Is there a way to get "timedif" just like "datedif"

[cuihee]

https://hyperformula.handsontable.com/guide/built-in-functions.html#list-of-available-functions
there is DATEDIF, but no timedif?

maybe like
=timedif(45616.0, 45616.3333, "H")
=8

[AMBudnik]

We do not have timedif among the list of available functions.

However, if you would like to create one for your own purpose, I recommend reading our custom functions guide at https://hyperformula.handsontable.com/guide/custom-functions.html#working-demo.

[cuihee]

yes, I create it , mainly copy from your DATEDIF
thank you for your efficiency reply

this is my mainly code
`private timeNumber2timedif(timeNumber: number): timediftype {
const 小数部分 = timeNumber - Math.trunc(timeNumber);
const totalSeconds = Math.trunc(小数部分 * 86400);
const h = Math.trunc(totalSeconds / 3600);
const m = Math.trunc((totalSeconds % 3600) / 60);
const s = totalSeconds % 60;
return { h, m, s };
}
timeDif(ast: any, state: any) {
return this.runFunction(
ast.args,
state,
this.metadata("TIMEDIF"),
(startTime: number, endTime: number, unit: string) => {
debugger
if (startTime > endTime) {
return new CellError(ErrorType.NUM, ErrorMessage.StartEndDate);
}
const start = this.timeNumber2timedif(startTime);
const end = this.timeNumber2timedif(endTime);
switch (unit) {
case "H":
if (end.m > start.m || (end.m === start.m && end.s >= start.s)) {
return end.h - start.h;
} else {
return end.h - start.h - 1;
}
break;
case "M":
return (
(end.h - start.h) * 60 +
(end.m - start.m) -
(end.s < start.s ? 1 : 0)
);
break;
case "S":
const end_type = this.timeNumber2timedif(endTime)
const start_type = this.timeNumber2timedif(startTime)
const endTimeTotalSeconds = end_type.h * 3600 + end_type.m * 60 + end_type.s
const startTimeTotalSeconds = start_type.h * 3600 + start_type.m * 60 + start_type.s
return endTimeTotalSeconds - startTimeTotalSeconds
break;

      default:
        return new CellError(ErrorType.NUM, ErrorMessage.BadMode);
        break;
    }
  }
);

}`

[AMBudnik]

Thank you for sharing. This may help other users who have the same need.