一个简单的JS问题~~

function enter_key_down(e)
{
    var keycode;
    if (window.event) keycode = window.event.keyCode;
    else if (e) keycode = e.which;
    else return true;
    if (keycode == 13)
    {
       SearchBtn_B_Click();
       return false;
    }
    else
       return true;
}<body onkeydown = " return enter_key_down(this.event);">
是这个意思吗

<html>
<head>
<script language="javascript">
function enter_key_down(e)
{
    var keycode;
    if (window.event) keycode = window.event.keyCode;
    else if (e) keycode = e.which;
    else return true;
    if (keycode == 13)
    {
       SearchBtn_B_Click();
       return false;
    }
    else{
       return true;
}
alert("success");
}
</script>
</head>
<body>
<input type="text" id="username"><br>
<input type="text" id="password"><br>
<input type="button" onkeydown = " return enter_key_down(this.event);" value="登陆">
</body>
</html>
我是这样写的，可以没反映？？

假如你页面有个登陆按钮客户端ID是： button1，输入框控件加客户端键盘事件根据键值（13是回车），触发登陆按钮的点击事件。
代码基本如楼上的。function enter_key_down(e)
{
    var keycode;
    if (window.event) keycode = window.event.keyCode;
    else if (e) keycode = e.which;
    else return true;
    if (keycode == 13)
    {
       var button = document.getElementByID("button1");
       button.click();
       return false;
    }
    else
       return true;
}

<html>
<head>
<script language="javascript">
function enter_key_down(e)
{
    var keycode;
    if (window.event) keycode = window.event.keyCode;
    else if (e) keycode = e.which;
    else return true;
    if (keycode == 13)
    {
       SearchBtn_B_Click();
       return false;
    }
    else{
       return true;
    }
alert("success");
}
</script>
</head>
<body>
<input type="text" id="username"><br>
<input type="text" id="password" onkeydown = " return enter_key_down(this.event);"><br>
<input type="button"  value="登陆">
</body>
</html>

<form id="form1" runat="server" defaultbutton="loginButton">
用这个不行嘛。

不好意思啊~~  可能是我蛮笨的
   代码没用你自己试了的？

     我想补充的是，当输完用户名后按enter光标跳到密码框，密码输完了，按enter登陆~~

那个光标跳到下个控件的代码我有，就是每次登陆要按2次enter  很不爽~~    <script language='javascript' for='document' event='onkeydown'>
      if(event.keyCode==13   &&   event.srcElement.type!='button'   &&   event.srcElement.type!='submit'   &&   event.srcElement.type!='reset'   &&   event.srcElement.type!='textarea'   &&   event.srcElement.type!=''   &&   event.srcElement.type!='image')
        {
            event.keyCode=9;
        }
    </script>

<html>
<head>
<title>asdfas</title>
<script type= "text/javascript" language="javascript">
function sayHello(aa)
{
alert(aa);
}
function test(e)
{
var keycode;
if(window.event) keycode = window.event.keyCode;
else if(e) keycode = e.which;
       else return true;
if(keycode == 13)
{
   var button = document.getElementById("login");
   if(typeof(button) == undefined || button == null)
    alert("button is not available.");
    else
    button.click();
   return false;
  }
  else
   return true;
}
</script></head>
<body>
<input id="username" type="text" />
<input id="psd" type="password" onkeydown="test(event)"/>
<input id="login"   type="button" onclick="sayHello('GO!')" />
</body>
</html>

<html>
<head>
<title>asdfas</title>
<script type= "text/javascript" language="javascript">
function sayHello(aa)
{
alert(aa);
}function test(e)
{
var keycode;
if(window.event) keycode = window.event.keyCode;
else if(e) keycode = e.which;
       else return true;
if(keycode == 13)
{
    var id;
   if(e.srcElement)
      id = e.srcElement.id;
      else
      id = e.target.id;
      if(id == "username")
      {
       var input = document.getElementById("psd");
       input.focus();
       }
       if(id == "psd")
       {
            var btn = document.getElementById("login");
            btn.click();
        }
  }
  else
   return true;
}
</script></head>
<body onload="document.getElementById('username').focus();">
<input id="username" type="text" onkeydown="test(event)"/>
<input id="psd" type="password" onkeydown="test(event)"/>
<input id="login"   type="button" value="登陆" onclick="sayHello('GO!')" />
</body>
</html>这个是完善过的代码。应该可以符合你的需求~~~~~

调试易

一个简单的JS问题~~

解决方案 »