未调用onreadystatechange

Question

我期望在调用show_Message函数时出现一个警报框，但onreadystatechange不是working.The，其他警报框工作正常

下面是我的js函数

 function send_Message(){
  var msg=document.getElementById("msg").value;
if(msg.length===0||msg===""){
    alert("please enter some message");
    return;
}

var sender=document.getElementById("username").value;
var sendto=document.getElementById("chat_id").options[document.getElementById("chat_id").selectedIndex].value;
alert(sender+" "+sendto);
var xhttp=new XMLHttpRequest();
xhttp.onreadystatechange=function(){
    alert('hello');
    if(xhttp.readyState==4 && xhttp.status==200){
        document.getElementById("chat_logs").innerHTML=xhttp.responseText;
    }
    xhttp.open('GET','send_messages.php?sender='+sender+'sendto='+sendto+'message='+msg,true);
    xhttp.send(null);
}
 }

Answer 1

xhttp.open('GET','send_messages.php?sender='+sender+'sendto='+sendto+'message='+msg,true); xhttp.send(null);

应该在onreadystatechange之外

将其更改为

function send_Message(){
  var msg=document.getElementById("msg").value;
if(msg.length===0||msg===""){
    alert("please enter some message");
    return;
}

var sender=document.getElementById("username").value;
var sendto=document.getElementById("chat_id").options[document.getElementById("chat_id").selectedIndex].value;
alert(sender+" "+sendto);
var xhttp=new XMLHttpRequest();
xhttp.onreadystatechange=function(){
    alert('hello');
    if(xhttp.readyState==4 && xhttp.status==200){
        document.getElementById("chat_logs").innerHTML=xhttp.responseText;
    }
}
xhttp.open('GET','send_messages.php?sender='+sender+'sendto='+sendto+'message='+msg,true);
    xhttp.send(null);
 }

Answer 2

你犯了结束语的错误。xhttp.onreadystatechange=function(){

上面的函数应该在if语句之后关闭。

检查下面更正的代码：

function send_Message(){
  var msg=document.getElementById("msg").value;
if(msg.length===0||msg===""){
    alert("please enter some message");
    return;
}

var sender=document.getElementById("username").value;
var sendto=document.getElementById("chat_id").options[document.getElementById("chat_id").selectedIndex].value;
alert(sender+" "+sendto);
var xhttp=new XMLHttpRequest();
xhttp.onreadystatechange=function(){
    alert('hello');
    if(xhttp.readyState==4 && xhttp.status==200){
        document.getElementById("chat_logs").innerHTML=xhttp.responseText;
    }
}; 
    xhttp.open('GET','send_messages.php?sender='+sender+'sendto='+sendto+'message='+msg,true);
    xhttp.send(null);

 }

Answer 3

这可能是因为您在onstatechange内部使用了打开和发送函数。

function send_Message(){
  var msg = document.getElementById("msg").value;
  if( msg.length === 0 || msg === "" ){
    alert("please enter some message");
    return;
  }

  var sender=document.getElementById("username").value;
  var sendto=document.getElementById("chat_id").options[document.getElementById("chat_id").selectedIndex].value;
  alert(sender+" "+sendto);

  var xhttp=new XMLHttpRequest();
  xhttp.onreadystatechange=function(){
    alert('hello');
    if(xhttp.readyState==4 && xhttp.status==200){
        document.getElementById("chat_logs").innerHTML=xhttp.responseText;
    }
  }

  xhttp.open('GET','send_messages.php?sender='+sender+'sendto='+sendto+'message='+msg,true);
  xhttp.send(null);
}