什么是 android 的严格性及其使用

Question

我是 android 新手，我正在关注本教程，我找到了下面的代码，他正在将 json 字符串转换为 StringEntity。如果我错了，请纠正我 StringEntity 用于传递数据，像 Accept 这样的标头，Content-type 到服务器。

            // 1. create HttpClient
        HttpClient httpclient = new DefaultHttpClient();

        // 2. make POST request to the given URL
        HttpPost httpPost = new HttpPost(url);



        String json = "";

        // 3. build jsonObject
        JSONObject jsonObject = new JSONObject();
        jsonObject.accumulate("name", person.getName());
        jsonObject.accumulate("country", person.getCountry());
        jsonObject.accumulate("twitter", person.getTwitter());

        // 4. convert JSONObject to JSON to String
        json = jsonObject.toString();

        // ** Alternative way to convert Person object to JSON string usin Jackson Lib
        // ObjectMapper mapper = new ObjectMapper();
        // json = mapper.writeValueAsString(person);

        // 5. set json to StringEntity
        StringEntity se = new StringEntity(json);

        // 6. set httpPost Entity
        httpPost.setEntity(se);

        // 7. Set some headers to inform server about the type of the content   
        httpPost.setHeader("Accept", "application/json");
        httpPost.setHeader("Content-type", "application/json");

        // 8. Execute POST request to the given URL
        HttpResponse httpResponse = httpclient.execute(httpPost);

        // 9. receive response as inputStream
        inputStream = httpResponse.getEntity().getContent();
.
.
.

我如何获取 servlet/jsp 中的数据？我应该使用 getStream() 还是 request.getParameter()

Answer 1

顺便说一句，我在 3 个步骤 中使用 Jackson 在 Netbeans/Glashfish 中解决了同样的问题。

1)要求：

我使用的一些罐子：

 commons-codec-1.10.jar
 commons-logging-1.2.jar
 log4j-1.2.17.jar
 httpcore-4.4.4.jar
 jackson-jaxrs-json-provider-2.6.4.jar
 avalon-logkit-2.2.1.jar
 javax.servlet-api-4.0.0-b01.jar
 httpclient-4.5.1.jar
 jackson-jaxrs-json-provider-2.6.4.jar
 jackson-databind-2.7.0-rc1.jar
 jackson-annotations-2.7.0-rc1.jar
 jackson-core-2.7.0-rc1.jar

如果我错过了上面的任何 jar，你可以从这里的 Maven 下载 http://mvnrepository.com/artifact/com.fasterxml.jackson.core

2) 示例：您发送帖子的 Java 类。 首先，使用 Jackson 将实体用户转换为 Json，然后将其发送到您的 Rest 类。

import com.fasterxml.jackson.databind.ObjectMapper;
import ht.gouv.mtptc.siiv.model.seguridad.Usuario;
import java.io.IOException;
import java.io.UnsupportedEncodingException;
import org.apache.http.HttpResponse;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.entity.StringEntity;
import org.apache.http.impl.client.DefaultHttpClient;
import org.json.simple.JSONObject;


public class PostRest {


    public static void main(String args[]) throws UnsupportedEncodingException, IOException {

         // 1. create HttpClient
        DefaultHttpClient httpclient = new DefaultHttpClient();

        // 2. make POST request to the given URL
        HttpPost httpPost 
        = new HttpPost("http://localhost:8083/i360/rest/seguridad/obtenerEntidad");


        String json = "";
        Usuario u = new Usuario();
        u.setId(99L);

        // 3. build jsonObject
        JSONObject jsonObject = new JSONObject();
        jsonObject.put("id", u.getId());

        // 4. convert JSONObject to JSON to String
        //json = jsonObject.toString();

        // ** Alternative way to convert Person object to JSON string usin Jackson Lib
         //ObjectMapper mapper = new ObjectMapper();
         //json = mapper.writeValueAsString(person);
        ObjectMapper mapper = new ObjectMapper();
       json = mapper.writeValueAsString(u);

        // 5. set json to StringEntity
        StringEntity se = new StringEntity(json,"UTF-8");


        // 6. set httpPost Entity
        httpPost.setEntity(se);

        // 7. Set some headers to inform server about the type of the content   
        httpPost.setHeader("Accept", "application/json");
        httpPost.setHeader("Content-type", "application/json");

        // 8. Execute POST request to the given URL
        HttpResponse httpResponse = httpclient.execute(httpPost);

        // 9. receive response as inputStream
        //inputStream = httpResponse.getEntity().getContent();

}


}

3) 示例：Java Class Rest 您希望在其中接收实体 JPA/Hibernate 。 在这里，您的 MediaType.APPLICATION_JSON) 以这种方式接收实体：

""id":99,"usuarioPadre":null,"nickname":null,"clave":null,"nombre":null,"apellidos":null,"isLoginWeb" :null,"isLoginMovil":null,"estado":null,"correoElectronic":null,"imagePerfil":null,"perfil":null,"urlCambioClave":null,"telefono":null,"celular":null ,"isFree":null,"proyectoUsuarioList":null,"cuentaActiva":null,"keyUser":null,"isCambiaPassword":null,"videoList":null,"idSocial":null,"tipoSocial":null," idPlanActivo":null,"cantidadMbContratado":null,"cantidadMbConsumido":null,"cuotaMb":null,"fechaInicio":null,"fechaFin":null}"

    import javax.ws.rs.Consumes;
    import javax.ws.rs.GET;
    import javax.ws.rs.POST;

    import javax.ws.rs.Path;
    import javax.ws.rs.PathParam;
    import javax.ws.rs.Produces;
    import javax.ws.rs.core.MediaType;
    import org.json.simple.JSONArray;

    import org.json.simple.JSONObject;
    import org.apache.log4j.Logger;

    @Path("/seguridad")
    public class SeguridadRest implements Serializable {



       @POST
        @Path("obtenerEntidad")
        @Consumes(MediaType.APPLICATION_JSON)
        public JSONArray obtenerEntidad(Usuario u) {
            JSONArray array = new JSONArray();
            LOG.fatal(">>>Finally this is my entity(JPA/Hibernate) which 
will print the ID 99 as showed above :" + u.toString());
            return array;//this is empty

        }
       ..

一些提示：如果您使用此代码后运行网页出现问题可能是因为@Consumes in XML ...您必须将其设置为@Consumes(MediaType.APPLICATION_JSON)

Answer 2

从字符串中检索内容的实体。

StringEntity 是您在请求中发送的原始数据。

服务器使用 JSON 进行通信，JSON 字符串可以通过 StringEntity 发送，服务器可以在请求正文中获取它，对其进行解析并生成适当的响应。

我们只设置了我们所有的 unicode 样式、内容类型

StringEntity se = new StringEntity(str,"UTF-8");
    se.setContentType("application/json");
    httpPost.setEntity(se); 

如需更多帮助，您可以参考此 http://developer.android.com/reference/org/apache/http/entity/StringEntity.html

根据您的要求，我将其编辑为 post 方法

HttpPost httpPost = new HttpPost(url_src);
HttpParams httpParameters = new BasicHttpParams();
httpclient.setParams(httpParameters);
StringEntity se = new StringEntity(str,"UTF-8");
se.setContentType("application/json");
httpPost.setEntity(se); 



try
{
    response = httpclient.execute(httpPost);

    StatusLine statusLine = response.getStatusLine();
    int statusCode = statusLine.getStatusCode();


    if(statusCode==200)
    {
        entity = response.getEntity();
        String responseText = EntityUtils.toString(entity);
        System.out.println("The response is" + responseText);   

    }
    else
    {
        System.out.println("error");;
    }
}
catch(Exception e)
{

    e.printStackTrace();

}

Answer 3

StringEntity 是您在请求中发送的原始数据。

大部分服务器使用 JSON 进行通信，JSON 字符串 可以通过 StringEntity 发送，服务器可以在请求正文中获取它，对其进行解析并生成适当的响应。

Accept、Content-type 等作为请求的标头发送，但 StringEntity 是它的内容。

标头未传入StringEntity。