java中post请求url拼接参数

ゞ浴缸里的玫瑰 2024-04-07 12:27 83阅读 0赞

#### 实现 ####

@GetMapping("/get001")
        public AjaxResult get001() throws IOException {
        
            System.out.println("开始进入接口");
            StringBuffer stringBuffer = new StringBuffer();
            stringBuffer.append( Post请求地址 )
                    .append("?")
                    .append("orgGridEvent.id="+"6")
                    .append("&address="+"dizhi")
                    .append("&contentText="+"contentText")
                    .append("&orgGridEvent.departmentNo=" + "371423102");
            // 如果不填写此转换格式，地址中包含 空格内容 转换失败，会出现下面的异常
            // java.lang.IllegalArgumentException: Illegal character in query at index
    		// 说是我输入的数据有问题
    		/*
    		？     %3F
    		&      %26
    		|      %124
    		=     %3D
    		#     %23
    		/      %2F
    		+     %2B
    		%      %25
    		空格   %20
    		*/
            String url = stringBuffer.toString().replaceAll(" ", "%20");
            System.out.println("开始进入post了");
            JSONObject jsonObject = EonUtils.doPostForUrl(url);
            System.out.println(jsonObject);
            return null;
        }

#### Post工具类 ####

public static JSONObject doPostForUrl(String url){
        
            JSONObject jsonObject = null;
            String result = "";
            CloseableHttpClient httpClient = HttpClients.createDefault();
            HttpPost httpPost = new HttpPost(url);
            //每个接口的Content-Type可能不同
            httpPost.setHeader("Content-Type", "application/x-www-form-urlencoded");
            try {
        
                HttpResponse response = httpClient.execute(httpPost);
                 result = EntityUtils.toString(response.getEntity());
                jsonObject = JSONObject.fromObject(result);
                return jsonObject;
            }catch (Exception e){
        
                e.printStackTrace();
                jsonObject = JSONObject.fromObject(result);
                return jsonObject;
            }
        }