ajax怎么根据界面标题来获取数据库数据！

// JavaScript Document
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');
        }
    }
}

------解决方案--------------------

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