使用HTTP客户端发送和解析响应以获取JSON列表

Question

在我的Java代码中，我需要向具有3个标头的特定URL发送http发布请求：

URL: http://localhost/something
Referer: http://localhost/something 
Authorization: Basic (with a username and password)
Content-type: application/json

这将返回一个带有JSON“键”：“值”对的响应，然后我需要解析以某种方式将键/值（Alan /
72）存储在MAP中。响应为（使用SOAPUI或Postman Rest时）：

    {
    "analyzedNames": [
        {
            "alternate": false               
        }
    ],
    "nameResults": [
        {
            "alternate": false,            
            "givenName": "John",           
            "nameCategory": "PERSONAL",
            "originalGivenName": "",
            "originalSurname": "",           
            "score": 72,
            "scriptType": "NOSCRIPT",            
        }
    ]
}

我可以使用SOAPUI或Postman Rest来执行此操作，但是当出现错误时，如何在Java中执行此操作：

****DEBUG main org.apache.http.impl.conn.DefaultClientConnection - Receiving response: HTTP/1.1 500 Internal Server Error****

我的代码是：

    public class NameSearch {

        /**
         * @param args
         * @throws IOException 
         * @throws ClientProtocolException 
         */
        public static void main(String[] args) throws ClientProtocolException, IOException {
            // TODO Auto-generated method stub
            DefaultHttpClient defaultHttpClient = new DefaultHttpClient();          
            StringWriter writer = new StringWriter();

            //Define a postRequest request
            HttpPost postRequest = new HttpPost("http://127.0.0.1:1400/dispatcher/api/rest/search");

            //Set the content-type header
            postRequest.addHeader("content-type", "application/json");
 postRequest.addHeader("Authorization", "Basic ZW5zYWRtaW46ZW5zYWRtaW4=");

            try {

                //Set the request post body
                StringEntity userEntity = new StringEntity(writer.getBuffer().toString());
                postRequest.setEntity(userEntity);

                //Send the request; return the response in HttpResponse object if any
                HttpResponse response = defaultHttpClient.execute(postRequest);

                //verify if any error code first
                int statusCode = response.getStatusLine().getStatusCode();                
            }
            finally
            {
                //Important: Close the connect
                defaultHttpClient.getConnectionManager().shutdown();
            }    
        }    
    }

我们将不胜感激任何帮助（带有一些示例代码，包括要导入的库）。

谢谢

Answer 1

是的，您可以使用Java做到这一点

您需要apache HTTP客户端库http://hc.apache.org/和commons-io

HttpClient client = new DefaultHttpClient();
HttpPost post = new HttpPost("http://localhost/something");


post.setHeader("Referer", "http://localhost/something");
post.setHeader("Authorization", "Basic (with a username and password)");
post.setHeader("Content-type", "application/json");

// if you need any parameters
List<NameValuePair> urlParameters = new ArrayList<NameValuePair>();
urlParameters.add(new BasicNameValuePair("paramName", "paramValue"));
post.setEntity(new UrlEncodedFormEntity(urlParameters));

HttpResponse response = client.execute(post);

HttpEntity entity = response.getEntity();
Header encodingHeader = entity.getContentEncoding();

// you need to know the encoding to parse correctly
Charset encoding = encodingHeader == null ? StandardCharsets.UTF_8 : 
Charsets.toCharset(encodingHeader.getValue());

// use org.apache.http.util.EntityUtils to read json as string
String json = EntityUtils.toString(entity, StandardCharsets.UTF_8);

JSONObject o = new JSONObject(json);