Cannot call sendError() after the response has been committed - baiyangliu

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;
    }
}

 

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");
        }

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

 以上，转自http://www.cnblogs.com/wang3680/p/2c5f28ecfb8c2e2772474f3fab4d771c.html