DateDiff

Similar Questions

• support of TimeStamp in datediff function

  Hello

  I run the Script even DIAdem Version 2012 and 2014. The Script includes the DateDiff() function of 2 channels that are identified as a date string. In the 2012 Version the Script runs without error and the 2014 Version the Script will throw an error due to overflow. In the 2014 Version it seems that the DateDiff() function does not support the stamp, because if you convert the date to a string, it works.

  Is it perhaps that in the new Version of Diadem 2014 the DateDiff() function does not support the timestamp format?

  For me, it would be interesting because of the performance of huge data sets.

  Thanks in advance,

  Thomas

  Hi Thomas,

  The variable subtype that is returned by a datetime string has changed a little, through the DIAdem versions.  Today, you have 3 options for how do retrieve value of a particular line of a datetime string:

  Set TimeChannel = Data.Root.ChannelGroups (1). Channels (1)
  
  VbsDateTime = TimeChannel.Values (1)
  
  DIA_Seconds = TimeChannel.dValues (1)
  
  Set TimeObj = TimeChannel.oValues (1)

  If you just want to compare the difference between two datetime in seconds values, I would use the Middle option above.

  Brad Turpin

  Tiara Product Support Engineer

  National Instruments

• using datediff, creating table

  Hi all

  I'm trying to calculate the number of years of a product when a new record is inserted.

  That's what I tried

  CREATE TABLE WaterMeter)

  meterID VARCHAR (25) NOT NULL,

  readingID VARCHAR (25) NOT NULL,

  meterSerialNum VARCHAR (25) NOT NULL,

  installationDate DATE NOT NULL,

  noOfYears AS (DATEDIFF (YEAR, sysdate, installationDate));

  CONSTRAINT watermeter_Pkey PRIMARY KEY (meterID).

  Watermeter_Altkey of UNIQUE CONSTRAINT (meterSerialNum)

  );

  and the following error appeared.

  Error in the command line column: 6:27

  Error report:

  SQL error: ORA-00904: "YEAR": invalid identifier

  00904, 00000 - '% s: invalid identifier '.

  * Cause:

  * Action:

  so, how can I go around to solve this issue?

  Virtual column is... virtual. It is not stored in the table. Oracle calculates the value of virtual column every time that reference you it. This is why you cannot use functions such as SYSDATE in a virtual column expression. I understand the logic underlying your table, but this is done using views. Simply create the table without column of noOfYears, then

  create or replace view WaterMeter_VW as select t.*, MONTHS_BETWEEN (sysdate, installationDate) / 12 WaterMeter noOfYears t;

  SY.

• DateDiff confusion

  Hello, everyone,

  I'm traveling all the days of the current month, using DateDiff to view the difference in days between the current date and the date of the loop, and I'm TWO days showing a difference of 0.

  <cfloop from="1" to="#DaysInMonth('01-Nov-14')#" index="ndx">
  #DateDiff('d',DateFormat(ndx & '-Nov-14'),now())#<br />
  </cfloop>
  

  The output will be

  6
  5
  4
  3
  2
  1
  0 <!--- (today, Nov 7) --->
  0 <!--- (tomorrow, Nov 8) --->
  -1
  -2
  -3
  -4
  

  How can there be two days with a difference of 0?

  V/r,

  ^_^

  DateDiff determines the number of units complete "datepart' between the two dates. In your code you are comparing two differently formatted dates. now() will include the time that well, which, once compared to the other date you formatted (which would be an hour of midnight) a difference between the dates of 7.5 hours (my time). As you've specified ' and this difference is not a full 24 hours the difference is 0.

  If you follow these steps:

  #DateDiff('d',DateFormat(ndx & '-Nov-14'), DateFormat(now()) )#
  

  You get the result you expect. You will be comparing things like for date formats.

• DATEDIFF, work in the database Oracle or not function?

  Hai all,
  
  I would like to know if DATEDIFF function work in the Oracle database or not.

  To add to what said nordine, ENDDATE - STARTDATE will give you a number that matches the number of days between two dates.

  If you want the results in hours, multiply by 24; If minutes, multiply by 24 * 60 and so on.

  You can also convert the result of an INTERVAL. There are two types of intervals:
  NUMTODSINTERVAL (ENDDATE - STARTDATE, 'DAY') or NUMTOYMINTERVAL (ENDDATE - STARTDATE, 'DAY')

• DateDiff w / incorrect results (huh?)

  Run the attached code and tell me I'm not crazy.
  
