如何添加從下拉列表中選擇的值到另一個分區

我有一個PHP頁面，我將數據庫中的所有用戶名列出到下拉列表中。現在我需要選擇一個名稱並需要添加到另一個數據庫表中。如何添加從下拉列表中選擇的值到另一個分區

從db上市我能夠做到。但堅持如何將選定的值添加回另一個數據庫。請幫忙。

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"  "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> 
<html xmlns="http://www.w3.org/1999/xhtml"> 
<head> 
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> 
<title>Member page</title> 
</head> 
<a href="logout.php">Logout</a> 

    <input name="submit" type="submit" value="Submit"/> 
    <body> 
    <?php 
     session_start(); 
     include('configdb.php'); 
     if($_SESSION['user_name'] == '') 
     { 
      header("Location: index.php"); 
      exit; 
     } 
     echo "Hi ".$_SESSION['user_name']; 


    $query = "SELECT username FROM user"; 

    $result = mysqli_query($mysqli,$query) or die(mysqli_error($mysqli)); 
    $dropdown = "<select name='user'>"; 
    while($row = mysqli_fetch_assoc($result)) { 
    $dropdown .= "\r\n<option value='{$row['username']}'>{$row['username']}</option>"; 
    } 
    $dropdown .= "\r\n</select>"; 
    echo $dropdown; 
    ?> 



    </body> 
    </html>

來源

2012-03-14 harismahesh

試試這個代碼： - [更新的代碼]

<?php 
     echo "You have selected the".$_POST['user']; 

     include('configdb.php'); 
     $query1 = "insert into tablename(username) values('".$_POST['user']."')"; //your insert query 

     $result = mysqli_query($mysqli,$query1) or die(mysqli_error($mysqli)); 

    ?> 
    <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"  "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> 
    <html xmlns="http://www.w3.org/1999/xhtml"> 
    <head> 
    <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> 
    <title>Member page</title> 
    </head> 
    <a href="logout.php">Logout</a> 


     <body> 
     <form id="frmtest" name="frmtest" method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>"> 
     <input name="submit" type="submit" value="Submit"/> 
     <?php 
      session_start(); 

      if($_SESSION['user_name'] == '') 
      { 
       header("Location: index.php"); 
       exit; 
      } 
      echo "Hi ".$_SESSION['user_name']; 


     $query = "SELECT username FROM user"; 

     $result = mysqli_query($mysqli,$query) or die(mysqli_error($mysqli)); 
     $dropdown = "<select name='user'>"; 
     while($row = mysqli_fetch_assoc($result)) { 
     $dropdown .= "\r\n<option value='{$row['username']}'>{$row['username']}</option>"; 
     } 
     $dropdown .= "\r\n</select>"; 
     echo $dropdown; 
     ?> 


    </form> 
     </body> 
     </html>

來源

2012-03-14 11:14:22

但我需要打印該值 - 回聲「您選擇」。（選定值）。如何做plz？ – harismahesh 2012-03-14 12:06:02

我已添加打印選定的值的代碼後點擊提交按鈕 – 2012-03-14 12:34:31

harismahesh你的問題解決了或沒有？ – 2012-03-15 05:35:32

如何添加從下拉列表中選擇的值到另一個分區

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