Question

我有一个php页面，我将所有用户名从db列出到下拉列表。现在我需要选择一个名称，并需要添加到datbase中的另一个表。

从db我可以做到的列表。但坚持如何将选定的值添加回另一个数据库。请帮忙。

    <!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>

Answer 1

试试此代码： - [更新代码]

 <?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>

如何将选定的值从下拉列表添加到另一个数据库

1 个答案: