eja*_*dra -1 javascript jquery typeerror
我是jQuery的新手,我试图创建一个登录表单,当用户输入一个简短的用户名时切换文本.所以这是我的代码,当我点击按钮时没有任何反应.我检查了控制台,它说在我的code.I've特定的行标题在错误可能的拼写错误检查过,但我没有看到任何.所以这是我的代码:
<!DOCTYPE HTML>
<html>
<head>
<script type="text/javascript" src = "jquery-2.1.1.min.js"></script>
<script type="text/javascript">
var usr = document.getElementById("textbox1").value;//error is over this line.
if(usr.length<=6){
$(document).ready(function(){
$("#press").click(function(){
$("#warning").toggle();
});
});
}
</script>
</head>
<body>
<div id = "middleBlock">
<form>
<p id = "username"> Username </p>
<p id = "password"> Password </p>
<input type="text" name="user" id="textbox1">
<input type="password" name="pass" id="textbox2">
<input type = "button" value = "LOG IN" id = "press">
<p id = "note"> Note: Use the username and password provided by your department. </p>
<div id = "warning" style = "display:none;padding:3%;">Username too short.</div>
</form>
</div>
</body>
移动script到结束body或将其包装在DOM就绪函数中.您getElementById在呈现页面之前执行,因此抛出空错误.
例:
$(document).ready(function(){
var usr = document.getElementById("textbox1").value;//error is over this line.
if(usr.length<=6){
$("#press").click(function(){
$("#warning").toggle();
});
}
});