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我正在创建一个html表单并使用它的值来动态地过滤来自mysql数据库的搜索结果。我有几个下拉的和一个复选框字段的多个值,我无法获得检查值通过AJAX发送到我的PHP页面。有没有办法根据这个复选框字段更新我的查询结果,其方式与其他选择字段相同?通过ajax为php/mysql查询发送复选框值
HTML代码:
<form name="searchForm">
<input type="checkbox" name="services" value="twic" onclick="get_check_value()">TWIC</input><br/></li>
<input type="checkbox" name="services" value="enclosedTrucking" onclick="get_check_value()">Enclosed Trucking</input><br/>
<input type="checkbox" name="services" value="flatBedTrucking" onclick="get_check_value()">Flat Bed Trucking</input><br/>
</form>
Java脚本的:
<script type="text/javascript">
function get_check_value()
{
var c_value = "";
for (var i=0; i < document.searchForm.services.length; i++) {
if (document.searchForm.services[i].checked)
{
c_value = c_value + document.searchForm.services[i].value + "\n";
}
}
return = c_value;
//alert(c_value);
}
</script>
<script language="javascript" type="text/javascript">
//Browser Support Code
function ajaxFunction(){
var ajaxRequest; // The variable that makes Ajax possible!
try{
// Opera 8.0+, Firefox, Safari
ajaxRequest = new XMLHttpRequest();
}catch (e){
// Internet Explorer Browsers
try{
ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
}catch (e) {
try{
ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
}catch (e){
// Something went wrong
alert("Your browser broke!");
return false;
}
}
}
// Create a function that will receive data
// sent from the server and will update
// div section in the same page.
ajaxRequest.onreadystatechange = function(){
if(ajaxRequest.readyState == 4){
var ajaxDisplay = document.getElementById('results');
ajaxDisplay.innerHTML = ajaxRequest.responseText;
}
}
// Now get the value from user and pass it to
// server script.
var os = document.getElementById('originState').value;
var c = document.getElementById('commodity').value;
var ds = document.getElementById('destState').value;
var ser = c_value;
var queryString = "?os=" + os ;
queryString += "&c=" + c + "&ds=" + ds + "&ser=" + ser;
//ajaxRequest.open("GET", "php/search2.php" +
// queryString, true);
ajaxRequest.open("GET", "php/testqueries.php" + queryString, true);
ajaxRequest.send(null);
}
</script>
PHP接收这些变量:
$ser = $_GET['ser'];
'收益率= c_value; ''使用'return c_value'.And你在做什么?你正在返回值onclick事件..为什么这是这样的? ..你应该处理'onsubmit'事件的形式和所有选中的复选框列表传递给ajax调用并获得结果 –
为什么你想(糟糕/不正确)返回一个字符串到onclick处理程序?你已经用jquery标记了这个问题,但绝对没有jQuery代码。 –