Question

我想在tomcat中建立Jquery函数和servlet之间的通信。

Servlet代码：

import java.io.BufferedReader;
import java.io.File;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;

import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;



public class Test extends HttpServlet {
public static String getBody(HttpServletRequest request) throws IOException {

    String body = null;
    StringBuilder stringBuilder = new StringBuilder();
    BufferedReader bufferedReader = null;

    try {
        InputStream inputStream = request.getInputStream();
        if (inputStream != null) {
            bufferedReader = new BufferedReader(new InputStreamReader(inputStream));
            char[] charBuffer = new char[128];
            int bytesRead = -1;
            while ((bytesRead = bufferedReader.read(charBuffer)) > 0) {
                stringBuilder.append(charBuffer, 0, bytesRead);
            }
        } else {
            stringBuilder.append("");
        }
    } catch (IOException ex) {
        throw ex;
    } finally {
        if (bufferedReader != null) {
            try {
                bufferedReader.close();
            } catch (IOException ex) {
                throw ex;
            }
        }
    }
    body = stringBuilder.toString();
    return body;
}
  public void doPost(HttpServletRequest request, HttpServletResponse response) throws IOException, ServletException {

        response.setContentType("text/html");
        PrintWriter out = response.getWriter();
        System.out.println(getBody(request));
        out.println("Success Call Ajax POST");

      }
    public void doGet( HttpServletRequest request, HttpServletResponse
        response ) throws ServletException, IOException{
    response.setContentType("text/html");
    response.setCharacterEncoding( "UTF-8" );
    PrintWriter out = response.getWriter();
    out.println("Get Method");

  }

}

servlet标识在web.xml中定义

的web.xml：

<?xml version="1.0" encoding="UTF-8"?>
<web-app
xmlns="http://java.sun.com/xml/ns/javaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee
http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd"
version="3.0">
  <servlet>
    <servlet-name>Test</servlet-name>
    <servlet-class>com.servlets.Test</servlet-class>
  </servlet>
  <servlet-mapping>
    <servlet-name>Test</servlet-name>
    <url-pattern>/toto</url-pattern>
  </servlet-mapping>
</web-app>

以下HTML包含JQuery函数：

Jquery代码：

<head>
    <meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1">
    <script type="text/javascript" src="jquery.js"></script>
    <script type="text/javascript">
    function login(){  

      $.ajax({  
        type: "POST",  
        url: "http://localhost:8080/test/toto",  
        data: "POST Call",
        success: function(result){  
          alert("success call"+result);
        },
        error: function(XMLHttpRequest, textStatus, errorThrown) { 
            alert("Status: " + textStatus); alert("Error: " + errorThrown); 
        }                 
      });  
    }        
    </script>
    <title>My AJAX</title>
</head>
<body>
    <button type="button" onclick="login()">Click Me!</button>
</body>

我用浏览器测试了servlet，没关系。当我尝试使用HTML / js应用程序并单击按钮时，我在eclipse控制台中显示了指令System.out.println（getBody（request））显示的消息; “POST Call”但我在浏览器中收到了错误警报。所以ajax函数成功调用了servlet中的方法post，但servlet无法成功返回对浏览器的响应。似乎servlet中存在问题。有人能帮帮我吗？

Answer 1

在doPost()尝试添加请求标头：

response.setContentType("text/html"); 
response.setHeader("Access-Control-Allow-Origin", "*");

我确信这与描述here

的Allow Origin问题有关

Answer 2

这是浏览器特定的。

它在Safari 8.0.3上工作正常，它会在Chrome和Firefox上抛出错误No 'Access-Control-Allow-Origin' header is present。

在Firefox中查看控制台：

阻止跨源请求：同源策略禁止在http://localhost:8080/toto读取远程资源。这可以通过将资源移动到同一域或启用CORS来解决。

可以通过在浏览器中禁用web security或已经提到过来修复它：

response.setHeader("Access-Control-Allow-Origin", your_domain);

Jquery和servlet与ajax的通信

2 个答案: