所以我有一个包含2个表的数据库和一个包含两个选择列表的页面。根据第一次选择时选择的值，我想在第二次选择时显示相应的项目。
没有任何错误或任何东西，但第二个选择从来没有显示任何东西，所以我可能错过了一些东西。有人能帮我吗？
连接.php
<?php
$dbhost = "localhost";
$dbuser = "root";
$dbpass = "";
$dberror1 = "Could not connect to the database";
$dberror2 = "Could not find the table";
$Conn = mysqli_connect($dbhost, $dbuser, $dbpass) or die ($dberror1);
$Select_db = mysqli_select_db($Conn, 'swimming') or die ($dberror2);
?>
结果.php
<?php
include ("Connections/connection.php");
$query_comp = mysqli_query($Conn, "Select * FROM comp");
?>
<!doctype html>
<html><head>
<meta charset="utf-8">
<title>Swimming Results</title>
<link href="file:///D|/Websites/Swimming/style.css" rel="stylesheet" type="text/css">
</head>
<body>
<div id="wrapper">
<!-- Div for Title -->
<div id="title">
<p>Swimming Results</p>
</div>
<form name="results" >
<!-- Div to Choose competition -->
<div id="competition">
<p>Competition</p>
<select name="sel_comp">
<option value="">---Select one competition---</option>
<?php
while ($fetch = mysqli_fetch_assoc($query_comp)){
echo "<option value='{".$fetch['id']."}'>".$fetch['comp_name']."</option>";
}
?>
</select>
</div>
<!-- Div to Chose event -->
<div id="event">
<p>Event</p>
<select name="sel_even">
<option value="">---Select one event---</option>
<?php
$chosen=$_POST["sel_comp"] ;
$query_even = mysqli_query($Conn, "Select comp.id, events.cid, events.event_name FROM events, comp WHERE events.cid = $chosen");
while ($fetch = mysqli_fetch_assoc($query_even)){
echo "<option value='{".$fetch['event_id']."}'>".$fetch['event_name']."</option>";
}
?>
</select>
</div>
<!-- Div to Show results -->
<div id="results">
<p>Resultados</p>
</div>
</form>
</div>
</body>
</html>
您的PHP代码在显示到用户浏览器之前会在服务器端进行解释。所以这段代码对用户将要做的选择一无所知
如果你想要一个"动态子选择"，你需要添加一些javascript代码。该javascript代码将在用户的浏览器上运行，并能够对用户在第一个选择框上的操作做出反应。