Question

这是客户端代码：

function save(calc){
    var request = new XMLHttpRequest();

    request.open("POST","/Calculator/Controller?action=saveCalc",true);
    //request.send("calc=calc")   ---this does not work also
    request.send(calc);
}

这是servlet：

if (action.equals("saveCalc")) {
        String parameter = request.getParameter("calc");
        System.out.println(parameter);
}

此处的参数为null ...

我也尝试过这样：

function save(calc){
        var request = new XMLHttpRequest();
        request.open("POST","/Calculator/Controller?action=saveCalc",true);
        var data = new FormData();
        data.append('calc', 'calc');
        request.send(data);
    }

也行不通。

如何获得此....参数。：d

我取消了。这是解决方案： - 内容类型......

function save(calc){
    var request = new XMLHttpRequest();
    request.open("POST","/Calculator/Controller?action=saveCalc",true);
    request.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
    request.send("calc=" + calc);
}


protected void doPost(HttpServletRequest request,
            HttpServletResponse response) throws ServletException, IOException {

        String action = request.getParameter("action");

        if (action.equals("saveCalc")) {
            String calc = request.getParameter("calc");
            System.out.println(calc);
        }
}

Answer 1

首先你做错了，因为来自GET的参数通过URL传播，所以如果你使用POST并通过URL发送参数，这不应该在请求中传播，POST意味着参数由请求主体传播所以你必须做类似的事情：

var form = document.getElementById('subscribe_frm');
form.setAttribute("method", method);
form.setAttribute("action", path);
form.submit();

您可以先创建属性。

让代码在其中工作的另一种方法是：

request.open("GET","/Calculator/Controller?action=saveCalc",true);

将POST移至GET

Answer 2

获取表单数据

  var parameter = $("input#CountryVal").val();
   dataString = "calc=" + parameter ;

进行Ajax调用

 $.ajax({
                type: "POST",
                url: "/Calculator/Controller?action=saveCalc",
                data: dataString,
                dataType: "json",

                //if received a response from the server
                success: function( data, textStatus, jqXHR) {...}

如何从ajax请求获取servlet中的参数？

2 个答案: