Variable number of points in visualization.arraytodatatable

[devel...@nctcog.org]

Is it possible to have a variable array of points in a line or bar
chart?
Your code example shows

var data = new google.visualization.arrayToDataTable ([

['heading1','heading2'],

[data point, datapoint],
[data point, datapoint]

]);

All of the points are hard coded, but I don't know how many points I'm
going to have. It could be any where from 0-96.

I have successfully passed in arrays so it reads as

['heading1', 'heading2'],
[arr1[0],arr2[0]],
[arr1[1],arr2[1]]
]);
and so forth but I'm having to build ridiculously large case
statements to handle the arrays so i don't get errors.

I tried building in a for next loop for the array elements, but I'm
getting syntax errors on that as well.

Here is my code with the loop that is giving syntax errors:

var data = new google.visualization.arrayToDataTable([
['Time','Count'],
for (var n=0;n<96;n++)
{
[arr2[n], parseInt(arr[n]) ],
}
[arr2(mycount),parseInt(arr[mycount])]

]);

[asgallant]

Here's one way of doing it (assumes you want everything in arr2):

var dataArray = [
    ['Time','Count']
];

// interate over all of arr2
for (var n=0; n < arr2.length; n++) { 
    dataArray.push([arr2[n], parseInt(arr[n])]), 
}
dataArray.push([arr2(mycount),parseInt(arr[mycount])]);

var data = new google.visualization.arrayToDataTable(dataArray); ​ 

[devel...@nctcog.org]

That worked with two slight modifications.  First, there isn't any need to put the final element outside the loop as I was initially doing so the line

 

dataArray.push([arr2(mycount),parseInt(arr[mycount])]);

can go away.  Also removed the comma after the array element.  Works great. Final code looks like

 

 

  var dataArray = [['Date','Count']];
 
 for (var n =0; n < arr2.length; n++)
 {
  dataArray.push ([arr2[n], parseInt(arr[n])]) }

[asgallant]

Oops, sorry, that comma was supposed to be a semi-colon  >;o)

[devel...@nctcog.org]

It actually is working without even the semicolon, though that might be bad coding syntax. 

[asgallant]

Leaving it out is bad syntax, but javascript has some limited tolerance of bad syntax, so it works in that case (but you should not count on it working).