Find nearest date before/after today in table/child form

[Chris Ju]

Hi,

has anyone ever tried to read the last date / next date (from the point of view of the current date) from a table / child form? Unfortunately, I have no idea and I don’t have the necessary programming skills. Maybe someone has a code snippet to achieve that.

The best alternative would be to get the nearest from the array with Math.min/Math.max, but i don’t know how to do that.

I tried the following, but can’t filter the max/min values:

function findBeforeAfter() {

	var buchungen_id = 'fld-fae5113e45f44af090693ccc0dd65ffe';
	var datum_id = 'fld-98e4dff9f7d74050a6580f44bebf4e9a';
	var buchungen = record.getFieldValue(buchungen_id);
	var now = new Date();
	var before = [];
    var after = [];
	
	console.log(now);
	
	for (var i = 0, count = buchungen.length; i < count; i++){
     	var tar = buchungen.getFieldValue(datum_id);
       	var diff = (tar - now) / (3600 * 24 * 1000);
       	
     	console.log(tar);
     	
       	if (diff > 0) {
			before.push({diff: diff, index: i});
			console.log(i);
			console.log(diff); 
        } else {
            after.push({diff: diff, index: i});
            console.log(i);
            console.log(diff);           
        }
  	}
    before.sort(function(a, b) {
        if(a.diff < b.diff) {
            return -1;
        }
        if(a.diff > b.diff) {
            return 1;
        }
        return 0;
    });

    after.sort(function(a, b) {
        if(a.diff > b.diff) {
            return -1;
        }
        if(a.diff < b.diff) {
            return 1;
        }
        return 0;
    });
    return {buchungenBefore: before, buchungenAfter: after};
}

findBeforeAfter();

Thanks in advance.
Chris

[Sam Moffatt]

You’re already iterating through everything, I’d just keep track of the lowest value and just update as you find better candidates. Here’s a quick example with artisan crafted random numbers to show what I mean:

var candidates = [ -32, 8, 21, 34, -3, 34, -10, -11, -12, 48, 42, 2, 3 ];

function findBeforeAfter() {
	var beforeValue = -Number.MAX_VALUE;
	var afterValue = Number.MAX_VALUE;
	for (var i = 0, count = candidates.length; i < count; i++) {
		var diff = candidates;
		if (diff > 0) {
			if (diff < afterValue) {
				afterValue = diff;
			}
		} else {
			if (diff > beforeValue) {
				beforeValue = diff;
			}
		}
	}
	return {"before": beforeValue, "after": afterValue};
}

JSON.stringify(findBeforeAfter());

The trick we’re using here is to initialise the before value to be the largest negative number and the after value to be the largest positive number. Then as we iterate through the array we update the value if it’s lower than the last value we saw. The first positive or negative value we see will naturally make the if condition pass (they’ll generally be closer to zero) and this also gives you a sentinel value to validate against, for example if beforeValue == -Number.INT_MAX is true then you didn’t find a before value. Eventually we traverse the full array and we’re left with the two values nearest to the zero point. Output for my artisan crafted random numbers:

{"before":-3,"after":2}

Linear traversal, constant memory usage and you can use extra variables to track other items you might care about like a reference to the record in a similar way (e.g. beforeRecord = buchungen or afterRecord = buchungen in the relevant if statement).

[Chris Ju]

That’s it, what i was searching for. I’ll implement it and publish my results here … when I get it working ;-/ …

Thanks again for your great support, Sam!

[Brendan]

So, I haven’t worked out how this would be done exactly in JavaScript, but in general what I would do is first insert the current date value into my array, then sort the array. Then find the index of the current date (in objective-c it would be something like NSInteger currentDateIndex = [array indexOfObject:currentDate]), then to get the before date, just index into the array with array[currentDateIndex - 1] and the after date array[currentDateIndex + 1]. Doing appropriate bounds checks first of course.

[Chris Ju]

Thanks, Brendan. That is also a similar but probably “simpler” and thus interesting approach… I’ll post my results!

[Author]

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