  DateDiff has not calculated the correct number starting with this date > > > (03/04/2006)... once he hit the of that date the counter does not increment and then the next date continuous incrementing... but with all the results now turned off by one.
  
  What is going on... I need to use DateDiff to get the correct number?
  
  Is there another calculation that I can use to get the difference between two dates... gives me the real number between those two dates? Don't you see the same crazy behavior on your end?
  
  Here's what I see on my side:
  
  (87)? -88
  (86)? -87
  86 - 86
  85 - 85
  
  
  Thanks for the comments.

  > I use ColdFusion Server Standard 6,1,0,63958 on Windows Server 2003.

  Shouyld refresh you your minor version of CF (ie: Install FC)
  updaters as they come out).

  ESP in this case, as the CFMX6.1 update fixes a few bugs on the dates
  (including those in dateDiff()).

  First try.

  --
  Adam

• DateDiff in query

  I need to do a DateDiff in the context of a CFQUERY rather than check in the output because I also need a CNT() of records returned as well. DateDiff results are compared to a field that determines the number of days to search results (7 days, 30 days, etc.) I tried the 2 methods listed below. Any thoughts?
  
  SELECT s.Store_City, (sc. Store_ID) COUNT AS TNCS, sc. Cook_Date
  Stores store_cookies sc, s
  WHERE s.Store_ID = sc. Store_ID
  AND (#DateDiff ("d", Now(), LSParseDateTime (STC Cook_Date)) # < = #form.) Days #)
  GROUP BY s.Store_City
  
  SELECT s.Store_City, (sc. Store_ID) COUNT AS TNCS, DateDiff ("d", Now(), LSParseDateTime (STC Cook_Date)) AS dDiff
  Stores store_cookies sc, s
  WHERE s.Store_ID = sc. Store_ID
  AND (dDiff < = #form.) Days #)
  GROUP BY s.Store_City
  
  Thank you
  Larry Moss

  How about you cfset var before the date your request #form. Days days #.
  Today, the bafore (Now()), then comparison of simple date to use in your query
  on this date?

  example code:

  CFSET datback = dateadd ("d", now(),-#form.days #) [you may have to use
  some formatting ls and the cratedatetime() function here, too...]

  SELECT s.Store_City, (sc. Store_ID) COUNT AS TNCS, sc. Cook_Date
  Stores store_cookies sc, s
  WHERE s.Store_ID = sc. Store_ID
  AND LSParseDateTime (STC Cook_Date) > = #datback #.
  GROUP BY s.Store_City

  --

  Azadi Saryev
  SABAI - Dee.com
  http://www.SABAI-Dee.com

• where datediff

  GDay, I have a question I want to select all when the date is 28 days old
  
  but I can't make it work, can someone tell me what is wrong with my code?
  
  

  Well it turns out in MySQL that we Now() to get the current date, then you wrote originally would have worked if you had used the correct syntax for dateDiff().

  Check it for all functions of time date: -.
  http://dev.MySQL.com/doc/refman/5.0/en/date-and-time-functions.html

  Here, which is what you need: -.
  DateDiff (Now (), InDate) > = 28

• Formula to determine if a specific day of the week between two dates?

  I am creating a spreadsheet to track my time PD & holiday. I usually work weekends, and my vacation allowance includes a specific maximum number of Sundays a year. Is there a way to calculate how much Sunday fall between a date range (or any day of the week, by the way)?

  (i.e. If my vacation starts February 1, 2016 and ends March 2, 2016, Sunday how are included?)

  I know that I can calculate how many weeks have passed, using = Quotient (DateDiff (Start_Date, End_Date), 7), but it is possible to include two Sundays in a period of 7 days.

  Hi Eric,.

  Here is one approach.

  

  C2 = A2 + (7−WEEKDAY(A2,2))

  This concludes the first Sunday following the start date.

  

  D2 = STRIPDURATION ((B2−C2) ÷7)

  I formatted this column without decimal places to see full weeks of the first numbers Sunday.

  Quinn

• Random date - combining forms

  Numbers ' 09-

  We will generate a random date between 1752 and 2099!

  A line, three columns over a single column as space reserved.

  A1 - C1 and E1 (or other) for the placeholder that will be hidden at any given time.

  A1 Gets the with year = HAZARD (1752, 2099).

  B1 Gets the using month = MONTHNAME (ALEA (1: 12).

  E1 determines if A1 is a leap year using = IF (OR (MOD(A1,400) = 0, AND (MOD(A1,4) = 0, MOD(A1,100) <>0)), "LY", "NOT of LY").

  It was easy... now the fun part...

  Here are the options I can not understand how to combine.

