【发布时间】:2019-05-30 21:38:43
【问题描述】:
这些按钮是可见的并且可以工作,但是当点击提交时它应该将选中的单选按钮的值(对于 3 个组中的每一个)分配给一个变量,但返回的值始终是未定义的。
由于一组按钮,我希望输出是值或包含值的对象
severityFormValue.on("click", (e) => {
const loggedVia = $('input[name=loggedVia]:checked').val()
const allStable = $('input[name = allStable]:checked').val()
const EnvDown = $('input[name = EnvDown]:checked').val()
});
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<div class="dropdown">
<button class="dropbtn" id="CCR_DACEbtn">CCR DACE ▼</button>
<div id="drowdownCCR_DACEID" class="dropdown-content">
Submitted by:
<input type="radio" name="loggedVia" value="phone" checked> phone
<input type="radio" name="loggedVia" value="email"> email<br>
All services stable?
<input type="radio" name="allStable" value="Yes"> Yes
<input type="radio" name="allStable" value="No" checked> No
<input type="radio" name="allStable" value="N/A"> N/A <br>
Environment Down?
<input type="radio" name="EnvDown" value="Yes"> Yes
<input type="radio" name="EnvDown" value="No" checked> No
<input type="radio" name="EnvDown" value="N/A"> N/A <br>
<button type="button" id="severityFormSubmit" >Submit</button>
</div>
</div>`
【问题讨论】:
-
什么是
severityFormValue? -
分配给按钮 severityFormSubmit 的变量
-
如果你做
$('input[name="loggedVia"]:checked').value怎么办?allStable和envDown相同。 -
@StephanSchrijver 然后你会得到
undefined因为 jQuery 对象没有value属性 -
@StephanSchrijver 之前试过,没有变化
标签: jquery html typescript radio-button undefined