ajax怎么根据界面标题来获取数据库数据！

ajax如何根据界面标题来获取数据库数据！ - Web 开发 / Ajax
如题！我的方法是下面的！可是获取不了数据！
function cLaw(id){

title=document.getElementById(id).innerText;
//alert("test:"+title);
  url=("/yjOutGate/Policy?title="+title);  
  cLawRequest(url);
 
}

//浏览器判断
function xhttp() 
{ var http_request = null;
if (window.XMLHttpRequest) {
http_request = new XMLHttpRequest();
if (http_request.overrideMimeType) {
http_request.overrideMimeType("text/xml");
}
} else if (window.ActiveXObject) {
try {
http_request = new ActiveXObject("Msxml2.XMLHTTP");
} catch (e) {
try {
http_request = new ActiveXObject("Microsoft.XMLHTTP");
} catch (e) {}
}
}
if (!http_request) {
alert("初始化XMLHttpRequest对象失败");
return false;
}
return http_request;
}

var http_law = xhttp() ;
function cLawRequest(url) {
http_law .onreadystatechange = alertC_Law;
http_law .open("GET",url, true);
//alert(url);
http_law .send(null);

}

function alertC_Law() {

if (http_law.readyState == 4) {
if (http_law.status == 200){
var info =http_law.responseText;
var infojson = eval("("+info+")");
alert(info);//这里就获取不到需要的数据了！
} else {
alert('请与工作人员联系或者检查您的网络');
}
}
}

[解决办法]
楼主太不厚道了格式化代码后你的代码

JScript code

// JavaScript Documentfunction cLaw(id) {    title = document.getElementById(id).innerText;    // alert("test:"+title);    url = ("/yjOutGate/Policy?title=" + title);    cLawRequest(url);}// 浏览器判断function xhttp() {    var http_request = null;    if (window.XMLHttpRequest) {        http_request = new XMLHttpRequest();        if (http_request.overrideMimeType) {            http_request.overrideMimeType("text/xml");        }    } else if (window.ActiveXObject) {        try {            http_request = new ActiveXObject("Msxml2.XMLHTTP");        } catch (e) {            try {                http_request = new ActiveXObject("Microsoft.XMLHTTP");            } catch (e) {            }        }    }    if (!http_request) {        alert("初始化XMLHttpRequest对象失败");        return false;    }    return http_request;}var http_law = xhttp();function cLawRequest(url) {    http_law.onreadystatechange = alertC_Law;    http_law.open("GET", url, true);    // alert(url);    http_law.send(null);}function alertC_Law() {    if (http_law.readyState == 4) {        if (http_law.status == 200) {            var info = http_law.responseText;            var infojson = eval("(" + info + ")");            alert(info);// 这里就获取不到需要的数据了！        } else {            alert('请与工作人员联系或者检查您的网络');        }    }}
[解决办法]
一个是注释掉var infojson = eval("("+info+")");  
二是在function cLawRequest(url) {
        http_law .onreadystatechange = alertC_Law;  
        http_law .open("GET",url, true);
    //alert(url);
     http_law .send(null);

}
改成
function cLawRequest() {
      var url="Default.aspx";         http_law .onreadystatechange = alertC_Law;  
        http_law .open("GET",url, true);
    //alert(url);
     http_law .send(null);

}
我没有你的那个url值,不知道你的是否写错了?????
其他没变