Question

我正在尝试从我从数据库中的另一个表填充的下拉菜单中填充数据库中的表。问题是每当我提交查询时，它都会给我同样的错误＆＃34;注意：未定义的索引：＆＃34;并且不会填满桌子。我是编码的新手，所以请保持温和。这是填充下拉菜单的部分

<?php
@mysql_connect("localhost", "root","") or die(mysql_error());
mysql_select_db("motocikli") or die(mysql_error());

$query = "SELECT kategorija_ime FROM kategorija";
$result = mysql_query($query) or die(mysql_error()."[".$query."]");
?>

<select name="kateg">
<?php 
while ($row = mysql_fetch_array($result))
{
    echo "<option value='".$row['kategorija_ime']."'>'".$row['kategorija_ime']."'</option>";
}
?>        
</select>

 <form action="insert.php" method="post">
 <input type="submit">
</form>

这就是insert.php

<?php

$dsn = 'mysql:dbname=motocikli;host=127.0.0.1';
$user = 'root';
$password = '';
$pdo = new \PDO($dsn, $user, $password);



function unesiPoruku($kateg)
{
    global $pdo;
    $upit = $pdo->prepare("INSERT INTO test (kateg) VALUES (:kateg)");
    $upit->bindParam('kateg',$kateg);

    $upit->execute();
}

$kateg = $_REQUEST['kateg'];
unesiPoruku($kateg);

?>

错误显示为$ kateg = $ _REQUEST [＆＃39; kateg＆＃39;];，＆＃39; kateg＆＃39;标签

Answer 1

您的选择框必须位于表单内，以便将值正确发布到服务器

即

<form action="insert.php" method="post">
      <select name="kateg">
      <?php 
         while ($row = mysql_fetch_array($result))
         {
             echo "<option value='".$row['kategorija_ime']."'>'".$row['kategorija_ime']."'</option>";
         }
      ?>        
      </select>

      <input type="submit">
</form>

继续收到通知：未定义的索引：无论我做什么

1 个答案: