In subtracting two stamps, and indicating the number of hours between them.

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• Count the number of hours between the dates off holiday/weekend

  Hi all
  
  I've not worked with dates and was recently asked to create a report where I am looking for the number of hours between two dates only count business days.
  
  So for example I have given as follows
  
  Created on 2011-03-30 15:00
  Treaty of 2011-03-30 15:03:46
  Filled 2011-04-01 17:25:02
  Posted 2011-04-01 17:45
  
  For a total of looking for the CREATION and dates
  
  50 hours and 45 minutes
  
  IM also trying to exclude weekends and holidays, I read around and was able to also find an array of dates where I main date, the HOLIDAY_IND column, and the column WEEKDAY_IND
  
  If the calendar table I have looks like
  
  CALENDER_DATE HOLIDAY_IND WEEKDAY_IND
  2011-03-31 YY
  2011-04-01 N Y
  2011-04-02 N N
  
  IM really very puzzled to know where to start
  
  I thought I'd try to write with PL/SQL, but I don't have access of the user to create procedures/functions, so looks like right up to SQL
  
  Any help is appreciated!

  Hello

  Depeneding on your data and your needs, you can do something like this:

  SELECT     created_DATE,
  ,     mailed_date
  ,     24 * ( (mailed_date - created_date)
            - (
              SELECT  COUNT (*)
              FROM        table_o_dates
              WHERE    main_date > created_date
              AND        main_date < TRUNC (mailed_date)
              AND        (   holiday_ind = 'Y'
                       OR  weekday_ind = 'N'
                    )
            )
             )               AS hours_between
  FROM     table_x
  ;
  

  I hope that answers your question.
  If not, post a small example of data (CREATE TABLE and only relevant columns, INSERT statements) for all tables and also post the results desired from these data.
  Explain, using specific examples, how you get these results from these data.
  Always tell what version of Oracle you are using.

  What do you do if created_date or mailed_date is not a working day? Examples in you data and results.

• Subtract two time_clock and get the result minutes

  Hi Experts,

  Nicely, I would subtract two type of time_clock that varcahr2 to the database and get the minutes of the result.

  Data:

  Select ' 08:00 ' as start_time_clock, 15:00 ' as the double end_time_clock

  Union of all the

  Select ' 09:00 ', 16:30 ' of the double

  Expected result:

  start_time_clock end_time_clock count_minutes

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

  08:00                      15:00                       420

  09:00                      16:30                       450

  Oracle Database 10g Enterprise Edition Release 10.2.0.4.0 - production

  
  

  Thanks in advance
  

  WITH dataset

  AS (SELECT ' 08:00 ' AS start_time_clock, 15:00 ' AS end_time_clock)

  OF THE DOUBLE

  UNION ALL

  SELECT ' 09:00 ', 16:30 ' DOUBLE)

  SELECT (to_date (end_time_clock,'HH24:MI')-to_date(start_time_clock,'HH24:MI')) * 24 * 60))

  Of THE dataset

• Dynamically calculate the number of days between two dates and amounts of split

  Hello
  
  I have searched for a solution for this, but had no success.
  I need to show the amounts broken down by days.
  
  I have a table that has an amount column and start and end dates.
  
  I need to write a query so that the amounts will be broken evenly based on the number of days between the start date and end date.
  
  For example, for this line.
  insert into my_test values (' 1, '' 3-mar-2010, ' 7 - mar - 2010 ", 1000);
  
  the query returns this (split $1,000 over 5 days)
  
  
  ID Date amount
  1 ' 3-mar-2010' 200,00
  1 ' 4-mar-2010' 200,00
  1 ' 5-mar-2010' 200,00
  1 ' 6-mar-2010' 200,00
  1 ' 7-mar-2010' 200,00
  
  
  
  create table my_test)
  ID number (10),
  start_date date,
  End_date date,
  amount number (10.2)
  );
  
  
  Select * from my_test
  
  insert into my_test values (' 1, '' 3-mar-2010, ' 7 - mar - 2010 ", 1000);
  insert into my_test values (2, 10-mar-2010 ", 19-mar-2010", 2000);
  insert into my_test values (3, 20-mar-2010 ',' 21-mar-2010, 5000);
  
  
  
  Thanks in advance.

