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我知道这是一个非常基本的问题,但我无法确定如何在过去1小时内爬网之后修复代码。 我有一个无序列表,其中包含有关数据库中类别的信息,cat_id作为主键。和一个以cat_id作为外键的主题表,所以我想通过给定类别ID的ajax请求访问主题表。下面是我用来生成类别的代码。我在哪里卡住是,我不知道它的DOM元素,以便发送的唯一ID的URL参数.. 感谢取..通过ajax传递参数到php
<ul id="search_form">
<?php
$cat = Category::find_all();
foreach($cat as $category) {
echo '<li id="';
echo $category->cat_id;
echo '"><a href="subject.php?id=';
echo $category->cat_id;
echo'">';
echo $category->category;
echo '</a></li>';
}
?>
</ul>
<div id="results">
<!-- ajax contents goes here -->
</div>
阿贾克斯文件
window.onload = init;
function init() {
if (ajax) {
if (document.getElementById('results')) {
document.getElementById('search_form').onclick = function() {
ajax.open('get', 'subject.php?id='+id); // subject.php?id=
// how will i pass the variable
ajax.onreadystatechange = function() {
handleResponse(ajax);
}
ajax.send(null);
return false;
}
}
}
}
function handleResponse(ajax) {
if (ajax.readyState == 4) {
if ((ajax.status == 200) || (ajax.status == 304)) {
var results = document.getElementById('results');
results.innerHTML = ajax.responseText;
results.style.display = 'block';
}
}
}
和该主题.php
<?php
//include("tpl/header.php");
include("includes/initialize.php");
?>
<h2></h2>
<?php
if (isset($_GET['id'])) {
$id= mysql_real_escape_string($_GET['id']);
$subject = Subject::find_subject_for_category($id);
foreach($subject as $subj) {
echo $subj->subject_title;
}
} else {
echo "No ID Provided";
}
?>
好,我不熟悉jquery的事情..我应该呼吁成功哪个函数? – XoR 2011-12-15 21:11:17