嗨,我是php的新手,可以帮忙吗。我正在制作一个网站,它有一个菜单,我需要它,这样如果点击"链接1"这样的链接,page1.php将加载到mainSectiondiv中,如果点击链接2,page2.php将加载在mainSection等中,所以所有页面:page1、page2、page3等都将加载到这个页面中,这取决于点击了什么链接。这可能吗?我不知道从哪里开始。感谢
<body>
<?php
<ul>
<li><a href="#" name="link1">link 1</a></li>
<li><a href="#" name="link2">link 2</a></li>
<li><a href="#" name="link3">link 3</a></li>
<li><a href="#" name="link4">link 4</a></li>
</ul>
?>
<div id="mainSection">
<?php
if (link1 == true){
include 'page1.php';
}
if (link2 == true){
include 'page2.php';
}
if (link3 == true){
include 'page3.php';
}
if (link4 == true){
include 'page4.php';
}
?>
</div>
</body>
以下是您可以从开始的内容
<body>
<ul>
<li><a href="?link=1" name="link1">link 1</a></li>
<li><a href="?link=2" name="link2">link 2</a></li>
<li><a href="?link=3" name="link3">link 3</a></li>
<li><a href="?link=4" name="link4">link 4</a></li>
</ul>
<div id="mainSection">
<?php
$link=$_GET['link'];
if ($link == '1'){
include 'page1.php';
}
if ($link == '2'){
include 'page2.php';
}
if ($link == '3'){
include 'page3.php';
}
if ($link == '4'){
include 'page4.php';
}
?>
</div>
</body>
除了majid的代码外,您还必须检查链接是否已设置,否则会引发未定义$link的错误
- 链接1
- 链接2
- 链接3
- 链接4
<div id="mainSection">
<?php
if(isset($_GET['link'])){
$link=$_GET['link'];
if ($link == '1'){
include 'page1.php';
}
if ($link == '2'){
include 'page2.php';
}
if ($link == '3'){
include 'page3.php';
}
if ($link == '4'){
include 'page4.php';
}
} ?>
</div>
</body>
将链接的格式更改为:
<a href="/?1" name="link1">link 1</a>...
然后将您的PHP更改为:
<?php
if ($_SERVER['QUERY_STRING'] == 1){
include 'page1.php';
}
if ($_SERVER['QUERY_STRING'] == 2){
include 'page2.php';
}
if ($_SERVER['QUERY_STRING'] == 3){
include 'page3.php';
}
if ($_SERVER['QUERY_STRING'] == 4){
include 'page4.php';
}
?>