如何在php中将下拉列表中的选定值保存到mysql

这是我的密码。我想将从4下拉列表中选择的选项保存到mysql。当我提交表单时，从最后一个框中选择的值将保存在mysql表的所有行中。现在，请告诉我该怎么办？

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我正在正确地从数据库表中获取下拉框值，所以有什么问题吗？

这与其他文本框类似

<?php 

  $status=&$_POST['status'];
  $from=&$_POST['date_from'];
  $to=&$_POST['date_to'];
  $conn=mysqli_connect('localhost','root','','punbus') or die("Database not connected".mysqli_error());

  if(isset($_POST['sub'])){
    $ins="insert into driver_status(driver_name,status,date_from,date_to)                    
          select Driver_name,'$status','$from','$to' from driver_master";    
    if(mysqli_query($conn,$ins)){
      echo "added";
    }
    else{
      echo "NOT".mysqli_error($conn);
    }
  }

  $sel='select Driver_name from driver_master';
  $query=mysqli_query($conn,$sel);

  echo "<form action='driver_status.php' method='post'>";
  echo "<table cellpadding=5>";
  echo "<tr>";
  echo "<th>Driver Name</th>";
  echo "<th>Status</th>";
  echo "<th>From</th>";
  echo "<th>To</th>";
  echo "</tr>";

  while($row=mysqli_fetch_assoc($query)){    
    echo "<tr>";
    echo "<td>".$row['Driver_name']."</td>";
    $sel1='select d_status from status';
    $query1=mysqli_query($conn,$sel1);
?>

  <td>
    <select name="status">
      <?php
        while($row1=mysqli_fetch_assoc($query1)){  
          $st=$row1['d_status'];
          echo "<option value='$st'>$st</option>";
        }
      ?>
    </select>
  </td>
  <?php
    echo "</tr>"; 

  }

  echo "</table>";
  echo '<input type="submit" name="sub" value="Update"/>';
  echo "</form>";
?>

$op=“从活动=0的客户端选择*”； $op1=mysql\u查询（$op）

<?php 

  $status=&$_POST['status'];
  $from=&$_POST['date_from'];
  $to=&$_POST['date_to'];
  $conn=mysqli_connect('localhost','root','','punbus') or die("Database not connected".mysqli_error());

  if(isset($_POST['sub'])){
    $ins="insert into driver_status(driver_name,status,date_from,date_to)                    
          select Driver_name,'$status','$from','$to' from driver_master";    
    if(mysqli_query($conn,$ins)){
      echo "added";
    }
    else{
      echo "NOT".mysqli_error($conn);
    }
  }

  $sel='select Driver_name from driver_master';
  $query=mysqli_query($conn,$sel);

  echo "<form action='driver_status.php' method='post'>";
  echo "<table cellpadding=5>";
  echo "<tr>";
  echo "<th>Driver Name</th>";
  echo "<th>Status</th>";
  echo "<th>From</th>";
  echo "<th>To</th>";
  echo "</tr>";

  while($row=mysqli_fetch_assoc($query)){    
    echo "<tr>";
    echo "<td>".$row['Driver_name']."</td>";
    $sel1='select d_status from status';
    $query1=mysqli_query($conn,$sel1);
?>

  <td>
    <select name="status">
      <?php
        while($row1=mysqli_fetch_assoc($query1)){  
          $st=$row1['d_status'];
          echo "<option value='$st'>$st</option>";
        }
      ?>
    </select>
  </td>
  <?php
    echo "</tr>"; 

  }

  echo "</table>";
  echo '<input type="submit" name="sub" value="Update"/>';
  echo "</form>";
?>

echo'；
while（$row=mysql\u fetch\u数组（$op1））
{
echo$s=$row[“c_name”]；
如果（$company！=$s）
{
回显'.$row[“c_名称”].'；
}
}
$c=$_请求['c']；
$sql=“插入项目c_名称）值（“$c”）”；
$sql1=mysql\u查询（$sql）；

这是您的代码

<?php 

  $status=&$_POST['status'];
  $from=&$_POST['date_from'];
  $to=&$_POST['date_to'];
  $conn=mysqli_connect('localhost','root','','punbus') or die("Database not connected".mysqli_error());

  if(isset($_POST['sub'])){
    $ins="insert into driver_status(driver_name,status,date_from,date_to)                    
          select Driver_name,'$status','$from','$to' from driver_master";    
    if(mysqli_query($conn,$ins)){
      echo "added";
    }
    else{
      echo "NOT".mysqli_error($conn);
    }
  }

  $sel='select Driver_name from driver_master';
  $query=mysqli_query($conn,$sel);

  echo "<form action='driver_status.php' method='post'>";
  echo "<table cellpadding=5>";
  echo "<tr>";
  echo "<th>Driver Name</th>";
  echo "<th>Status</th>";
  echo "<th>From</th>";
  echo "<th>To</th>";
  echo "</tr>";

  while($row=mysqli_fetch_assoc($query)){    
    echo "<tr>";
    echo "<td>".$row['Driver_name']."</td>";
    $sel1='select d_status from status';
    $query1=mysqli_query($conn,$sel1);
?>

  <td>
    <select name="status">
      <?php
        while($row1=mysqli_fetch_assoc($query1)){  
          $st=$row1['d_status'];
          echo "<option value='$st'>$st</option>";
        }
      ?>
    </select>
  </td>
  <?php
    echo "</tr>"; 

  }

  echo "</table>";
  echo '<input type="submit" name="sub" value="Update"/>';
  echo "</form>";
?>

                   echo '<select name="c"  id="c" style="width:160px;"  required>';

                         while ($row = mysql_fetch_array($op1))
                        {
                       echo  $s=$row["c_name"]; 
                          if($company!=$s)
                          {
                          echo '<option  value="'.$row["c_name"].'">'.$row["c_name"].'</option>';
                          }

                        }

            $c=$_REQUEST['c'];
             $sql="insert into project c_name) values('$c')";
   $sql1=mysql_query($sql); 

您的表单中是否有四个选择框？感谢您发布此代码，但它对我没有帮助。它也不起作用。选择标记中的status_$numb是否正确？

status_$numb

不是标记，它是select name属性的值。。它将在您的选择中给出名称

select\u 1、select\u 2、select\u 3、select\u 4

。在php中，您可以使用

$\u POST['select\u 1']..等
访问这些值。因此，您的4个选择没有相同的名称，但名称不同。请在此主题上帮助我。如果我想在另一个输入类型文本中使用一个或多个输入类型文本，请如何使用foreach循环。只需添加类似于
的内容并放入标记中即可。在插入查询中，您可以访问类似于
driver\u name
post的内容。