Question

我试图在servlet中获取参数，但我收到的所有东西都是null值。我的jsp代码：

<form id="loginForm" name="loginForm" class="loginForm"  method="POST" >
        <table cellpadding="10px" border="0">
            <tr>
                <td class="textLogin">Username :</td>
                <td class="inputColumn">
                    <input id="inputUserName" class="inputText" type="text" name="username" value="Username" 
                     onfocus="if(this.value=='Username'){this.value='';}" onblur="if(this.value==''){this.value='Username';}"/>

                </td>
            </tr>
            <tr>
                <td class="textLogin">Password :</td>
                <td class="inputColumn">
                    <input id="inputPassword" class="inputText"   type="password" name="password" value="" 
                    onfocus="" onblur=""/>
                </td>
            </tr>
            <tr>
                <td class="textLogin" colspan="2">
                    <span id="noityText" ></span>
                </td>
            </tr>
            <tr>
                <td class="resetColumn"><button class="basicButton" id="resetLogin" type="reset"  >Reset</button></td>
                <td><button class="basicButton" id="submitLogin" type="submit"  >Submit</button></td>
            </tr>
        </table>
</form>

JavaScrpit：

$(document).ready(function() {
$("#loginForm").submit(function(e){
     e.preventDefault();
});
$("#submitLogin").click(function(){
var userName=$("#inputUserName").val();
userName=userName.trim();
var password=$("#inputPassword").val();
password=password.trim();
//get the form data and then serialize that
dataString = $("#loginForm").serialize();
dataString ="username"+userName;
console.log(dataString);
if((userName=='')||(userName=='Username')){
    //$("#noityText").html("<span id='noityText' >Vui lòng điền tên đăng nhập</span>");
    $("#noityText").text("Vui lòng điền tên đăng nhập");
}
else{
    $("#noityText").text("");
}
$.ajax({
    type:"POST",
    url:"/MavenWebApp/login",
    data:dataString,
    dataType:"json",
    success:function(data,textStatus, jqXHR){
        if(data.success){
            $("#noityText").text("Đăng nhập thành công :D");
            console.log(data);
        }
        else{
            $("#noityText").text("Đăng nhập thất bại :D");
            console.log(data);
        }
    },
    error:function(jqXHR,textStatus, errorThrow){
        $("#noityText").text("Errors");
    },
    beforeSend:function(jqXHR,settings){
        $("#noityText").text("Start send");
    },
    complete:function(jqXHR,textStauts){}
    });
});
});

和servlet：

protected void service(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    System.out.println("In servlet :D ");
    String userName=request.getParameter("username");
    //userName=userName.trim();
    String password=request.getParameter("password");
    //password=password.trim();
    System.out.println(password);
    PrintWriter out = response.getWriter();
    response.setContentType("text/html");
    response.setHeader("Cache-control", "no-cache, no-store");
    response.setHeader("Pragma", "no-cache");
    response.setHeader("Expires", "-1");

    response.setHeader("Access-Control-Allow-Origin", "*");
    response.setHeader("Access-Control-Allow-Methods", "POST");
    response.setHeader("Access-Control-Allow-Headers", "Content-Type");
    response.setHeader("Access-Control-Max-Age", "86400");

    Gson gson = new Gson();
    JsonObject myObj = new JsonObject();
    if (this.getUser(userName, password)){
        myObj.addProperty("success", true);
    }else{
        myObj.addProperty("success", false);
    }
    out.println(myObj.toString());
    out.close();


}

这是eclipse中的文本显示：

In servlet :D - service 
null
null

servlet中的字符串userName和密码为null！我怎么能解决这个错误？：d 对不起，如果我的问题让你不舒服，因为我是初学者，我的英语不好：D

Answer 1

//get the form data and then serialize that
dataString = $("#loginForm").serialize();

看起来没问题（但检查这个字符串是什么），但是然后用

覆盖它

 dataString ="username"+userName;

为什么要覆盖它？如果这样做，您至少需要保持表单编码（例如，参数名称和值之间需要=）。

Answer 2

尝试使用serializeArray（） -

dataString = $("#loginForm").serializeArray();

删除此行 -

dataString ="username"+userName;

servlet中不能有getParameter吗？

2 个答案: