Php 如何修复下拉框中的选定项
我有一个编辑表单,从数据库中获取信息Php 如何修复下拉框中的选定项,php,html,mysql,Php,Html,Mysql,我有一个编辑表单,从数据库中获取信息 <select name="table"> <?php //fetch all tables from database $user = $con->query("SELECT * FROM table") or die(mysql_error()); while($row = $table->fetch_object()) { ?> <option val
<select name="table">
<?php
//fetch all tables from database
$user = $con->query("SELECT * FROM table") or die(mysql_error());
while($row = $table->fetch_object()) { ?>
<option value="<?php echo $row->tablename;?>">
<?php echo $row->tablename; ?>
</option> <?php }?> </select>
<label for="time">Time :</label>
<select name="time">
<option value="twotothree">2PM-3PM</option>
<option value="threetofour">3PM-4PM</option>
<option value="fourtofive">4PM-5PM</option>
<option value="fivetosix">5PM-6PM</option>
</select>
呈现选择列表时,可以添加选定属性,具体取决于可用的$\u POST变量。例如,对于表选择元素:
<select name="table">
<?php
$user = $con->query("SELECT * FROM table") or die(mysql_error());
while($row = $table->fetch_object()) { ?>
<option value="<?php echo $row->tablename;?>" <?php if (isset($_POST['table']) && $_POST['table'] == $row->tablename) echo 'selected'; ?> >
<?php echo $row->tablename; ?>
</option>
<?php }?>
</select>
如果($_POST){$tablename=trim(mysqli_real_escape_string($con,$_POST['fortables'));$user=trim(mysqli_real_escape_string($con,$_POST['user']);$time=trim(mysqli_real_escape_string($con time);$result_用户=$con(“从表中选择*,其中id='$tablename'”;$rowuseradd=$result\u user->fetch\u object();$current=$rowuseradd->$time;//更新表$actual=$current'.$user;$con->query(“更新表集$time='$actual'其中id='$tablename'”);”用户:另外,我还有其他用户的表格和时间,所以你的页面被刷新了,下拉列表中的选定项目丢失了?根据下面的评论