http 可以添加超时时间，防止超时问题

刺骨的言语ヽ痛彻心扉 2022-10-27 05:24 695阅读 0赞

1、http请求，很容易网络抖动，所以加上超时可以解决这种问题

public static String doHttpPost(String uri, JSONObject jsonObject) {
            CloseableHttpResponse response = null;
            try {
                HttpPost httpPost = new HttpPost(uri);
    
                RequestConfig config = RequestConfig.custom()
                        .setConnectTimeout(1000) //连接超时时间
                        .setConnectionRequestTimeout(1000) //从连接池中取的连接的最长时间
                        .setSocketTimeout(1000) //数据传输的超时时间
                        .build();
                //设置请求配置时间
                httpPost.setConfig(config);
    
                //装填参数
                StringEntity s = new StringEntity(jsonObject.toString(), "utf-8");
                s.setContentEncoding(new BasicHeader(HTTP.CONTENT_TYPE, "application/json"));
                //设置参数到请求对象中
                httpPost.setEntity(s);
                httpPost.setHeader("Content-type", "application/json");
                httpPost.setHeader("User-Agent", "Mozilla/4.0 (compatible; MSIE 5.0; Windows NT; DigExt)");
    
                response = httpClient.execute(httpPost);
    
                int statusCode = response.getStatusLine().getStatusCode();
                System.out.println("statusCode:" +statusCode );
                if (HttpStatus.SC_OK == statusCode) {
                    HttpEntity entity = response.getEntity();
                    if (null != entity) {
                        String resStr = EntityUtils.toString(entity, "utf-8");
                        return resStr;
                    }
                }
            } catch (Exception e) {
                System.out.println("CloseableHttpClient-post-请求异常:" + e.getMessage()+",case" +String.valueOf(e.getStackTrace()));
            } finally {
                try {
                    if (null != response) {
                        EntityUtils.consume(response.getEntity());
                        response.close();
                    }
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return null;
        }