分享程序员开发的那些事...
更新时间：2023-12-03 22:57:52
服务器不应该验证名称&密码？我想是的。
所以让服务器回答一个合适的返回值：
如果是真/通过你应该授予访问，如果是假/没通过，你应该出示另一页。
编辑：
所以你不要做服务器请求。那么你也不应该使用HTTPServlet。
Shouldn''t the server validate the name & password? I guess so.
So let the server answer with a suitable return value:
In case of true / "pass" you should grant access, in case of false / "no pass" you should present another page.
So you do NOT do Server-Requests. Then you should also not use a HTTPServlet.
public class SimpleServlet extends GenericServlet {
public void service(ServletRequest request, ServletResponse response)
throws ServletException, IOException {
// figure values from client's request
String strParam = request.getParameter("param");
String strNick = request.getParamter("nick");
String strPW = request.getParamter("pw");
if("login".equals(strParam) ){ // ask first, cause strNick & strPW could be NULL
// present login
}
else if("dolly".equals(strNick) && "dolly".equals(strPW) ){ // login values fine
response.getWriter().write("<html><body>Login granted</body></html>");
}
else { // login not granted
response.getWriter().write("<html><body>OOOPS!</body></html>");
}
}
}
您只需提供登录（就像您现在在代码中一样） ）点击登录调用内部函数检查静态缺口/密码。
之后，你出现
You simply present login (as you do in your code now) and call an internal function on click "login" to check the static nick/pw.
After that, you present