Javascript 未捕获的引用错误:';somevar';没有定义
我试图通过ajax调用将一些数据插入mysql,但当我试图从表单中获取数据时,javascript在第一个变量上抛出了一个错误“uncaughtreferenceerror:anjelis未定义”Javascript 未捕获的引用错误:';somevar';没有定义,javascript,php,jquery,mysql,ajax,Javascript,Php,Jquery,Mysql,Ajax,我试图通过ajax调用将一些数据插入mysql,但当我试图从表单中获取数据时,javascript在第一个变量上抛出了一个错误“uncaughtreferenceerror:anjelis未定义” <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8"> <title>aloooooooooooooo!!!!!!!!!</title>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
<title>aloooooooooooooo!!!!!!!!!</title>
<link rel="stylesheet" href="bootstrap/css/bootstrap.css">
<link rel="stylesheet" type="text/css" href="css.css">
<script type="text/javascript" src="js/jquery-1.11.2.min.js"></script>
<script type="text/javascript" src="bootstrap/js/bootstrap.js"></script>
<script src="js/main.js"></script>
<script src="js/ajax.js"></script>
<script>
$(document).ready(function(e){
$.ajaxSetup({cache:false});
setInterval(function(){$('#chatlog').load('logs.php');},1000);
setInterval(function(){$('#user_list').load('user_list.php');},1000);
});
function submitChat(){
var uname=<?php echo $_SESSION['username']?>;
var msg=form.msg.value;
var xmlhttp=new XMLHttpRequest();
xmlhttp.onreadystatechange=function(){
if(xmlhttp.readyState==4 && xmlhttp.status==200){
document.getElementById('chatlog').innerHTML=xmlhttp.responseText;
}
}
xmlhttp.open('GET','insert.php?username='+uname+'&msg='+msg,true);
xmlhttp.send();
}
</script>
</head>
你好,你好!!!!!!!!!
$(文档).ready(函数(e){
$.ajaxSetup({cache:false});
setInterval(function(){$('#chatlog').load('logs.php');},1000);
setInterval(function(){$('#user_list').load('user_list.php');},1000);
});
函数submitChat(){
var uname=;
var msg=form.msg.value;
var xmlhttp=new XMLHttpRequest();
xmlhttp.onreadystatechange=函数(){
if(xmlhttp.readyState==4&&xmlhttp.status==200){
document.getElementById('chatlog').innerHTML=xmlhttp.responseText;
}
}
open('GET','insert.php?username='+uname+'&msg='+msg,true);
xmlhttp.send();
}
身体
<body>
<nav class="navbar navbar-inverse">
<div class="container">
<div class="navbar-header">
<a class="navbar-brand" href="#">ChatCY</a>
</div>
<ul class="nav navbar-nav navbar-right">
<li><?php if(!isset($_SESSION['username'])){echo '<a href="login.php">Login/Signin</a>';}else{
echo'<a href="logout.php">Logout</a>';}?></li>
</ul>
</div>
</nav>
<div class="container">
<div class="row">
<div class="col-md-8">
<div class="panel panel-default">
<div id="chatlog" class="panel-body"></div>
</div>
</div>
<div class="col-md-2">
<div class="panel panel-default">
<div id="user_list" class="panel-body"></div>
</div>
</div>
</div>
<div class="row">
<div class="col-md-8">
<form class="form-inline">
<form name="form" class="form-group" onsubmit="return false;">
<input type="textarea" class="form-control" name="msg" style="width:90%;"<?php if(!isset($_SESSION['username'])){
echo 'disabled="disabled" placeholder="You must be logged in to chat .."';}?>></input>
<button type="button" class="btn btn-default" name="send" <?php if(!isset($_SESSION['username'])){
echo 'disabled="disabled"';}?> onClick="submitChat();">send</button>
</form>
</form>
</div>
</div>
</div>
onClick=“submitChat();”>发送
错过了引用-
var uname= "<?php echo $_SESSION['username']?>";
var form = $('form.form-group');
somevar
将是未定义的
您没有将
表单
传递或定义到函数或函数内部。代码中有很多问题
var uname = "<?php echo $_SESSION['username']?>";
echo
编写HTML标记
有结束标记,也没有您错过了代码
会话\u start()
,
你可以用另一种方法,在html中创建一个inpu标记
changevar uname=代码>此为var uname=“”代码>并请检查。您忘记在页面顶部启动会话会话\u start()
!!另外,也不是必需的。@dll\u onFire您不能使用jQuery吗?现在它将错误(表单未定义)带到下一行,即var msg=form.msg.value@哦,太好了!接下来的一行是var msg=form.msg.value,它表示(form未定义)@dll\u onFire您能用呈现的HTML更新问题吗?你在查看源代码时看到的那个?
var uname = "<?php echo $_SESSION['username']?>";
session_start();
$.ajax(function () {
"url": "/",
// Blah Blah
});