form里用GET提交参数到servlet里的话,用request.getParameter可以获得想要的参数,可是用POST提交后再用request.getParameter得到的就是NULL了,网上看好多人说POST用getParameter也可以得到,可我就是得不到啊谁来指点下啊。有人说用截取POST的流来获得参数,哪位大侠能举个小例子啊,在此谢过。新手分少,别见怪
(能顺便讲一下get和post的区别最好了,我只知道GET后面URL有参数,POST没有……………………)
(能顺便讲一下get和post的区别最好了,我只知道GET后面URL有参数,POST没有……………………)
<input type="text" value="10" name="proid"/>
<input type="submit" value="upload" >
</form>servlet里就一句话public void doGet(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException { response.setContentType("text/html");
System.out.println(request.getParameter("proid"));
}public void doPost(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
doGet(request, response);
}
你只能去得到流:ServletInputStream sis = request.getInputStream();
然后,你再去处理。
如果你不想自己去处理流,那最好下载一个smartupload.jar的包。来方便帮你处理。
一般来说,如果你不涉及到文件上传的话最好不要用enctype="multipart/form-data"改为用:
enctype="application/x-www-form-urlencoded"
public void doGet(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {response.setContentType("text/html");System.out.println(request.getParameter("proid"));
}public void doPost(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
System.out.println(request.getParameter("proid"));
}
form method="post" action="servlet/posttest">
<input type="text" value="10" name="proid"/> <input type="submit" value="upload" >
</form>enctype="multipart/form-data"这段去掉::
http://blog.csdn.net/darxin/archive/2010/01/02/5119767.aspx