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function timeDiff

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  • function timeDiff

    I hope you can help me with a problem on a function which should work properly but isnít.

    The function is:

    function timeDiff(time1, time2) {

    if(isNaN(Date.parse(time1)) || isNaN(Date.parse(time2))) return '';

    return ( ((Date.parse(time2) -Date.parse(time1))/1000)/60 )
    }

    The problem is that if I set time1 to say 25/05/11 and time2 to say 26/05/2011 the result should be 1440 (the number of minutes in I day) however the result I am receiving is 44640 (the number of minutes in 31 days)

    I have check the date formats on the server and the dbase and they are all set to the UK format of dd/mm/yyyy

    My data connection is SQL

    Any ideas ?

    Thanks
    Bobby123

  • #2
    Re: function timeDiff

    Hi Forum

    My problem with this function is the date format, the function is calulating the result based on the mm/dd/yyyy US format and my date format is the UK format dd/mm/yyyy, but i have as yet been unable to fine a working javascript answer to this which doesn't retun error.

    Has anyone found a way to format the date to mm/dd/yyyy for the calulation part of the function which would deliver the correct result, then change the date format back to the dd/mm/yyyy.

    I have looked at a lot of javascript sites and forums over the past week or so and have yet to find a workable solution to this

    thanks
    bobby123

    Comment


    • #3
      function timeDiff

      I am posting this question here as it is UK date related, I had posted it on the main forum and recieved no response or assistance. So I am now hoping someone in the UK will have come across the same issue and found a sulotion or can point me in the right direction.

      The function is:

      function timeDiff(time1, time2) {

      if(isNaN(Date.parse(time1)) || isNaN(Date.parse(time2))) return '';

      return ( ((Date.parse(time2) -Date.parse(time1))/1000)/60 )
      }

      The problem is that if I set time1 to say 25/05/11 and time2 to say 26/05/2011 the result should be 1440 (the number of minutes in I day) however the result I am receiving is 44640 (the number of minutes in 31 days)

      My problem with this function is the date format, the function is calulating the result based on the mm/dd/yyyy US format and my date format is the UK format dd/mm/yyyy, but i have as yet been unable to fine a working javascript answer to this which doesn't retun error.

      Has anyone found a way to format the date to mm/dd/yyyy for the calulation part of the function which would deliver the correct result, then change the date format back to the dd/mm/yyyy.

      I have looked at a lot of javascript sites and forums over the past week or so and have yet to find a workable solution to this

      My data connection is SQL

      Any ideas ?

      Thanks
      Bobby123

      Comment


      • #4
        Re: function timeDiff

        I sometimes take a sledgehammer approach to these things... specially with dates since formats can drive you crazy. So, I figure out the format the input is, then change it to what I need in code. You only need to get minutes back so you don't really need to reformat a full date back out again. You're passing dates into your function but I'm just testing here directly on a grid... other than the dates coming in, the JS should be the same for you.

        Code:
        var newDate1 = {grid.Object}.getValue('G','APPTDATE',{Grid.RowNumber});
        var newDate2 = {grid.Object}.getValue('G','APPTDATE2',{Grid.RowNumber});
        
        var date1 = new Date(newDate1.slice(3,5) + '/' + newDate1.slice(0,2) + '/' + newDate1.slice(6,10));
        var date2 = new Date(newDate2.slice(3,5) + '/' + newDate2.slice(0,2) + '/' + newDate2.slice(6,10));
        
        var yr1 = date1.getFullYear();
        var mth1 = date1.getMonth();
        var day1 = date1.getDate();
        
        var yr2 = date2.getFullYear();
        var mth2 = date2.getMonth();
        var day2 = date2.getDate();
        
        var t1 = new Date(yr1, mth1, day1, 0, 0, 0, 0);
        var t2 = new Date(yr2, mth2, day2, 0, 0, 0, 0);
        var dif = t1.getTime() - t2.getTime();
        
        var Minutes_from_T1_to_T2 = dif / 1000 / 60;
        var Minutes_Between_Dates = Math.abs(Minutes_from_T1_to_T2);
        
        alert(Minutes_Between_Dates);

        Comment


        • #5
          Re: function timeDiff

          Hi Bobby,

          I have done this with Xbasic Functions Declarations.

          After setting the variables, I get the value from the grid then Set the field value with the calculation. In my case all I wanted was the number of days. But for you to get the number of minutes multiply each day by 1440.
          Code:
          function calc_days as c (e as p)
           dim date1 as d
           dim date2 as d
           dim v_days as ndate1=e._currentRowDataNew.end_date
           date2=e._currentRowDataNew.start_date
           v_days=date1-date2
           
           e._set.days.value=v_days*1440
          end function
          Hope this gives you another point of view.
          Regards
          Keith Hubert
          Alpha Guild Member
          London.
          KHDB Management Systems
          Skype = keith.hubert


          For your day-to-day Needs, you Need an Alpha Database!

          Comment


          • #6
            Re: function timeDiff

            Thanks Keith

            I will give it a go, and get back to you

            Thanks again
            Bobby123

            Comment


            • #7
              Re: function timeDiff

              Thanks Davidk

              I will give it a try

              Thanks again

              Bobby123

              Comment

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