Cannot call sendError() after the response has been committed - baiyangliu - 博客频道 - CSDN.NET

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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：

1. import java.io.Writer;  
2.   
3. import javax.servlet.http.HttpServletResponse;  
4.   
5. import org.apache.struts2.ServletActionContext;  
6.   
7. import com.opensymphony.xwork2.ActionSupport;  
8.   
9. public class LoginAction extends ActionSupport {  
10.   
11.     /** 
12.      *  
13.      */  
14.     private static final long serialVersionUID = 1L;  
15.     private String userName;  
16.     private String pwd;  
17.     private String verifyCode;  
18.     private String ajax;  
19.   
20.     // 错误的写法  
21.     @Override  
22.     public String execute() throws Exception {  
23.         // 通过ajax登录  
24.         if (ajax != null) {  
25.             HttpServletResponse response = ServletActionContext.getResponse();  
26.             Writer writer = response.getWriter();  
27.             writer.write("登录成功!");  
28.             writer.flush();  
29.             writer.close();  
30.         }  
31.         return SUCCESS;  
32.     }  
33.   
34.     // 正确写法  
35.     public String login1() throws Exception {  
36.         if (ajax != null) {  
37.             HttpServletResponse response = ServletActionContext.getResponse();  
38.             Writer writer = response.getWriter();  
39.             writer.write("登录成功!");  
40.             writer.flush();  
41.             writer.close();  
42.             return null;  
43.         }  
44.         return SUCCESS;  
45.     }  
46.   
47.     // 正确写法  
48.     public String login2() throws Exception {  
49.         if (ajax != null) {  
50.             HttpServletResponse response = ServletActionContext.getResponse();  
51.             Writer writer = response.getWriter();  
52.             writer.write("登录成功!");  
53.             writer.flush();  
54.             writer.close();  
55.         }  
56.         return null;  
57.     }  
58.   
59.     public String getUserName() {  
60.         return userName;  
61.     }  
62.   
63.     public void setUserName(String userName) {  
64.         this.userName = userName;  
65.     }  
66.   
67.     public String getPwd() {  
68.         return pwd;  
69.     }  
70.   
71.     public void setPwd(String pwd) {  
72.         this.pwd = pwd;  
73.     }  
74.   
75.     public String getVerifyCode() {  
76.         return verifyCode;  
77.     }  
78.   
79.     public void setVerifyCode(String verifyCode) {  
80.         this.verifyCode = verifyCode;  
81.     }  
82.   
83.     public String getAjax() {  
84.         return ajax;  
85.     }  
86.   
87.     public void setAjax(String ajax) {  
88.         this.ajax = ajax;  
89.     }  
90. }  

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：

1. protected void doGet(HttpServletRequest request, HttpServletResponse response)   
2.                 throws ServletException, IOException {  
3.             Writer writer = response.getWriter();  
4.             writer.write("Hello");  
5.             writer.flush();  
6.             writer.close();  
7.   
8.             response.sendRedirect("http://blog.csdn.net/baiyanglu/article/details/8076104");  
9.         }  

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

来自为知笔记(Wiz)