Javascript 做第二个<；选择>；如果首先<；选择>；设置为某个选项

如果第一次选择“是”，如何进行第二次选择以进行验证

<div class="form-group" ng-class="{ 'has-error' : articleEditForm.pnp.$invalid && !articleEditForm.pnp.$pristine }">
    <label class="control-label col-sm-2">Working? (yes/no) *</label>
    <div class="col-sm-10">
      <select class="form-control" name='pnp' id='pnp' ng-model="article.pnp" ng-options="o.value as o.option for o in items" required></select>
    </div>
  </div>

  <div class="form-group" ng-class="{ 'has-error' : articleAddForm.idpnpgrupe.$invalid && !articleAddForm.idpnpgrupe.$pristine }">
    <label class="control-label col-sm-2">Pnp grupa </label>
    <div class="col-sm-10">
      <select class="form-control" name='idpnpgrupe' id='idpnpgrupe' ng-model="article.idpnpgrupe" ng-options="o.id as o.opis for o in catchpnpgrupe"></select>
    </div>
  </div>

工作（是/否）*
Pnp grupa

您可以使用指令并为输入元素定义表达式。如果表达式的计算结果为

true

，

输入将是
必需的

<select class="form-control" name='idpnpgrupe' id='idpnpgrupe' ng-model="article.idpnpgrupe" ng-options="o.id as o.opis for o in catchpnpgrupe" ng-required="article.pnp=='yes'"></select>


您可以使用指令并为输入元素定义表达式。如果表达式的计算结果为
true
，
输入将是
必需的

<select class="form-control" name='idpnpgrupe' id='idpnpgrupe' ng-model="article.idpnpgrupe" ng-options="o.id as o.opis for o in catchpnpgrupe" ng-required="article.pnp=='yes'"></select>


我很高兴它有帮助！编码快乐我很高兴它有帮助！快乐编码