制作一個新工具并需要一個下拉菜單，但它似乎不像我編碼的那樣工作。不確定到底是什么問題。已經在網上研究過，但我一直無法弄清楚。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>如果選擇了選項，則可以提交表單。如果沒有選擇選項，它會在 if 語句中給出錯誤 - 請選擇一個原因
1 回答
慕的地6264312
TA貢獻1817條經驗 獲得超6個贊
查看更正后的<select>元素，以及代碼縮進如何使 HTML 更易于閱讀和調試。
<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>
該<select>元素沒有value屬性，并且您不小心關閉了<select>beforename屬性。
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