<html>
<head>
<title>Classroom research</title>
</head>
<body>
<h1>CRMS - Classroom research</h1>
<form action="research.php" method="post">
<table border="0">
<tr>
<td>Classroom ID</td>
<td><input type="text" name="Cno" maxlength="13" size="13"></td>
</tr>
<tr>
<td>Course ID</td>
<td>
<select>
<option value=0>--请选择--</option>
<?php
$con = mysql_connect("localhost","root","");
$search_course = "SELECT CID FROM course2";
$result = mysql_query($search_course, $con);
while($row = mysql_fetch_array($result))
{
<option value="$row">$row</option>
}
?>
</select>
</td>
</tr>
<tr>
<td colspan="2"><input type="submit" value="Register"></td>
</tr>
</table>
</form>
</body>
</html>
1、我希望在第一个框输入课室号，第二框通过下拉框显示数据库中course2表的CID的内容。
这应该就是不对的，毫无头绪，不知道怎么实现。求详细代码！
2、怎样实现填写其中一个数据（即不用填两个）就可以查询数据？
------解决思路----------------------
仅供参考：
<?php
/* Created on [2012-5-16] */
#查询标题信息
$sql="select * from table";
$res=mysql_query($sql);
if(!$res) die("SQL: {$sql} <br>Error:".mysql_error());
if(mysql_affected_rows() > 0){
$titles = array();
while($rows = mysql_fetch_array(MYSQL_ASSOC)){
array_push($titles,$rows);
}
}
?>
<table border=1>
<?php foreach($titles as $row_Recordset_task){ ?>
<tr>
<td>
<a href="javascript:void(0)" onclick="record(<?=$row_Recordset_task['TID']?>)" >
<?=$row_Recordset_task['csa_title']?>
</a>
</td>
</tr>
<?php } ?>
</table>
<div id="show"></div>
<form name="frm">
<select name="s1" onChange="redirec(this.value)">
<option selected>请选择</option>
<option value="1">内科</option>
<option value="2">内科</option>
<option value="3">内科</option>
</select>
<div id="s2"></div>
</form>
<script>
//Ajax
var xmlHttp;
function createXMLHttpRequest() {
if(window.XMLHttpRequest) {
xmlHttp = new XMLHttpRequest();
} else if (window.ActiveXObject) {
xmlHttp = new ActiveXObject("Microsoft.XMLHTTP");
}
}
function record(id){
createXMLHttpRequest();
url = "action.php?id="+id+"&ran="+Math.random();
method = "GET";
xmlHttp.open(method,url,true);
xmlHttp.onreadystatechange = show;
xmlHttp.send(null);
}
function show(){
if (xmlHttp.readyState == 4){
if (xmlHttp.status == 200){
var text = xmlHttp.responseText;
document.getElementById("s2").innerHTML = text;
}else {
alert("response error code:"+xmlHttp.status);
}
}
}
</script>
<?php
#action.php
if(isset($_GET['id'])){
$sql="select * from table where id=".$_GET['id'];
$res=mysql_query($sql);
if(!$res) die("SQL: {$sql} <br>Error:".mysql_error());
if(mysql_affected_rows() > 0){
$arrMenu=array();
while($rows = mysql_fetch_array(MYSQL_ASSOC)){
array_push($arrMenu,$rows);
}
}
mysql_close();
if(!empty($arrMenu)){
echo "<select name='menu2'>";
foreach($arrMenu as $item2){
echo "<option value='{$item2['id']}'>{$item2['name']}</option>";
}
echo "</select>";
}
}
?>
------解决思路----------------------
就这么简单，不需要上面那么复杂！！伱直接复制过去就可以用了。。。。
<?php
//require_once('conn.php'); //伱最好写个连接数据库的文件 每次包含一下就行了！ 而且要写在最上面
//最好把下面三行写在conn.php文件里 以后每次用时 向上面那样包含一下就OK了！！
$con = mysql_connect("localhost","root","***") or die("错误信息：".mysql_error()); //连接
$db = mysql_select_db("表course2所在的数据库名"); //这个要不写就取不着数据 但不会报错
mysql_query("set names gb2312");
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=gb2312" />
<title>无标题文档</title>
</head>
<body>
<select>
<option>-请选择-</option>
<?php
$sql="select CID from course2";
$result=mysql_query($sql);
while($row=mysql_fetch_assoc($result)){
?>
<option value="<?php echo $row['CID'] ?>"><?php echo $row['CID'] ?></option> //这个值要用php的方法取出来
<?php
}
?>
</select>
</body>
</html>
------解决思路----------------------
搞这么复杂？script标签在代码底部加也行啊，只要有：
<script>
//Ajax
var xmlHttp;
function createXMLHttpRequest() {
if(window.XMLHttpRequest) {
xmlHttp = new XMLHttpRequest();
} else if (window.ActiveXObject) {
xmlHttp = new ActiveXObject("Microsoft.XMLHTTP");
}
}
function record(id){
createXMLHttpRequest();
//指定目标地址及参数
url = "action.php?id="+id+"&ran="+Math.random();
method = "GET";
xmlHttp.open(method,url,true);
xmlHttp.onreadystatechange = show;
xmlHttp.send(null);
}
function show(){
if (xmlHttp.readyState == 4){
if (xmlHttp.status == 200){
//回调函数，返回的后端结果
var text = xmlHttp.responseText;
document.getElementById("s2").innerHTML = text;
}else {
alert("response error code:"+xmlHttp.status);
}
}
}
</script>
能运行就行