function timeDiff

[bobby123]

I hope you can help me with a problem on a function which should work properly but isn’t.

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

[bobby123]

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

[bobby123]

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

[Davidk]

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.

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);

[Keith Hubert]

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.

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.

[bobby123]

Thanks Keith

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

Thanks again
Bobby123

[bobby123]

Thanks Davidk

I will give it a try

Thanks again

Bobby123