java.lang.IllegalStateException：PWC1227：提交响应后无法转发……为什么要发送？
[英]java.lang.IllegalStateException: PWC1227: Cannot forward after response has been committed…why it was comming?
问题描述
to add some row data into a table, affter submmiting the button i have to show the details(data) in the next page of that regarding table. 
要将某些行数据添加到表中，请在单击按钮后在该表的下一页中显示详细信息（数据）。 when i am using RequestDispather class i am getting the java.lang.IllegalStateException: ........ it was also comming while using response.sendRedirect("View.jsp"); 当我使用RequestDispather类时，我正在获取java.lang.IllegalStateException: ........在使用response.sendRedirect("View.jsp"); ..... i am sending the code what i used in my page. .....我正在发送我在页面中使用的代码。
if(msg.equals("Values Added")){
RequestDispatcher rd = request.getRequestDispatcher("View.jsp");
rd.forward(request, response);
}
(OR) （要么）
if(msg.equals("Values Added")){
response.sendRedirect("View.jsp");
}
3 个解决方案
解决方案1
3 2010-12-09 12:10:46
JSP is part of the response. JSP是响应的一部分。 You cannot change the response like that from inside a JSP. 您无法从JSP内部更改响应。 It's too late then. 那时为时已晚。 This piece of code should have been placed in a servlet class. 这段代码应该放在servlet类中。
Change your form to submit to a servlet instead: 更改您的表单以提交给servlet：
<form action="servleturl" method="post">
Create a servlet class which is mapped on an url-pattern of /servleturl/* and move all the Java code you have there in JSP into the doPost() method. 创建一个Servlet类，该类映射到/servleturl/*的url-pattern ， /servleturl/* JSP中存在的所有Java代码移到doPost()方法中。
See also: 也可以看看：
解决方案2
0 2010-12-09 12:07:10
使用else代替if
解决方案3
0 已采纳 2011-01-27 20:42:17
The following is not true per se: 以下内容本身是不正确的：
"You cannot change the response like that from inside a JSP. It's too late then." “您不能从JSP内部更改响应。那时为时已晚。”
Just place your postback check and redirect before the html tag in your jsp...then everything will be fine. 只需将您的回发检查并重定向到jsp中的html标记之前即可。
So: 所以：
<% if(msg.equals("Values Added")){
response.sendRedirect("View.jsp");
} %>
<html > ... </html>
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