Warning: file_get_contents(/data/phpspider/zhask/data//catemap/9/javascript/405.json): failed to open stream: No such file or directory in /data/phpspider/zhask/libs/function.php on line 167

Warning: Invalid argument supplied for foreach() in /data/phpspider/zhask/libs/tag.function.php on line 1116

Notice: Undefined index: in /data/phpspider/zhask/libs/function.php on line 180

Warning: array_chunk() expects parameter 1 to be array, null given in /data/phpspider/zhask/libs/function.php on line 181
Javascript html中的java脚本代码,用于使用java中的Web服务_Javascript - Fatal编程技术网

Javascript html中的java脚本代码,用于使用java中的Web服务

Javascript html中的java脚本代码,用于使用java中的Web服务,javascript,Javascript,我得到{错误:请使用post请求}。以下是我尝试的代码: $(document).ready(function () { $("#btn").click(function (event) { var uid = document.getElementById("fpsid").value; var pwd = document.getElementById("Pass").value; var dataString = "{ 'fpsid'

我得到{错误:请使用post请求}。以下是我尝试的代码:

$(document).ready(function () {
    $("#btn").click(function (event) {
        var uid = document.getElementById("fpsid").value;
        var pwd = document.getElementById("Pass").value;
        var dataString = "{ 'fpsid' : '" + uid + "', 'Pass' : '" + pwd + "'}";

        $.ajax({
            ServiceCallID: 1,
            url: "http://epds.nic.in/PossWebService/services/Poss?wsdl",
            type: 'POST',
            data: dataString,
            contentType: "application/json; charset=utf-8",
            dataType: "json",

            success: function (result) {
                returnVal = result.d;
                alert(returnVal);
            },

            error: function (XMLHttpRequest, textStatus, errorThrown) {
                returnVal = '';
            }
        });
    });
});

首先,
dataString
不可用。为此,有一种比字符串串联更好的方法来生成给定对象的JSON。使用:

或者让jQuery来操心一下:

var postData = {
    fpsid: uid,
    Pass: pwd
};

$.ajax({
    // ...
    data: postData,
    // ...
});

将其编辑到您的帖子中。这不属于注释。你有js小提琴吗?什么是
ServiceCallID
应该是什么?var dataString。。。这不是创建json对象的方法
var postData = {
    fpsid: uid,
    Pass: pwd
};

$.ajax({
    // ...
    data: postData,
    // ...
});