尝试在 html 和 php 中创建下拉列表时出现未知问题

Question

制作一个新工具并需要一个下拉菜单，但它似乎不像我编码的那样工作。不确定到底是什么问题。已经在网上研究过，但我一直无法弄清楚。if(!isset($_POST["ReasonList"])) {  $error .= '<p><label class="text-danger">Please select a reason</label></p>'; }<div class="form-group"><label>Select Reason for Request</label><select id="ReasonList" name="ReasonList" class="form-control" value="<?php echo $ReasonList; ?>" />  <option value = "">Select...</option>  <option value = "1">Original Engineer has left the company</option>  <option value = "2">Actively involved in field work on customer site</option>  <option value = "3">No capacity due to customer mandated deadlines</option>  <option value = "4">Exception request by manager</option></select></div>如果选择了选项，则可以提交表单。如果没有选择选项，它会在 if 语句中给出错误 - 请选择一个原因

Answer 1

查看更正后的<select>元素，以及代码缩进如何使 HTML 更易于阅读和调试。

<div class="form-group">

    <label>Select Reason for Request</label>

    <select id="ReasonList" name="ReasonList" class="form-control">

        <option value = "">Select...</option>

        <option value = "1">Original Engineer has left the company</option>

        <option value = "2">Actively involved in field work on customer site</option>

        <option value = "3">No capacity due to customer mandated deadlines</option>

        <option value = "4">Exception request by manager</option>

    </select>

</div>

该<select>元素没有value属性，并且您不小心关闭了<select>beforename属性。