D3.js | d3.timeYear Function

The d3.timeYear function in D3.js is used to return all the years with date in the given range of start and end date.
Syntax:
d3.timeYear.range(start, end, step);
Parameters: This function accept three parameters which are given below:
- Start: This parameter holds the given start date.
- end: This parameter holds the given end date.
- step: It is the optional parameter which holds the value used to skip years.
Return Value: This function returns all the possible years in the given range.
Below programs illustrate the d3.timeYear function in D3.js:
Example 1:
| <!DOCTYPE html> <html>    <head>       <title>           D3.js | d3.timeYear Function       </title>       <script src =       </script>    </head>   Â   <body>         Â      <script>         Â         // Initialising start and end date          varstart = newDate(2015, 01, 01);          varend = newDate(2020, 01, 01);           Â         // Calling the timeYear function          // without step value          vara = d3.timeYear.range(start, end);           Â         // Getting the years values          console.log(a);       </script>    </body> </html>  | 
Output:
["2015-12-31T18:30:00.000Z","2016-12-31T18:30:00.000Z", "2017-12-31T18:30:00.000Z","2018-12-31T18:30:00.000Z", "2019-12-31T18:30:00.000Z"]
Example 2:
| <!DOCTYPE html> <html>    <head>       <title>           D3.js | d3.timeYear Function       </title>       <script src =       </script>    </head>   Â   <body>         Â      <script>         Â         // Initialising start and end date          varstart = newDate(2015, 01, 01);          varend = newDate(2020, 01, 01);           Â         // Calling the timeYear function          // with step value          vara = d3.timeYear.range(start, end, 2);           Â         // Getting the years values          console.log(a);       </script>    </body> </html>  | 
Output:
["2015-12-31T18:30:00.000Z","2017-12-31T18:30:00.000Z", "2019-12-31T18:30:00.000Z"]
Reference: https://devdocs.io/d3~5/d3-time#timeYear
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