【发布时间】:2013-05-11 17:36:45
【问题描述】:
我正在尝试使用 jquery 创建一个动态菜单,并且我想使用 Ajax 数据(取自数据库)填充后续菜单。
然后我希望能够添加其他下拉菜单并保留相同的功能。我让它适用于单个下拉菜单并且 ajax 调用正在运行,但它不适用于通过 jquery(重复)函数添加的其他下拉菜单。
我不认为我需要在 php 中添加一个计数器变量,因为我正在处理 jquery 中的增量,但我无法从其他下拉列表中获取输入(进入 (.additionalsubj')进入 Ajax - 我什至无法让它们在 Firebug 中显示。 我有点卡住了 - 任何想法都会受到极大的赞赏。
HTML:
<h3>Primary Subject</h3>
<div class="subjselect">
<div class="select">
<select id="subject">
<option value="">Subject</option>
<option value="math">Math</option>
<option value="science">Science</option>
<option value="languages">Languages</option>
<option value="humanities">Humanities</option>
<option value="econ">Economics/Finance</option>
<option value="gmat">GMAT</option>
<option value="sat">SAT</option>
</select>
<select id="topic">
<option value="">Topic</option>
<option value="math">Math</option>
<option value="science">Science</option>
<option value="languages">Languages</option>
<option value="humanities">Humanities</option>
<option value="econ">Economics/Finance</option>
<option value="gmat">GMAT</option>
<option value="sat">SAT</option>
</select>
</div>
<a href="#" id="another" onclick="Repeat(this)"></br>Add Another Subject</a>
</div>
jquery:
var counter=1;
$(document).on('change', 'select#subject'+counter+'', function(){
var subject = $("select#subject"+counter+">option:selected").text();
var selector=$("select#subject"+counter+"");
console.log(selector);
console.log(subject);
$.ajax({
type: 'GET',
url: 'tutorprofileinput.php',
data: {"subject": subject},
dataType:'json',
success:function(data){
console.log(data);
var options = [];
$.each(data, function (key, val) {
options += '<option value="' + val.topic + '">' + val.topic + '</option>';
console.log(options);
});
$("select#topic").html(options);
},
error:function(){
// failed request; give feedback to user
$('#ajax-panel').html('<p class="error"><strong>Oops!</strong> Try that again in a few moments.</p>');
}
});
});
function Repeat(obj){
counter++;
console.log(counter);
var selecoptions = '<div class="select"><select id="subject'+counter+'"><option value="">Subject</option><option value="math">Math</option><option value="science">Science</option><option value="languages">Languages</option><option value="humanities">Humanities</option><option value="econ">Economics/Finance</option><option value="gmat">GMAT</option><option value="sat">SAT</option></select></div><div class="select"><select id="topic'+counter+'"><option value="">Topic</option></select></div>';
$('.additionalsubj').append(selecoptions);
console.log($('.additionalsubj'));
}
和 PHP 从数据库中获取数据:
<?php
include("php_includes/db_conx.php");
if (isset($_GET['subject'])){
$subject = $_GET['subject'];
$query = ("SELECT subject.id, topic FROM topics, subject WHERE subject='$subject' AND subject.id=topics.subjID ORDER BY subject");
$result = mysqli_query($db_conx, $query);
$rows = array();
while($r = mysqli_fetch_array($result, MYSQLI_ASSOC)){
$rows[] = $r;
}
echo json_encode($rows);
exit();
}
?>
【问题讨论】:
标签: php javascript jquery mysql ajax