java解决跨域问题

news2024/11/15 15:28:28

浏览器js在访问服务器中的资源时，会出现同一页面或者不同域名(协议，IP，端口)不可访问
例如:file://d://test.html页面(file协议)中通过ajax访问服务器api.test.com的接口(http协议)，由于协议不同，此时会出现浏览器访问不了后台接口的问题
此时需要在服务器接口的请求头配置好，当浏览器访问时就会根据请求头判断是否可以访问

前端代码

<!DOCTYPE html>
<html>
<head>
    <meta charset="UTF-8">
    <title>跨域请求示例</title>
</head>
<body>
    <button onclick="sendRequest()">发送请求</button>
    <div id="result"></div>
    <script>
        function sendRequest() {
            var xhr = new XMLHttpRequest();
            xhr.open('GET', 'http://localhost:8888/user/queryById?id=1', true);
            xhr.onreadystatechange = function() {
                if (xhr.readyState == 4 && xhr.status == 200) {
                    document.getElementById('result').innerText = xhr.responseText;
                }
            };
            xhr.send();
        }
    </script>
</body>
</html>

调用后台接口时报错

补充跨域过滤

package com.example.demo.config;

import org.springframework.stereotype.Component;

import javax.servlet.Filter;
import javax.servlet.FilterChain;
import javax.servlet.ServletException;
import javax.servlet.ServletRequest;
import javax.servlet.ServletResponse;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import java.io.IOException;


@Component
public class CorsFilter implements Filter {
    @Override
    public void doFilter(ServletRequest req, ServletResponse res, FilterChain chain) throws IOException, ServletException {
        HttpServletResponse response = (HttpServletResponse) res;
        HttpServletRequest request = (HttpServletRequest) req;
        response.setHeader("Access-Control-Allow-Origin", "*");
        response.setHeader("Access-Control-Allow-Methods", "*");
        response.setHeader("Access-Control-Allow-Credentials", "true");
        response.setHeader("Access-Control-Allow-Headers", "Authorization,Origin,X-Requested-With,Content-type,Accept,Access-Token");
        response.setHeader("Access-Control-Max-Age", "3600");
        chain.doFilter(request, response);
    }
}

再次请求，可以正常得到结果