HTML5输入类型仅显示年份

请试一试 select and show year only guide is here

如何解析输入年份 <input date="year">

  <select class="form-control" name="startyear">
   <?php
   for ($year = (int)date('Y'); 1900 <= $year; $year--): ?>
     <option value="<?=$year;?>"><?=$year;?></option>
   <?php endfor; ?>
  </select>

JavaScript版本:

    <select name="yearpicker" id="yearpicker"></select>

    <script type="text/javascript" src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.4.1/jquery.min.js"></script>
    <script type="text/javascript">
    let startYear = 1800;
    let endYear = new Date().getFullYear();
    for (i = endYear; i > startYear; i--)
    {
      $('#yearpicker').append($('<option />').val(i).html(i));
    }
    </script>

为了提高效率,我想对Rhalp的回答做一个小小的修正。不要初始化对象219次。一般来说,如果在循环浏览之前获取成员,那么在浏览器引擎上的工作量就要小得多。

    <select name="yearpicker" id="yearpicker"></select>

    <script type="text/javascript"
            src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.4.1/jquery.min.js"> 
    </script>
    <script type="text/javascript">
    let startYear = 1800;
    let endYear = new Date().getFullYear();
    for (i = endYear; i > startYear; i--)
    {
      $('#yearpicker').append($('<option />').val(i).html(i));
    }
    </script>