1.当输入用户名和密码为空的时候,需要判断。这时候就用到了校验用户名和密码,这个需要在jsp的前端页面写;有两种方法,一种是用submit提交。一种是用button提交。
方法一:
在jsp的前端页面的头部插入一个js方法:
function checkUser(){
var result = document.getElementById("userid").value;
var password = document.getElementById("userpassid").value;
if(result == "" ){
alert("用户名不能为空");
return false;
}
if(password == "" ){
alert("密码不能为空");
return false;
}else{
return true;
}
}
在form表单里写成这样:
<form id="formid" name= "myform" method = ‘post‘ action = ‘user_login_submit.action‘ onsubmit = "return checkUser();" >
<table width="100%" border="0">
<tr>
<td width="60" height="40" align="right">用户名 </td>
<td><input type="text" value="" class="text2" name = "username" id = "userid"/></td>
</tr>
<tr>
<td width="60" height="40" align="right">密 码 </td>
<td><input type="password" value="" class="text2" name = "userpass" id = "userpassid"/></td>
</tr>
<tr>
<td width="60" height="40" align="right"> </td>
<td><div class="c4">
<input type="submit" value="" class="btn2" />
方法二:
function checkUser(){
var result = document.getElementById("userid").value;
var password = document.getElementById("passid").value;
if(result == "" ){
alert("用户名不能为空");
return false;
}
if(password == "" ){
alert("密码不能为空");
return false;
}
document.getElementById("formid").submit();
}
form表格的写法,需要写id
<form id="formid" method = ‘post‘ action = ‘user_login_submit.action‘ >
button按钮的写法如下:
<input type="button" value="" class="btn2" onclick = "checkUser();" />