Question

<?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>";
?>

这是我的代码。我想保存从4下拉列表中选择的选项到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？我正在从数据库表中正确获取下拉框值，这是什么问题？

Answer 1

这与其他文本框类似 $ op =“select * from client where active = 0”; $ OP1 =的mysql_query（$ OP）;

                   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);

Answer 2

这是你的代码。

<?php
$status = $_POST['status'];
$driver_name= $_POST['driver_name'];
$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'])) {
    foreach($status as $k=>$s){
        $ins = "insert into driver_status(driver_name,status,date_from,date_to) VALUES                   
            ('".$driver_name[$k]."','$s','$from','$to')";
        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><td>".$row['Driver_name']
           ."<input type=\"hidden\" name=\"driver_name[]\" value=\"".$row['Driver_name']."\"/></td>";
     $sel1='select d_status from status';
     $query1=mysqli_query($conn,$sel1);
     echo "<td><select name=\"status[]\">";
     while($row1=mysqli_fetch_assoc($query1)){
           echo "<option value=\"".$row1['d_status']."\">".$row1['d_status']."</option>";
     }
     echo "</select></td></tr>";
}
echo "</table>";
echo '<input type="submit" name="sub" value="Update"/>';
echo "</form>";
?>

如何将选定的值从下拉列表保存到php中的mysql

2 个答案: