Need to increment values in js array

Question

I am building a dynamic svg calendar bar. In order to calculate how many divisions the bar needs (based in months) I divide the total width of the bar by the date diff in months to figure out how many pixels should be in each division.

Basic example: There are 3 months between the two provided dates; The calendar bar is 935 pixels wide. 935/3(rounded) = 312. Each section of the calendar bar should be approx 312 pixels wide.

Now for the problem. I am using a basic loop to run the math and push the values in to a javascript array. Which works fine. What I would like to do is increment each subsequent value in the array by adding it to the previous value.

Example: 1st value is 312, second value is 614 (312+312), and the final value is 936(312+614) or (312+312+312) however the math has to work.

For some reason, I cannot for the life of me figure out how to make it push the values in this manner.

Here is the code I am currently working with. (note that some of the variables that are being updated may be declared elsewhere in the script)

function divideCalendarBar(){

    var calBarWidth = 935;  
    var calBarDivisions = Math.round(calBarWidth/cb_dateDiff);
    var leftPositions = calBarDivisions;
    var leftPositionsArray = new Array();

    for (var i = 1; i < cb_dateDiff; i++) {     
        leftPositionsArray.push(leftPositions);
    };

    cb_leftPositionsArray = leftPositionsArray;
}

Any thoughts you fine folks might have are much appreciated!

This worked as well. Ok guys, I'm gonna make sure I drink more coffee before I start posting stupid questions. Thanks for humoring me! — Joshua Fort
– Joshua Fort, Commented Feb 21, 2014 at 16:56

Matt Burland · Accepted Answer · 2014-02-21 16:54:06Z

4

1st value is 312, second value is 614 (312+312), and the final value is 936(312+614) or (312+312+312)

So you could say your first value is 312 x 1, your second is 312 + 312 = 312 x 2 and your third is 312 + 312 + 312 = 312 x 3.

Since you already have a loop with an index running from 1 to (presumably) 3:

for (var i = 1; i < cb_dateDiff; i++) {     
    leftPositionsArray.push(leftPositions * i);
};

Should work

answered Feb 21, 2014 at 16:54

Matt Burland

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3 Comments

njwags Over a year ago

I don't think this will work since you never get to i === 3, because you're using the less-than instead of less-than-or-equal

Matt Burland Over a year ago

@jwags: Since we have no idea what cb_dataDiffs value actually is, then no.

njwags Over a year ago

From the OP, you can see that cb_dateDiff is the number of elements that should end up in the leftPositionArray. Since leftPositions is defined as Math.round(calBarWidth/cb_dateDiff); we can see that the loop should run cb_dateDiff times, and since i starts at 1 it should be <=

njwags · Accepted Answer · 2014-02-21 16:53:23Z

2

So I think you should be able to do the following

for (var i = 1; i <= cb_dateDiff; i++) {     
    leftPositionsArray.push(leftPositions * i);
};

Assuming cb_dateDiff is 3 in your example from above

answered Feb 21, 2014 at 16:53

njwags

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Comments

Suman Bogati · Accepted Answer · 2014-02-21 16:52:50Z

0

If we add the last value of array to leftPositions then it would be solved your problem

for (var i = 1; i < cb_dateDiff; i++) {    
     if(leftPositionsArray.length > 0){
         leftPositionsArray.push(leftPositions+[leftPositionsArray.length-1]); 
      }else{
           leftPositionsArray.push(leftPositions); 
      } 
};

answered Feb 21, 2014 at 16:52

Suman Bogati

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Comments

Joshua Fort · Accepted Answer · 2014-02-21 17:00:19Z

0

The answer was originally provided by Matt Burland:

leftPositionsArray.push(leftPositions * i);

Thanks Matt and everyone else.

answered Feb 21, 2014 at 17:00

Joshua Fort

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