XmlHttpRequest的post，参数为空？

Question

function doRequestUsingPOST() {
    createXMLHttpRequest();    
    var url = "dogp.aspx?timeStamp=" + new Date().getTime();
    var queryString = createQueryString();    
    xmlHttp.open("POST", url, true);
    xmlHttp.onreadystatechange = handleStateChange;
    xmlHttp.setRequestHeader("Content-Type", "application/x-www-form-urlencoded");    
    xmlHttp.send(queryString);
}
服务器端：
string firstname = this.Request.QueryString["FirstName"];
        string middlename = this.Request.QueryString["MiddleName"];
        string birthday = this.Request.QueryString["Birthday"];        
        StringBuilder sb = new StringBuilder();
        sb.Append("Hello ").Append(firstname).Append(" ").Append(middlename);
        sb.Append(".Your birthday is ").Append(birthday).Append(". [Method ").Append(this.Request.RequestType).Append("]");
        this.Response.Write(sb.ToString());
        this.Response.Flush();
        this.Response.Close();
为什么参数传过来都是空的？

Answer 1

createQueryString();    函数什么内容啊,怎么都差不得贴出来呢,
post方式的send()方法,URL中的参数是在send方法里传过去的,你检查一下queryString的内容.

Answer 2

function createQueryString() {
    var firstName = document.getElementById("firstName").value;
    var middleName = document.getElementById("middleName").value;
    var birthday = document.getElementById("birthday").value;
    
    var queryString = "FirstName=" + firstName + "&MiddleName=" + middleName
        + "&Birthday=" + birthday;
    
    return queryString;
}
客户端的应该不会有错，我从另一个文件copy过来的

Answer 3

<script type="text/javascript">
var xmlHttp;function createXMLHttpRequest() {
    if (window.ActiveXObject) {
        xmlHttp = new ActiveXObject("Microsoft.XMLHTTP");
    } 
    else if (window.XMLHttpRequest) {
        xmlHttp = new XMLHttpRequest();
    }
}
    
function createQueryString() {
    var firstName = document.getElementById("firstName").value;
    var middleName = document.getElementById("middleName").value;
    var birthday = document.getElementById("birthday").value;
    
    var queryString = "FirstName=" + firstName + "&MiddleName=" + middleName
        + "&Birthday=" + birthday;
    
    return queryString;
}function doRequestUsingGET() {
    createXMLHttpRequest();
    
    var queryString = "dogp.aspx?";
    queryString = queryString + createQueryString() 
        + "&timeStamp=" + new Date().getTime();
    xmlHttp.onreadystatechange = handleStateChange;
    xmlHttp.open("GET", queryString, true);
    xmlHttp.send(null);
}function doRequestUsingPOST() {
    createXMLHttpRequest();
    
    var url = "dogp.aspx?timeStamp=" + new Date().getTime();
    var queryString = createQueryString();
    
    xmlHttp.open("POST", url, true);
    xmlHttp.onreadystatechange = handleStateChange;
    xmlHttp.setRequestHeader("Content-Type", "application/x-www-form-urlencoded");    
    xmlHttp.send(queryString);
}
    
function handleStateChange() {
    if(xmlHttp.readyState == 4) {
        if(xmlHttp.status == 200) {
            parseResults();
        }
    }
}function parseResults() {
    var responseDiv = document.getElementById("serverResponse");
    if(responseDiv.hasChildNodes()) {
        responseDiv.removeChild(responseDiv.childNodes[0]);
    }
    
    var responseText = document.createTextNode(xmlHttp.responseText);
    responseDiv.appendChild(responseText);
}</script>

Answer 4

好像这么写：string firstname = this.Request.Parms["FirstName"];

Answer 5

正在寻找XMLHttpRequest对象open()  函数的POST用法，学习啦～