这个代码运行时跳转不到success.jsp或者fail.jsp页面,高手帮忙解决下!小妹不胜感激……
success.jsp<%@ page contentType="text/html;Charset=GB2312" %>
<HTML>
<BODY>
SUCCESS!
</BODY>
</HTML>fail.jsp<%@ page contentType="text/html;Charset=GB2312" %>
<HTML>
<BODY>
fial!
</BODY>
</HTML>以下是我的login.jsp
<%@ page contentType="text/html;Charset=GB2312" %>
<%@ page import="java.io.*" %><HTML>
<BODY>
<h1>登录界面</h1>
<script language="javascript">
function unkonow_click()
{document.frm1.action="success.jsp";
document.frm1.submit();
}
function know_click()
{
Connection con;
Statement sql;
String name;
try{
Class.forName("sun.jdbc.odbc.JdbcOdbcDriver");
}
catch(ClassNotFoundException e)
{
out.print(e);
}
try{
con=DriverManager.getConnection("jdbc:odbc:messagedb","sa","sa");
sql=con.createStatement();
name=request.getParameter(username);
if(name.length()>0)
{
int rs=sql.executeQuery("SELECT * FROM user WHERE username="name"");
if(rs!=0)
{ document.frm1.action="success.jsp";
document.frm1.submit();
}
else
{
document.frm1.action="fail.jsp";
document.frm1.submit();
}
}
else
{
</script>
<form method="post" name="frm1">
用户名:<input type=text name="username">
<BR>
密码:<input type=text name="password">
<BR>
<input type="button" name="guest" value="匿名登陆" onclick="unkonow_click()">
<input type="button" name="username" value="登录" onclick="know_click()">
</form>
<script language="javascript">
}
}
catch(SQLException e)
{
out.print(e);
}
} </script>
</BODY>
</HTML>
success.jsp<%@ page contentType="text/html;Charset=GB2312" %>
<HTML>
<BODY>
SUCCESS!
</BODY>
</HTML>fail.jsp<%@ page contentType="text/html;Charset=GB2312" %>
<HTML>
<BODY>
fial!
</BODY>
</HTML>以下是我的login.jsp
<%@ page contentType="text/html;Charset=GB2312" %>
<%@ page import="java.io.*" %><HTML>
<BODY>
<h1>登录界面</h1>
<script language="javascript">
function unkonow_click()
{document.frm1.action="success.jsp";
document.frm1.submit();
}
function know_click()
{
Connection con;
Statement sql;
String name;
try{
Class.forName("sun.jdbc.odbc.JdbcOdbcDriver");
}
catch(ClassNotFoundException e)
{
out.print(e);
}
try{
con=DriverManager.getConnection("jdbc:odbc:messagedb","sa","sa");
sql=con.createStatement();
name=request.getParameter(username);
if(name.length()>0)
{
int rs=sql.executeQuery("SELECT * FROM user WHERE username="name"");
if(rs!=0)
{ document.frm1.action="success.jsp";
document.frm1.submit();
}
else
{
document.frm1.action="fail.jsp";
document.frm1.submit();
}
}
else
{
</script>
<form method="post" name="frm1">
用户名:<input type=text name="username">
<BR>
密码:<input type=text name="password">
<BR>
<input type="button" name="guest" value="匿名登陆" onclick="unkonow_click()">
<input type="button" name="username" value="登录" onclick="know_click()">
</form>
<script language="javascript">
}
}
catch(SQLException e)
{
out.print(e);
}
} </script>
</BODY>
</HTML>
如果不会servlet,用jsp也可以,把连接数据库那段写在另一个jsp里,然后判断再跳转
当然<form>里必须要有<form action="你的jsp或servlet">
报错的内容如下:
org.apache.jasper.JasperException: An exception occurred processing JSP page /login.jsp at line 2522: con=DriverManager.getConnection("jdbc:odbc:messagedb","sa","sa");
23: sql=con.createStatement();
24: name=request.getParameter("username");
25: if(name.length()>0)
26: {
27: rs=sql.executeQuery("SELECT * FROM user WHERE username="+name);
28: if(rs!=null)
那就照我说的第二种方法做啊,另写一个jsp,把连数据库那些放jsp里
然后你目前的jsp里的form的action指向那个jsp