<tr>
<td>Crew Name:</td>
<td><input type="text" name="CrewName" id ="CrewName"required></td>
</tr>
<tr>
<td>Crew Rank:</td>
<td>
<input type="hidden" name="CrewRank" id="CrewRank" required>
<select>
<?php
$con = getDbConnect();
if (!mysqli_connect_errno($con)) {
$queryStr = "SELECT rank "."FROM rankpage";
}
$result = mysqli_query($con, $queryStr);
while ($row = mysqli_fetch_array($result)) {
echo "<option>" . $row['rank'] . "</option>";
}
?>
</select>
</input>
</td>
</tr>
当我尝试使用下拉列表选项为我的输入表单,我可以得到下拉显示,但选择的结果不进入数据库。我不确定如何链接我的选择选项作为输入类型或id到我选择的结果
您不需要链接选择选项作为输入类型。您所需要的只是为select元素添加名称,并为option元素添加值:
<tr>
<td>Crew Name:</td>
<td>
<input type="text" name="CrewName" id ="CrewName" required />
</td>
</tr>
<tr>
<td>Crew Rank:</td>
<td>
<select name="CrewRank" id="CrewRank">
<?php
$con = getDbConnect();
if (!mysqli_connect_errno($con)) {
$queryStr = "SELECT rank " .
"FROM rankpage";
}
$result = mysqli_query($con, $queryStr);
while ($row = mysqli_fetch_array($result)) {
echo "<option value='"" . $row['rank'] . "'">" . $row['rank'] . "</option>";
}
?>
</select>
</td>
</tr>
检查option标签的value属性。您需要将值设置为POST
应该是这样的:
echo "<option value='"" . $row['rank'] . "'">" . $row['rank'] . "</option>";
传递一个name
属性(如rank
)给select
标签。你应该得到$_REQUEST['rank']
中的值。您可以使用$_GET
或$_POST
,取决于<form>