我有3 jsp pages(index.jsp, result1.jsp,result2.jsp)
和一個servlet
。處理正在被如下實現:在jsp中添加後退按鈕
的index.jsp
<form method="post" action="MYSERVLET" >
<input type="text" name="studentname" id="studentname"/>
<input type="submit" value="click" />// goes to a servlet named as MYSERVLET
</form>
MYSERVLET
String t=request.getParameter("studentname");
------fetching data from database and sending to result1.jsp------
String nextJSP = "/result1.jsp";
RequestDispatcher dispatcher = getServletContext().getRequestDispatcher(nextJSP);
dispatcher.forward(request,response);
result1.jsp
//檢索從servlet的數據和在顯示result1.jsp
數據,並且還通過相同的值到result2.jsp
out.println(retrieved result from servlet);
<a href="result2.jsp?somestring=<%out.println(retrieved result from servlet);%>"> <%out.println(retrieved result from servlet);%></a>// after clicking on "retrieved result from servlet" i am passing this retrieved value to another jsp result2.jsp and also retrieving some data from database as follows:
result2.jsp
String w=request.getParameter("somestring");
out.println(w);
我想給一個back button here (in result2.jsp)
,使點擊back button
後,它會帶我去哪裏result1.jsp
我可以看到之前顯示的數據(out.println(retrieved result from servlet);)
。但是,當我點擊瀏覽器後退按鈕(在result2.jsp),然後每次警報即將"Confirm Form Resubmission"
然後我重新加載該頁面看到(out.println(retrieved result from servlet);)
,我想避免此警報,我該如何避免它?我怎樣才能直接看到沒有任何警報的頁面?