Question

我在Javascript中有一个函数正在检查我的servlet：

function validate() {
var idField = document.getElementById("name1");
var data = "name1=" + encodeURIComponent(idField.value);
if (typeof XMLHttpRequest != "undefined") {
    req = new XMLHttpRequest();
} else if (window.ActiveXObject) {
    req = new ActiveXObject("Microsoft.XMLHTTP");
}
var url = "Validator"
req.open("POST", url, true);        
req.onreadystatechange = inserter
req.setRequestHeader("Content-Type", "application/x-www-form-urlencoded");
   req.send(data);
}

function inserter() {
if (req.readyState == 4) {
    if (req.status == 200) {
        var msg1 = req.responseText
    if (msg1 == "") {
        document.getElementById("msg").innerHTML = "<div style=\"color:red\">Wadliwa nazwa</div>";
        document.getElementById("org").value = ''
    }
    else
        document.getElementById("org").value = msg1
    }
}

我发现：指令

req.onreadystatechanged = inserter

设置负责处理服务器响应的函数。

动态更改页面内容的插入器函数将在服务器响应到达时异步调用。

在我的servlet上，我从我的网站获取参数：

protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    System.out.println(request.getParameter("name1"));
    if(!request.getParameter("name1").equals("")) {
        serviceConnection(request, response);
    }
}

不幸的是，当我尝试做同样的事情，但作为doGet，我根本无法做到......：

protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    System.out.println(request.getParameter("name2"));
    if(!request.getParameter("name2").equals("")) {
        serviceConnection(request, response);
    }
}

我试图将ajax的代码更改为：

var url = "Validator/?" + data2;
req.open("GET", url, true);        
req.onreadystatechange = inserter2;

但它不起作用。我怎么解决这个问题？我应该做些什么而不是

req.onreadystatechanged = inserter 有一个doGet功能？

Answer 1

尝试使用此代码..我纠正了它，因为你错了关闭括号，现在应该可以正常工作..

function validate() {
    var idField = document.getElementById("name1");
    var data = "name1=" + encodeURIComponent(idField.value);
    if (typeof XMLHttpRequest != "undefined") {
        req = new XMLHttpRequest();
    } else if (window.ActiveXObject) {
        req = new ActiveXObject("Microsoft.XMLHTTP");
    }
    var url = "Validator"
    req.open("GET", url, true);
    req.onreadystatechange = inserter
    req.setRequestHeader("Content-Type", "application/x-www-form-urlencoded");
    req.send(data);
}


function inserter() {
    if (req.readyState == 4) {
        if (req.status == 200) {
            var msg1 = req.responseText
            if (msg1 == "") {
                document.getElementById("msg").innerHTML = "<div style=\"color:red\">Wadliwa nazwa</div>";
                document.getElementById("org").value = '';
            } else {
                document.getElementById("org").value = msg1;
            }
        }
    }
}

使用servlet的Ajax - 我如何设法制作doGet？

1 个答案: