我正在創建幾個下拉菜單,根據選定的上一個下拉值來填充。如何在HTML下拉菜單中顯示AJAX結果
到目前爲止,我能夠成功地使用ajax從php腳本中獲取數據,但是,我不確定如何將該數據附加到名爲AREAS的第二個下拉列表中。
HTML/Javascript/AJAX
<html>
<head>
<title>IPSLA Report</title>
<script type="text/javascript">
function changeSelections() {
var departments = document.selections.department;
var areas = document.selections.areas;
var months = document.selections.months;
var years = document.selections.years;
var s = document.getElementById("department");
switch(departments.selectedIndex) {
case 0:
areas.options.length = 0;
months.options.length = 0;
years.options.length = 0;
areas.options[0] = new Option("Select an Area");
months.options[0] = new Option("Select a Month");
years.options[0] = new Option("Select a Year");
departments.options[0].selected = true;
break;
default:
months.options.length = 0;
years.options.length = 0;
months.options[0] = new Option("Select a Month");
years.options[0] = new Option("Select a Year");
var pass = s.options[s.selectedIndex].text;
ajaxFunction(pass);
}
}
function ajaxFunction(pass) {
var ajaxRequest;
try {
ajaxRequest = new XMLHttpRequest();
}
catch(e) {
try {
ajaxRequest = new ActiveXObjext("Msxml2.XMLHTTP");
}
catch(e) {
try {
ajaxRequest = new ActiveXObjext("Microsoft.XMLHTTP");
}
catch(e) {
alert("Please use another browser");
return false;
}
}
}
ajaxRequest.onreadystatechange = function() {
if (ajaxRequest.readyState == 4) {
var ajaxDisplay = document.getElementById("areas");
if (ajaxDisplay != null) {
ajaxDisplay.options.selectedIndex.text = ajaxRequest.responseText;
}
else {
document.write("NULL!!!");
}
}
}
if (pass == "CRAN") {
var active_id = '0';
}
var queryString = "?active_id=" + active_id;
ajaxRequest.open("GET","db_connect.php" + queryString, true);
ajaxRequest.send(null);
}
</script>
</head>
<body>
<div id="wrapper">
<div id="select">
<form name="selections" id="selections" action="">
<select name="department" id="department" onChange="changeSelections()">
<option value="none">Select Department</option>
<option value="none">CRAN</option>
<option value="none">BackBone</option>
<option value="none">Datacenter</option>
<option value="none">Enterprise</option>
</select>
<select name="areas" id="areas" onChange="changeMonth()">
<option value="none">Select an Area</option>
</select>
<select name="months" id="months" onChange="changeYear()">
<option value="none">Select a Month</option>
</select>
<select name="years" id="years" onChange="go(this)">
<option value="none">Select a Year</option>
</select>
</form>
</div>
<div id="image">
</div>
<div id="incidents">
</div>
</div>
</body>
</html>
PHP Script
<?php
$host = "";
$dbName = "";
$username = "";
$password = "";
$conn = mysql_connect($host,$username,$password);
$db = mysql_select_db($dbName,$conn);
$dept_id = $_GET['id'];
$area_query = "SELECT area_name FROM incident_area WHERE FK_dept= '$dept_id'";
$area_result = mysql_query($area_query);
$options = "";
while ($area_row = mysql_fetch_assoc($area_result)) {
#$options .= '<option value="'.$area_row['area_name'].'">'.$area_row['area_name'].'</option>';
$options .= $area_row['area_name'];
}
echo $options;
?>
~
所以,基本上我只需要知道如何將由AJAX收集的值附加到下拉菜單「區域」。
no raw'XMLHttpRequest',強烈推薦jQuery。 – 2012-04-23 00:36:19
@SiweiShen - jQuery中的等價物是什麼?你可以指點我的資源? – jmg0880 2012-04-23 00:38:57
谷歌官網,有很多文件。請參閱:http://api.jquery.com/jQuery.ajax/。首先,看看它的「選擇器」是如何工作的,然後檢查它的'ajax'方法。很簡單。 – 2012-04-23 01:34:26