  Hello

  One way is to join a Meter of Table , a table, or (more often) a set of results includes a line for eery number 1, 2, 3,... until the maximum number of times you need to divide a line.
  For example:

  WITH     cntr     AS
  (
       SELECT     LEVEL - 1     AS n
       FROM     (  SELECT  MAX (end_date - start_date)     AS max_day_cnt
               FROM        my_test
            )
       CONNECT BY     LEVEL <= 1 + max_day_cnt
  )
  SELECT       t.id
  ,       t.start_date + c.n                    AS dt
  ,       t.amount / (t.end_date + 1 - t.start_date)     AS amt
  FROM       my_test      t
  JOIN       cntr            c     ON     c.n <= t.end_date - t.start_date
  ORDER BY  id
  ,            dt
  ;
  

  This assumes that all dates have the same number of hours, minutes, and seconds, as is the case in your sample data.
  If this isn't the case, then use TRUNC (start_date) and TRUNC (end_date) instead of start_date and end_date or post some sample data and results if some lines do not represent a whole number of days.

• Align the two signals and measure the Phase Shift

  Hello

  I do an experiment in which I use the NI USB-6221 DAQ card. The jury is able to make 250 k samples/second. I want to measure two voltages in a circuit and find the phase shift between them at frequencies between 1 and 10000. First I ouputted a wave sinusoidal frequency variable through the Commission and applied to a test circuit. Then I used the Board to measure the two tensions consecutively (thus reducing the maximum sampling frequency at 125 k). I used the signals align VI and measured the two phases and then calculates the phase shift (VI attached in Phase 1). It worked well for the test circuit I built in which the phase shift went way logarithmique.20 degrees ~84.5 degrees and then stabilized. At frequencies above 5 000 Hz phase shift must have remained constant, but it varies more or less 1 degree. When the phase shift is 84.5 degrees, present a degree of variability is not particularly explicit. When I asked my program on the circuit that I really wanted to measure, the phase shift went from-. 5 degrees up to about 1.2 degrees. The change in the values of phase shift at high frequencies (> 3000) was environ.2 degrees. Given the small phase shift, this variation is unacceptable. Now I tried to use a sequence to each blood individually (increase the maximum sampling frequency to 250 k) and then align the two signals and measure the phase of each shift. When I use align it and re - sample Express VI to realign the two signals, I get the message "error 20333 analysis: cannot align two waveforms with dt even if their samples are not clocked in phase." Is it possible to align two signals I describe here? I enclose the new VI as Phase 2

  Matthew,

  I think I have an idea for at least part of the problem.

  I took your program data and deleted stuff DAQ.  I have converted the Signal on the chart control and looked then what was going on with the signal analysis.

  The output of the Waveforms.vi line has two waveforms, like the entry.  However, arrays of Y in the two waveforms are empty!  It does not generate an error. After some head scratching, reading the help files and try things out, that's what I think is happening: the time t0 two input signals are 1,031 seconds apart. Since the wavefoms contains 1,000 seconds of data, there is no overlap and may not align them.

  I changed the t0 on two waveforms are the same, and it lines up.  The number of items in the tables is reduced by one. Then I increased the t0 of 0.1 seconds on the first element. The output had both greater than the entry by dt t0 t0 and the size of the arrays was 224998.  Reversing the t0 two elements shifts the phase in the opposite direction.

  What that tells me, is that you can not reliably align two waveforms which do not overlap.

  I suggest that you go to 2-channel data acquisition and that it accept the reduced sample rate.  You won't get the resolution you want, but you should be able to tell if something important happens.

  You may be able to improve the equivalent resolution by taking multiple steps with a slight phase shift. This is similar to the way that old oscilloscopes of sampling (analog) worked. Take a series of measures with the signal you are currently using.  The make enough average to minimize changes due to noise. Then pass the phase of the signal of excitement to an amount that is smaller than the resolution of phase of sampling rate and repeat the measurements.  Recall that I calculated that for a 5 kHz signal sampled at 125kHz, you get a sample every 14.4 degrees. If shift you the phase of 1 degree (to the point/mathematical simulation), you get a different set of samples for excitement.  They are always separated by 14.4 degrees.  Take another series of measures. Transfer phase another degree and repeat.  As long as your sampling clocks are stable enough so that frequency does not drift significantly (and it shouldn't with your equipment), you should be able to get near resolution of what you need.  The trade-off is that you need to perform more measurements and may need to keep track of the phase shifts between the various measures.

  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
  

• 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 count the number of Sundays between the two dates

  Hello
  
  I want the number of Sundays between the two dates
  
  example of
  
  number of number of Sundays between 4 January 2013 ' and April 30, 2013 "in a select query, I have to include this as a sub query in my select statement.

  nordine B wrote:
  Hi Frank,.
  Have 1 doubt...

  In many countries the week could me "Monday". How the application handles it?
  Or did I get something wrong?

