d3.js d3.timeYear用法及代码示例

D3.js中的d3.timeYear函数用于返回日期在给定开始日期和结束日期范围内的所有年份。

用法:

d3.timeYear.range(start, end, step);

参数：该函数接受以下三个参数：

Start:此参数保存给定的开始日期。
end:此参数保存给定的结束日期。
step:它是可选参数，其中包含用于跳过年份的值。

返回值：此函数返回给定范围内的所有可能年份。

以下程序说明了D3.js中的d3.timeYear函数：

示例1:

<!DOCTYPE html> 
<html> 
   <head> 
      <title> 
          D3.js | d3.timeYear Function 
      </title> 
      <script src = 
          "https://d3js.org/d3.v4.min.js"> 
      </script> 
   </head> 
   
   <body> 
         
      <script> 
         
         // Initialising start and end date 
         var start = new Date(2015, 01, 01); 
         var end = new Date(2020, 01, 01); 
           
         // Calling the timeYear function 
         // without step value 
         var a = d3.timeYear.range(start, end); 
           
         // Getting the years values 
         console.log(a); 
      </script> 
   </body> 
</html>

输出：

["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"]

示例2:

<!DOCTYPE html> 
<html> 
   <head> 
      <title> 
          D3.js | d3.timeYear Function 
      </title> 
      <script src = 
          "https://d3js.org/d3.v4.min.js"> 
      </script> 
   </head> 
   
   <body> 
         
      <script> 
         
         // Initialising start and end date 
         var start = new Date(2015, 01, 01); 
         var end = new Date(2020, 01, 01); 
           
         // Calling the timeYear function 
         // with step value 
         var a = d3.timeYear.range(start, end, 2); 
           
         // Getting the years values 
         console.log(a); 
      </script> 
   </body> 
</html>

输出：

["2015-12-31T18:30:00.000Z","2017-12-31T18:30:00.000Z",
 "2019-12-31T18:30:00.000Z"]

参考： https://devdocs.io/d3~5/d3-time#timeYear

相关用法

注：本文由纯净天空筛选整理自Kanchan_Ray大神的英文原创作品 D3.js | d3.timeYear Function。非经特殊声明，原始代码版权归原作者所有，本译文未经允许或授权，请勿转载或复制。