这是一段 UI 代码
<select id="order_unit_line_rate_806782_is_addenda_enabled" class="selects_for_487886" onchange="select_addendum(806782, this);dateShowMemory(this.options[this.selectedIndex].value, '806782');" uniqueattr="Dynamic Site Accelerator / Dynamic Site Accelerator / Additional Usage Commitment / drop down" name="order_unit_line_rate[806782][is_addenda_enabled]">
<option value="0" uniqueattr="Dynamic Site Accelerator / Dynamic Site Accelerator / Additional Usage Commitment / Fee"> Fee </option>
<option value="1" uniqueattr="Dynamic Site Accelerator / Dynamic Site Accelerator / Additional Usage Commitment / See Attached Addendum"> See Attached Addendum </option>
</select>
<option>
标签嵌套在标签内的位置<select>
。我需要click()
在第二个<option>
元素上,它是下拉列表中的一个项目。当我尝试使用 id / uniqueattrclick()
在标签上时,下拉菜单是可点击的。<select>
如何遍历<option>
嵌套在下面的标签<select>
并单击正确的项目?