Question

吨我有关于一个匹配所选id的频段的数据，然后我使用band_id并让它在其他地方使用它。我想对Name字段做同样的事情，但是我遇到了一些问题，它只是将word数组放入var $ name。

码

<?php
require 'core/init.php';


$Band_id = $_GET['id'];
$name = ['Name']; // trying to get the band name
$result = mysql_query("SELECT * FROM bands WHERE Band_id = $Band_id");

echo "<table border = '1'>
<tr>
    <th>Band Name</th>
    <th>Venue</th>
    <th>Category</th>
    <th>Stock</th>
    <th>Buy Ticket</th>
</tr>";

 while($row = mysql_fetch_array($result))
    {
        echo "<tr><form name=\"myform\" action=\" order.php\"  method=\"post\">";
        echo " <input name=\"band\" type=\"hidden\" value=\"". $Band_id."\" >";
        echo " <input name=\"bandn\" type=\"hidden\" value=\"". $name."\" >";
        echo "<td>" .$row['Name']. "</td>";
        echo "<td>" .$row['Venue']. "</td>";
        echo "<td>" .$row['Category']. "</td>";
        echo "<td>" .$row['Stock']. "</td>";
        echo "<td><button>Buy Ticket</button></td>";
        echo "</tr> </form>";

    }

echo "</table>";

＆GT;

Answer 1

$name = ['Name']不会在查询之上工作，并且通常根本无法正常工作，因为它是您尝试执行的操作的无效语法。您曾使用$name使用$row['Name']

的地方

尝试从当前选定的数据中检索名称

1 个答案: