基本上我希望在选择DROPDOWNMENU1的选项时更新DROPDOWNMENU2选项。
下面是一些信息:
DROPDOWNMENU1:
Web开发 图像 视窗
DROPDOWNMENU2(用于Web开发):
PHP HTML CSS
DROPDOWNMENU2(代表图形):
PNG JPEG ICO PSD
DROPDOWNMENU2(适用于Windows):
七 远景 XP
请帮帮忙?
编辑:
我正在尝试使用此代码
$('#topic2').change(function() {
$('#topic3').empty();
switch($(this).val()) {
case 'Tutorials':
$('#topic3')
.append('<option>PHP</option>')
.append('<option>HTML</option>')
.append('<option>CSS</option>');
case 'Graphics':
// yada yada
break;
case 'Windows':
// yada yada
break;
}
});
HTML:
<form action="inc/post.php" method="POST">
Topic:
<select name="topic">
<option>Web Development</option>
<option>Windows</option>
<option>Graphics</option>
</select>
/
<div id="topic2">
<select name="topic2" id="topic2">
<option>Tutorials</option>
<option>Article</option>
</select>
</div>
/
<div id="topic3">
<select name="topic3" id="topic3">
<option>Seven</option>
<option>Vista</option>
<option>XP</option>
</select>
</div>
<br /><Br />
Name:
<input type="text" name="name" />
<Br /><br />
Icon:
<input type="file" name="icon2" id="icon2">
<Br /><Br />
Content:
<textarea name="info" size="400"></textarea></div>
</div>
</form>
答案 0 :(得分:2)
我会使用jQuery。像这样......
$('#DROPDOWNMENU1').change(function() {
$('#DROPDOWNMENU2').empty();
switch($(this).val()) {
case 'Web Development':
$('#DROPDOWNMENU2')
.append('<option>PHP</option>')
.append('<option>HTML</option>')
.append('<option>CSS</option>');
case 'Graphics':
// yada yada
break;
case 'Windows':
// yada yada
break;
}
});
答案 1 :(得分:-1)
<form name="x" id="x" action="x.php">
<select name="DROPDOWNMENU1" id="DROPDOWNMENU1"
onchange="javascript:document.x.action='dropmenu.php';
javascript:document.x.submit();">
<option>select web development graphics windows</option>
<?php
$q=select * from development graphics windows;
$result=msql_query($q);
while($row = mysql_fetch_array($result)) {
if($_POST["DROPDOWNMENU1"]==$row[web_development_id]){
?>
<option value='<?php echo $row["web_development_id"]; ?>'selected>
<?php echo $row["web_development_name"];?>
</option>
in DROPDOWNMENU2 select data where DROPDOWNMENU2.id=$_POST["DROPDOWNMENU1"];
我希望这个答案可以帮助你