Cannot call sendError() after the response has been committed – baiyangliu – 博客频道

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当response提交后，不能调用sendError()，什么意思？

出现这个错误，一定是多次response导致的。可以这么理解，承载客户端和服务器进行Http交互的Socket连接已经关闭了，而你还试图发送数据给客户端，显然会出错。就好比我俩打电话，我都挂电话了，你还在“喂喂喂”。

例如下面这段代码就会出现此错误：

import java.io.Writer;
import javax.servlet.http.HttpServletResponse;
import org.apache.struts2.ServletActionContext;
import com.opensymphony.xwork2.ActionSupport;
public class LoginAction extends ActionSupport {
/**
*
*/
private static final long serialVersionUID = 1L;
private String userName;
private String pwd;
private String verifyCode;
private String ajax;
// 错误的写法
@Override
public String execute() throws Exception {
// 通过ajax登录
if (ajax != null) {
HttpServletResponse response = ServletActionContext.getResponse();
Writer writer = response.getWriter();
writer.write(“登录成功!”);
writer.flush();
writer.close();
}
return SUCCESS;
}
// 正确写法
public String login1() throws Exception {
if (ajax != null) {
HttpServletResponse response = ServletActionContext.getResponse();
Writer writer = response.getWriter();
writer.write(“登录成功!”);
writer.flush();
writer.close();
return null;
}
return SUCCESS;
}
// 正确写法
public String login2() throws Exception {
if (ajax != null) {
HttpServletResponse response = ServletActionContext.getResponse();
Writer writer = response.getWriter();
writer.write(“登录成功!”);
writer.flush();
writer.close();
}
return null;
}
public String getUserName() {
return userName;
}
public void setUserName(String userName) {
this.userName = userName;
}
public String getPwd() {
return pwd;
}
public void setPwd(String pwd) {
this.pwd = pwd;
}
public String getVerifyCode() {
return verifyCode;
}
public void setVerifyCode(String verifyCode) {
this.verifyCode = verifyCode;
}
public String getAjax() {
return ajax;
}
public void setAjax(String ajax) {
this.ajax = ajax;
}
}

import java.io.Writer;import javax.servlet.http.HttpServletResponse;import org.apache.struts2.ServletActionContext;import com.opensymphony.xwork2.ActionSupport;public class LoginAction extends ActionSupport {/** *  */private static final long serialVersionUID = 1L;private String userName;private String pwd;private String verifyCode;private String ajax;// 错误的写法@Overridepublic String execute() throws Exception {// 通过ajax登录if (ajax != null) {HttpServletResponse response = ServletActionContext.getResponse();Writer writer = response.getWriter();writer.write("登录成功!");writer.flush();writer.close();}return SUCCESS;}// 正确写法public String login1() throws Exception {if (ajax != null) {HttpServletResponse response = ServletActionContext.getResponse();Writer writer = response.getWriter();writer.write("登录成功!");writer.flush();writer.close();return null;}return SUCCESS;}// 正确写法public String login2() throws Exception {if (ajax != null) {HttpServletResponse response = ServletActionContext.getResponse();Writer writer = response.getWriter();writer.write("登录成功!");writer.flush();writer.close();}return null;}public String getUserName() {return userName;}public void setUserName(String userName) {this.userName = userName;}public String getPwd() {return pwd;}public void setPwd(String pwd) {this.pwd = pwd;}public String getVerifyCode() {return verifyCode;}public void setVerifyCode(String verifyCode) {this.verifyCode = verifyCode;}public String getAjax() {return ajax;}public void setAjax(String ajax) {this.ajax = ajax;}}

以上为登录测试代码（Struts2），在以上示例中，如果判断为ajax!=null成立，那么一定会报如题所示的错误，原因就是：if子句里已经做了一次response，在writer.close();的时候，本次response已经完成；但是紧接着在return SUCCESS;的时候，相当于又做了一次response，所以就出错了~

类似的错误也会出现于以下代码中：

protected void doGet(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
Writer writer = response.getWriter();
writer.write(“Hello”);
writer.flush();
writer.close();
response.sendRedirect(“http://blog.csdn.net/baiyanglu/article/details/8076104”);
}

protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {Writer writer = response.getWriter();writer.write("Hello");writer.flush();writer.close();response.sendRedirect("http://blog.csdn.net/baiyanglu/article/details/8076104");}

出现本错误后，web前端能够接收到第一次response的内容（就是说，挂电话之前说的话，我还是可以听到的，挂电话后讲的，当然听不到咯~），不会报错，只会在后台显示出错了

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