  Help, please!

  For ' IW'(ISO week) early in the day is always Monday...

  It's so simple - calculate the weeks between two dates based on Monday... This is the number of Sundays...

  NEXT_DAY is another option...

  SQL> with dd as
    2  (
    3      select TO_DATE('01-04-2013','dd-mm-yyyy') fdt, TO_DATE('30-04-2013','dd-mm-yyyy') ldt from dual
    4  )
    5  SELECT       fdt,ldt,
    6            (next_day(ldt,'sunday')-next_day(fdt-1,'sunday'))/7 sdays
    7  FROM         dd;
  
  FDT       LDT            SDAYS
  --------- --------- ----------
  01-APR-13 30-APR-13          4
  

  Published by: JAC on May 2, 2013 12:20

• 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.

• Compare and get the number of days since the same date field

  Hello
  
  I need to get the results in days when comparing the same field, date
  
  For example,.
  
  Primary_key_field Date_field
  100000002 1 January 10
  100000004 February 1st, 10
  100000005 30 April 10
  100000006 April 18, 10
  100000007 29 April 10
  100000008 May 1st, 10
  
  extract the first two date_field records based on the primary_key_field. (January 1, 2010 and February 1, 2010) and compare and find the difference and who appear in number of days, the same fetch two disks and do the same math... How can I achieve this... is it possible to make the SQL thro? Please help me solve this problem...
  
  Thank you and best regards,
  KBG.

  This is the query using functions oracle Analytics

  Primary_key_field Date_field
  100000002 1 January 10
  100000004 February 1st, 10
  100000005 30 April 10
  100000006 April 18, 10
  100000007 29 April 10
  100000008 May 1st, 10

  Select rn, date_field, next, next - date_field "Diff".
  de)
  ROW_NUMBER() SELECT over (ORDER BY primary_key_field) rn,
  date_field, LEAD (date_field) OVER (ORDER BY date_field) next
  FROM table_name)
  /

• by pressing the arrow down 'quick search' opens and inserts the number 3

  pressing arrow opens 'fast search' and inserts the number 3. pressing the left or right arrows, the slider moves correctly but also inserts the number. To ti3mes the number 3 seems sel3f 333 "insert" as it does now as I'm typing

  You are welcome.

  Thanks for reporting on your results.
  You can mark your answer above as the solution by clicking on the "Solves the problem" button next to this response.

• My one year contract is about to expire. Can I extend the same price for two months and cancel the contract without redoing the accession of another full year?

  My one year contract is about to expire. Can I extend the same price for two months and cancel the contract without redoing the accession of another full year?

  Hello

  The plan of photography is an annual plan therefore would renew for another 12 months.

  Under the terms of subscription, an early termination fee would be applied if you have decided to cancel two months in the new contract - Adobe - General conditions of subscription

  Fill the cloud creative and plans unique app are available on a monthly basis, which can be another option - https://creative.adobe.com/plans

  Kind regards

  Bev

• I am trying to remove users from my team and decrease the number of licenses that I pay. Many places tell me how to buy more, but I can't find anywhere to remove licenses to reduce my costs. Where should I look?

  I am trying to remove users from my team and decrease the number of licenses that I pay. Many places tell me how to buy more, but I can't find anywhere to remove licenses to reduce my costs. Where should I look?

  Manage your account http://forums.adobe.com/thread/1460939?tstart=0 team can help

  or

  contact Adobe staff to help
  Chat/phone: Mon - Fri 05:00-19:00 (US Pacific Time)

  Creative cloud support (all creative cloud customer service problems)
  http://helpx.Adobe.com/x-productkb/global/service-CCM.html

• 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 ();

• Limit the number of calls between areas of VCS/subareas?

  Hello

  I wonder if there is a way that we can limit the number of calls between two subareas, areas or VCS.

  The script as below:

  VCS-E (owncompany) < -----neighbor/traversal/dns="" zone="" over="" internet------=""> > VCS-E (partner)

  My intention is to limit to 5 video calls from VCS-E (partner) so that the 6th call (dial in endpoint) will be blocked.

  I know that we can control the total bandwidth between areas using links & pipes but that does not stop the remote end to lower their rate of recall and deal with several endpoints.

  Thank you.

  Hi Hafiz,

  have you tried to disable the 'downspeed by mode of appeal"in a configuration of Bandwidth?

  try disabling the "downspeed" by call OFF mode under VCS config-. > bandwidth--> configuration.

  This will not allow the calls get the downspeed but it must block because of insufficient bandwidth.

  Thank you

  Alok