'Subtracting current hour’s value from previous hour’s value using javascript
I got my first touch with js yesterday and i got this small task to do a small script in bulding controller. I am reading values from certain location and calculating energy, and after 1 hour I should read the same values and subtract them to get get delta energy delta. The script should run constantly so Im using while(1) sleep(3,6 * 1000000)
Here is my code where I'm at
function summa() {
var E1 = (parseFloat(read("location1","value1")) *
parseFloat(read("location11","value11"))) / Math.pow(10,6)
executePropertyCommand("object1","Value","Write", E1)
var E2 = (parseFloat(read("location2","value2")) *
parseFloat(read("location22","value22"))) / Math.pow(10,6)
executePropertyCommand("object2","Value","Write", E2)
var IT_Sum = (E1 + E2)
return IT_Sum}
setTimeout(summa1,3.599 * 1000000);{
function summa1() {
var E1 = (parseFloat(read("location1","value1")) *
parseFloat(read("location1","value1"))) / Math.pow(10,6)
var E2 = (parseFloat(read("location2","value2")) *
parseFloat(read("location22","value2"))) / Math.pow(10,6)
var IT_Sum1 = (E1 + E2)
return IT_Sum1 }}
while(1) {
var sum1 = summa()
var sum2 = summa1()
var IT_delta = summa2 - summa1
sleep(3.6 * 1000000)}
I've tried to locate the settimeout in different locations like into the while loop but i cant seem to get the sum2 to wait for the delay.
Any ideas for better way to calculate the subtraction of same data in 1 hour loops?
Solution 1:[1]
You can add values to the array every hour and then calculate the difference of adjacent indexes.
To run code every hour use window.setTimeout and pass callback and time.
// array with values added each hour
var numbers = [];
function runHourly() {
var dateNow = new Date();
var mins = dateNow.getMinutes();
var secs = dateNow.getSeconds();
var interval = (60*(60-mins)+(60-secs))*1000;
if (interval > 0) {
window.setTimeout(runHourly, interval);
}
// your code for calculating delta goes here
numbers.push(summa());
if(numbers.length >= 2) {
var IT_delta = numbers[1] - numbers[0];
// do something with delta here
// shift array for getting delta in an hour for new values…
numbers.shift();
}
}
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | NazaRN |