  C1 Gets the date. We are not concerned by the day of the week because the point is to understand that mentally.

  Very well. Then... 30 days has September, April, June and November. All the rest have 31. . .

  = IF (OR (B1 = "September", B1 = "April", B1 = "June", B1 = "November"), ALEA (1, 30), ALEA (1: 31))

  Except February which has 28, unless it's a leap year and then is 29, and interprets the above formula February which have 31 days.

  We will temporarily to paste the formula down in C3 and change of B1 to the constant "February" for testing.

  = IF ((ET (B1 = "Février", E1 = «LY»)), ALEA (1, 29), ALEA (1: 28))

  During the test I also changed the true and false constants functions to see if it is working properly, and he does.

  Somehow, I need to combine the two so that a single number that satisfies all the arguments appear in C1. To do this, I have to 'trap' of February by nesting it perfectly in there. It seems that it should fall after the months of 30 days trial with the other seven months with 31 days. In other words: IF cell B1 is not one of the 04:30 day month THEN assess February and if it isn't wouldn't that be, SO it's a thirty and one day months.

  Thoughts. Alternatives.

  Thank you

  -Joe

  I think you can simplify the solution with something like this:

  

  This adds a random number of days between 0 and 127104 to the start date.

  You can get the number of days between two days simply by subtracting B2 - A2, here.  Or you can use = DATEDIFF (A2, B2, "D")

  SG

• Using the formula of numbers

  Hello.

  I seems to have hit a wall with a simple spreadsheet, I'm developing.

  I want to administer the spreadsheet on my mac, but I want to be able to edit the spreadsheet through my people, using ipads. I also want to synchronize the file to the cloud so all have an updated file to use. I'm testing number and Excel. Excel works fine but the numbers I think has a bug. I want to use the numbers for several reasons.

  I've attached a screenshot of a fake spreadsheet which basically highlights what I want to achieve below. The spreadsheet works fine in number for mac, but on the iPad, I have an annoying problem that it may take a more complex formula to make it work.

  Basically, this is how I want my spreadsheet to work:

  Cells C4 - C6 and D6 - D4 are the date cells through which we enter the date in and out. The column days Out (column E) subtracts basically two dates. The return last column (column F) subtracts the Date in the column (column D) with the date in H1. So, I want to be able to sort the F column in descending order (which is the most important function of the spreadsheet, numbers of thugs are so unwanted) the spreadsheet works fine on my desktop, but on my iPad numbers puts a large number of column F (usually 73000 + according to the effective date, see line 7 in the attached photo). Those big numbers seem to be the number of days since the year "0".

  

  Here are the formulas that I currently use:

  Column E: IF (OR(C6=0,D6=0), "", D6−C6)

  Column F: IF (OR (C6 = 0, D6 = 0),"", $H$ 1−D6 ")

  Column H: today()

  I used the function DateDiff but that does not work.

  Is this a bug? Or can I somehow generate a formula that will require to work?

  Fixing unfortunately it's become outside my experience base with spreadsheets.

  Hi Duke,

  Can't tell what's going on, but I get the correct results, recording the same formulas in column d a new document worksheet in numbers for iPad.

  Results in column D are the same as yours. Results in column E are one day less than yours for the first two items (it's always may 20 here) and d 187 to the third point.

  What is the (real, not "scheduled") formula in the iPad version F4?

  Kind regards

  Barry

• How can I show age based on a birthday and empty if no birthday is entered?

  I want to show the age of a person based on their birthday today. I use:

  DATEIF (K3, today, "Y")

  It is not big, but if I'm not birthday, or do not enter a birthday it shows "2016". I wish it were empty if there is no birthday. What I have to add to the formula to get this result?

  jcapune wrote:

  I want to show the age of a person based on their birthday today. I use:

  DATEIF (K3, today, "Y")

  It is not big, but if I'm not birthday, or do not enter a birthday it shows "2016". I wish it were empty if there is no birthday.

  You can try this:

  = IF (K2 = "","", DATEDIFF (K2, TODAY (), "Y"))

  SG

• Use numbers to determine the age of individuals within a group

  Hello!

  I'm doing a spreadsheet to help me to manage all of the cattle we have on a farm.  I have all their dates recorded, but need a second column to take these dates, then auto-calcualte their age in months and days.  Is it possible to do?

