问题截图

问题分析

今天，多个接口突然出现 block:mixed-content  错误，于是排查了一下发现：

错误：https页面去发送http请求报错(浏览器阻止https发送http请求)

原来是由于项目改成了https协议的缘故，出现了请求被拦截；

其实是浏览器不允许在https页面里嵌入http的请求，现在高版本的浏览器为了用户体验，都不会弹窗报错，只会在控制台上打印一条错误信息。

Mixed Content: The page was loaded over HTTPS，blocked the content must be served over HTTPS

如下图：

解决方法

方法1. 在主页面的head中加入下面代码（将调用的http请求升级成https请求并调用）：

<meta http-equiv="Content-Security-Policy" content="upgrade-insecure-requests">

方法2. 走一下本地后端（将本地后端当成service中间层），从后端再去调用其他服务器的http请求

下面为后端中转接口

	/**
	 * 删除应用App
	 */
	@RequestMapping(value = "/delVersionConfig", produces = "text/html;charset=UTF-8")
	@ResponseBody
	public JSONObject delVersionConfig(Model model, HttpServletRequest request, String appId) {
		JSONObject resultJSON = new JSONObject();
		String JSONURL = "http://" + CommonData.DATA_URL + "/app/xxx/xxxx";
		Map<String, Object> condMap = new HashMap<String, Object>();
		condMap.put("id", appId);
// 调用http封装类
		String result = HttpUtils.getHttpClient(JSONURL, condMap);
		LOGGER.info("selectAllApplication_info--result:" + result);
		resultJSON = JSON.parseObject(result);
		return resultJSON;
	}

HttpUtils封装类中HttpUtils.getHttpClient执行方法

	/**
     * 向指定URL发送GET方法的请求
     *
     * @param url   发送请求的URL 例如：http://localhost:8080/demo/login
     * @param param 请求参数 例：{ "userName":"admin", "password":"123456" }
     * @return URL 所代表远程资源的响应结果
     */
public static String getHttpClient(String url, Map<String, Object> param) {
StringBuilder result = new StringBuilder();
BufferedReader in = null;
try {
URL realUrl = createURL(url, param);
HttpURLConnection conn = (HttpURLConnection) realUrl.openConnection();
conn.setRequestProperty("accept", "*/*");
conn.setRequestProperty("connection", "Keep-Alive");
conn.setRequestProperty("Charset", "UTF-8");
conn.connect();
in = new BufferedReader(new InputStreamReader(
conn.getInputStream(),"UTF-8"));
String line;
while ((line = in.readLine()) != null) {
result.append(line);
}
} catch (Exception e) {
System.out.println("发送GET请求出现异常！" + e);
e.printStackTrace();
} finally {
try {
if (in != null) {
in.close();
}
} catch (Exception e2) {
e2.printStackTrace();
}
}
return result.toString();
}