (July 31, 2013 ',' dd-mon-yyyy ') d1, to_date (3 August 2013 ',' dd-mon-yyyy "") double d2

  Union all select to_date (3 July 2013 ',' dd-mon-yyyy "") d1, to_date (August 31, 2013 ',' dd-mon-yyyy ') d2 double

  Union all select to_date ('31-08-2013', ' dd-mm-yyyy'), to_date('05-10-2013','dd-mm-yyyy') of the double

  Union all select to_date ('05-02-2013', ' dd-mm-yyyy'), to_date('31-03-2013','dd-mm-yyyy') of the double

  Union all select to_date ('05-02-2013', ' dd-mm-yyyy'), to_date('05-03-2013','dd-mm-yyyy') of the double

  Union all select to_date ('05-02-2013', ' dd-mm-yyyy'), to_date('05-02-2013','dd-mm-yyyy') of the double

  )

  Select d1, d2,

  1 + 30 * trunc (months_between (d2, d1)) + LESS (extract (day of d2), 30)-LESS (excerpt (d1 day), 30)

  + CASE when extracted (d2 day)< extract(day="" from="" d1)="" then="" 30="" else="" 0="" end ="">

  d

  D1 D2 DAYSBETWEEN

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

  October 10, 2013 5 December 2013 56

  March 1, 2013 30 March 31, 2013

  5 February 2013 March 31, 2013 56

  February 25, 2013 March 23, 2013 29

  February 25, 2013 March 31, 2013 36

  August 2, 2013 29 October 2013 88

  February 1, 2013 may 31, 2013 120

  August 25, 2013 03 - Sep-2013 9

  July 30, 2013 31 August 31, 2013

  July 31, 2013 August 30, 2013 31

  July 31, 2013 3 August 2013 4

  July 3, 2013 August 31, 2013 58

  31 August 2013 5 October 2013 36

  5 February 2013 March 31, 2013 56

  5 February 2013 March 5, 2013 31

  February 5, 2013 February 5, 2013 1

  In my view, which corresponds to your rules.

• the list of numbers between two columns

  Hello
  
  I need the list of numbers between two columns.
  
  My source table contains the intervals with different ranges and I'm currently thinking dividing them by size.
  
  I was able to get a solution following the information contained in the Web page:
  http://StackOverflow.com/questions/3971981/get-list-of-numbers-in-between-two-columns
  
  The solution I've used is based on the following example:

  with t as (   
    select 10 startnum, 15 endnum from dual   
    union all   
    select 18, 22 from dual   
    union all   
    select 10000, 19999999999 from dual   
  )   
  select lvl   
    from (select level lvl  
             from dual  
          connect by level <= (select max(endnum) from t)) a  
    where exists (  
       select 1  
         from t  
        where lvl between t.startnum and t.endnum ) 

  But this doesn't work in my table if I have few and modest intervals (maximum 5 digits range).
  In the example, I was trying to understand why this might be failing, but I do not have a conclusion yet.
  There is something I noticed and that I try to understand, that is:
  If I change this query to:

  with t as (   
    select 10 startnum, 15 endnum from dual   
    union all   
    select 18, 22 from dual   
    union all   
    select 19999999990, 19999999999 from dual   
  )   
  select lvl   
    from (select level lvl  
             from dual  
          connect by level <= (select max(endnum) from t)) a  
    where exists (  
       select 1  
         from t  
        where lvl between t.startnum and t.endnum ) 

  It will increase from 2 to 3 seconds to return results to a few minutes at least (canceled after a few minutes!)
  
  Thus, using this example to understand if my problem is related to having many between start and end number or if it is linked with having small intervals between large numbers.
  
  Can someone explain to me why I have this speed of different treatment depending on the value of the interval? And why it takes longer with the small interval?
  
  I need to apply it to different beaches (from 2 ranges of numbers to 16 range) and in all ranges, I will have small intervals to check.
  
  For example:
  endnum startnum
  7341 7342
  7422-7423
  7439 7456
  
  2522200050 2522200054
  2522200146 2522200150
  
  4216969839880 4216969839884
  4216969839893 4216969839897
  
  Having this problem I can only ask the first values I have in the picture!
  
  Best regards
  Ricardo Tomas

  Hi, Ricardo,

  Here's one way:

  WITH       cntr     AS
  (
       SELECT     LEVEL - 1     AS n
       FROM     (
                SELECT  MAX ( endnum - startnum)     AS max_diff
                FROM    t
            )
       CONNECT BY     LEVEL <= max_diff + 1
  )
  SELECT  startnum + c.n
  FROM     t
  JOIN     cntr  c      ON   t.endnum  >= t.startnum + c.n
  ;
  

  In your original query, you generate all the numbers from 1 to the highest endnum. You only need enough numbers to cover the range of startnum to endnum. (This method also works if startnum is less than 1).
  Joins are usually faster than EXISTS subqueries.