  

  Select cell B2 and type (or copy and paste it here) the formula:

  = IF (A2≠"", INT (DATEDIFF (A2, TODAY (), "D") ÷365) & "Y" & MOD (DATEDIFF (A2, TODAY (), "D"), 365) & "D", "" "" ")

  shortcut for this:

  B2IF (A2≠' ", INT (DATEDIFF (A2, TODAY (),"D") ÷365) &" Y "& MOD (DATEDIFF (A2, TODAY (),"D"), 365) &"D"," "" "")

  Select cell B2, copy

  Select cell B2, at the end of the column (or), paste

• Need help with an expiration date

  I have as of today on my spreadsheet, need help to make a column of expiry date which will go to the Red 20 days before expiry. Not to find the formula, please help? Thank you.

  Hi Emtsrus,

  Conditional highlighting in column A. Select the cells and then click Conditional highlighting

  

  Add rule > Dates > in the next

  

  Another reminder, formula in B2 (and fill down)

  = DATEDIFF (TODAY (), A2, "D")

  Kind regards

  Ian.

• Get and set a Position of the pre-numbered WPF legend

  I'm creating a WPF application that displays a chart of data (several courses) and adds several annotations programmatically. The user has the ability to drag the annotation (InteractionMode = AnnotationInteractionModes.DragLabel) labels. I need to be able to record the new position of the label and load it later.

  I use the NationalInstruments.Controls.PointAnnotation a NationalInstruments.Controls.Graph object.

  I thought it would be reflected in the LabelAlignment, but that does not change the value that I set when I initially create the annotation of.

  Here's how I add the annotations on the graph:

  /// 
  
          /// Add the annotations to the graph
          /// 
          /// 
          /// 
          /// 
          private void DisplayAnnotations(DataHandler dataHandler, DateTime StartDate, DateTime EndDate)
          {
              PointAnnotation tmpAnnotation = new PointAnnotation();
              //Iterate through all the annotations in the source data set
              foreach (Post_Job_Charting.Annotation dataPointAnnotation in dataHandler.Annotations)
              {
                  TimeSpan DateDiff = EndDate - StartDate;
                  TimeSpan HalfWay = new TimeSpan(StartDate.AddSeconds(DateDiff.TotalSeconds / 2.0).Ticks);
  
                  //Only add annotations that are in the visible data range
                  if (dataPointAnnotation.Date >= StartDate && dataPointAnnotation.Date <= EndDate)
                  {
                      tmpAnnotation = new PointAnnotation();
                      tmpAnnotation.HorizontalPosition = dataPointAnnotation.Date;
                      tmpAnnotation.VerticalPosition = dataPointAnnotation.Value;
                      tmpAnnotation.Label = dataPointAnnotation.Text;
  
                      if (dataPointAnnotation.Date.Ticks < HalfWay.Ticks) { tmpAnnotation.LabelAlignment = new AnnotationLabelAlignment(BoundsAlignment.None, 0, -25); }
                      else { tmpAnnotation.LabelAlignment = new AnnotationLabelAlignment(BoundsAlignment.None, -75, -25); }
                      tmpAnnotation.InteractionMode = AnnotationInteractionModes.DragLabel;
  
                      //Show/Hide the annotation
                      if (dataPointAnnotation.Display) { tmpAnnotation.Visibility = System.Windows.Visibility.Visible; }
                      else { tmpAnnotation.Visibility = System.Windows.Visibility.Hidden; }
  
                      //Tag the annotation so I can find it later
                      tmpAnnotation.Tag = dataPointAnnotation.ID;
  
                      //Display annotation on the graph
                      niGraph.Children.Add(tmpAnnotation);
                  }
              }
          }
  

  Once I have the position of the label, I need to be able to reset the label to this post, but let's start first of all, I need to get this job.

  Any help you can offer would be greatly appreciated.

  Best regards

  -Marcel

  There is no property on the annotation of point that sets the position of the label. I created a task to look by adding this.

  However, you can monitor the changes using a custom BoundsAlignment application that registers the new offset as it is updated:

  public sealed class RecordingBoundsAlignment : BoundsAlignment {
        public Vector LastOffset { get; private set; }

  public override Point AdjustPosition( AlignmentArgs args ) {
            LastOffset = new Vector( args.XOffset, args.YOffset );

  return None.AdjustPosition( args );
        }
    }

  When you create the annotation, you use an instance of the alignment of the record, and when you want to store the position you will be able to read the value of the recorder:

  tmpAnnotation.LabelAlignment = new AnnotationLabelAlignment(
        new RecordingBoundsAlignment( ),
        initialXOffset,
        initialYOffset );

  // ...

  var recorder = (RecordingBoundsAlignment)tmpAnnotation.LabelAlignment.Alignment;
    Save( recorder.LastOffset );