dsqbkh3630 2015-03-19 15:36
浏览 23
已采纳

HTML表单下拉选择问题

I've got a login form which I'm trying to simplify. It all worked when you manually inputted your username and password. Now I am wanting a drop down for the username box from the MySql database.

This is the code that I have put into the form and it drops down and shows all the users but when you select it, put the password in and click login it doesn't pass the username.

<?php
mysql_connect('localhost', 'user', 'password.'); 
mysql_select_db('database');

$sql = "SELECT username FROM users";
$result = mysql_query($sql);
echo "<select username='sub1'>";
while ($row = mysql_fetch_array($result))   {
echo"<option value'" . $row['username'] ."'>" . $row['username'] ."</option>"; } echo "</select>";?>

Any Ideas Anyone

Thanks

  • 写回答

2条回答 默认 最新

  • dongluobei9359 2015-03-19 15:41
    关注

    Tom, always check your generated HTML to investigate errors.

    Change your code to:

    <?php
    mysql_connect('localhost', 'user', 'password.'); 
    mysql_select_db('database');
    
    $sql = "SELECT username FROM users";
    $result = mysql_query($sql);
    echo "<select name='username'>"; // Note name attribute
    while ($row = mysql_fetch_array($result))   {
      echo "<option value='" . $row['username'] ."'>" . $row['username'] ."</option>"; // Note `=` sign after value
     } 
    echo "</select>";
    ?>
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
查看更多回答(1条)

报告相同问题?

悬赏问题

  • ¥15 hexo+github部署博客
  • ¥15 求螺旋焊缝的图像处理
  • ¥15 blast算法(相关搜索:数据库)
  • ¥15 请问有人会紧聚焦相关的matlab知识嘛?
  • ¥15 网络通信安全解决方案
  • ¥50 yalmip+Gurobi
  • ¥20 win10修改放大文本以及缩放与布局后蓝屏无法正常进入桌面
  • ¥15 itunes恢复数据最后一步发生错误
  • ¥15 关于#windows#的问题:2024年5月15日的win11更新后资源管理器没有地址栏了顶部的地址栏和文件搜索都消失了
  • ¥100 H5网页如何调用微信扫一扫功能?