php - Javascript/ajax/php 问题 : sending from server to client works, 从客户端发送到服务器失败

Question

抱歉重新发帖(管理员，请删除另一张!)。 因为你们给了我很大的帮助，我有点希望你们能再次帮助我，同时有以下问题: 我目前正在尝试通过允许 PHP 中的管理器类通过 XmlHttpobject 与客户端的 javascript 进行通信来使用 AJAX。但是，我可以通过 JSON 向客户端发送一些内容，但我无法在客户端读取它。事实上，我收到的错误是“时间”是 session 中 undefined index 。所以我想知道:我做错了什么？

Ajax 的 JavaScript 代码:

<script type="text/javascript">
            var sendReq = GetXmlHttpObject();
            var receiveReq = GetXmlHttpObject();
            var JSONIn = 0;
            var JSONOut= 0;
            //var mTimer;   
//function to retreive xmlHTTp object for AJAX calls (correct)
function GetXmlHttpObject()
{
    var xmlHttp=null;
    try
     {
         // Firefox, Opera 8.0+, Safari
         xmlHttp=new XMLHttpRequest();
     }
    catch (e)
     {
         // Internet Explorer
        try
        {
        xmlHttp=new ActiveXObject("Msxml2.XMLHTTP");
        }
        catch (e)
        {
            xmlHttp=new ActiveXObject("Microsoft.XMLHTTP");
        }
     }
    return xmlHttp;
}

            //Gets the new info from the server
            function getUpdate() {
                if (receiveReq.readyState == 4 || receiveReq.readyState == 0) {
                    receiveReq.open("GET", "index.php?json="+JSONIn+"&sid=$this->session", true);
                    receiveReq.onreadystatechange = updateState; 
                    receiveReq.send(null);
                }           
            }
            //send a message to the  server.
            function sendUpdate(JSONstringsend) {
                JSONOut=JSONstringsend;
                if (sendReq.readyState == 4 || sendReq.readyState == 0) {
                    sendReq.open("POST", "index.php?json="+JSONstringsend+"&sid=$this->session", true);
                    sendReq.setRequestHeader('Content-Type','application/x-www-form-urlencoded');
                    alert(JSONstringsend);
                    sendReq.onreadystatechange = updateCycle;
                    sendReq.send(JSONstringsend);
                }                           
            }
            //When data has been send, update the page.
            function updateCycle() {
                getUpdate();
            }
            function updateState() {
                if (receiveReq.readyState == 4) {
                    // JSONANSWER gets here (correct):
                    var JSONtext = sendReq.responseText;
                    // convert received string to JavaScript object (correct)
                    alert(JSONtext);
                    var JSONobject = JSON.parse(JSONtext);
                    //   updates date from the JSONanswer (correct):
                     document.getElementById("dateview").innerHTML= JSONobject.date;        

                    }
                    //mTimer = setTimeout('getUpdate();',2000); //Refresh our chat in 2 seconds
                }
    </script>

实际使用ajax代码的函数:

//datepickerdata
        $(document).ready(function(){
                $("#datepicker").datepicker({
                    onSelect: function(dateText){
                    var JSONObject = {"date": dateText};
                    var JSONstring = JSON.stringify(JSONObject);
                    sendUpdate(JSONstring);
                    },  
                    dateFormat: 'dd-mm-yy' 
            });

        });

        </script>

以及 PHP 代码:

private function handleReceivedJSon($json){
    $this->jsonLocal=array();
    $json=$_POST["json"];
    $this->jsonDecoded= json_decode($json, true);
    if(isset($this->jsonDecoded["date"])){
        $_SESSION["date"]=$this->jsonDecoded["date"];
        $this->useddate=$this->jsonDecoded;

    }
    if(isset($this->jsonDecoded["logout"])){
        session_destroy();
        exit("logout");
    }
    header("Last-Modified: " . gmdate( "D, d M Y H:i:s" ) . "GMT" ); 
    header("Cache-Control: no-cache, must-revalidate" ); 
    header("Pragma: no-cache" );
    header("Content-Type: text/xml; charset=utf-8");
    exit($json);
}

Answer 1

这里只是一个问题，您是否正在将旧版应用程序移植到 jquery？

我首先要摆脱这里的“Ajax”调用，以在 jquery 中清理它

$(...).ajax(...)

或原型(prototype)

ajax = new Ajax.Request(...)

然后当响应返回时，请务必解析/评估`

response.responseText
response.responseText.evalJSON()

等等...