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所以我的問題是，我無法使用單選按鈕將數據插入選定的表。單選按鈕用於告訴我們我們需要訪問哪個表並添加數據。我認爲我說得對，但沒有奏效。這是我的。任何形式的幫助將不勝感激。使用單選按鈕插入功能
<h1 align="center">Event</h1>
<h1 align="center"><a href="stroage.html" target="new">Storage</a></h1>
<form action="set_event.php" method="post">
<p>
<p>
<input type="submit" name="submit" value="Set Items">
</p>
<p>
<br>
<label>
<input type="radio" name="storage" value="concourse_stairs_s" id="s1">
Concourse Stairs</label>
<br>
<label>
<input type="radio" name="storage" value="bat_cave_s" id="s2">
Bat Cave</label>
<br>
<label>
<input type="radio" name="storage" value="fireside_s" id="s3">
Fireside Storage</label>
<br>
</p>
<?php
if ($_SERVER['REQUEST_METHOD'] == 'POST') {
require ('mysqli_connect.php'); // Connect to the db.
$errors = array(); // Initialize an error array.
}
if (isset($_REQUEST['bat_cave_s']))
{
$q = "INSERT INTO bat_cave_s (item_name, check_in) VALUES ('test', NOW())";
$r = @mysqli_query ($dbc, $q); // Run the query.
}
else if(isset($_REQUEST['concourse_stairs_s']))
{
$q = "INTO concourse_stairs_s (item_name, check_in) VALUES ('test', NOW())";
$r = @mysqli_query ($dbc, $q); // Run the query.
}
else if(isset($_REQUEST['fireside_s']))
{
$q = "INSERT INTO fireside_s (item_name, check_in) VALUES ('test', NOW())";
$r = @mysqli_query ($dbc, $q); // Run the query.
}
NOW())";