pojo

Up-to-date Swing MVC example + Question [closed]

独自空忆成欢 提交于 2019-11-27 01:41:46
问题 As it currently stands, this question is not a good fit for our Q&A format. We expect answers to be supported by facts, references, or expertise, but this question will likely solicit debate, arguments, polling, or extended discussion. If you feel that this question can be improved and possibly reopened, visit the help center for guidance. Closed 7 years ago . I'm looking for an article or tutorial that gives an example of what an up-to-date MVC pattern (2.0?) should look like with the Swing

Why shouldn't I use immutable POJOs instead of JavaBeans?

十年热恋 提交于 2019-11-26 22:29:47
问题 I have implemented a few Java applications now, only desktop applications so far. I prefer to use immutable objects for passing the data around in the application instead of using objects with mutators (setters and getters ), also called JavaBeans. But in the Java world, it seems to be much more common to use JavaBeans, and I can't understand why I should use them instead. Personally the code looks better if it only deals with immutable objects instead of mutate the state all the time.

What is java pojo class, java bean, normal class? [duplicate]

↘锁芯ラ 提交于 2019-11-26 21:16:26
Possible Duplicate: Difference between DTO, VO, POJO, JavaBeans? Hi please don't say my question is duplicate :-) I saw all questions but didn't understand the exact difference. Can someone explain what is POJO , Bean , Normal Class in easy language? Kumar Vivek Mitra Normal Class : A Java class Java Beans : All properties private (use getters/setters) A public no-argument constructor Implements Serializable. Pojo : Plain Old Java Object is a Java object not bound by any restriction other than those forced by the Java Language Specification. I.e., a POJO should not have to Extend prespecified

Android:dynamically pass model class to retrofit callback

醉酒当歌 提交于 2019-11-26 19:59:21
问题 In retrofit to map json response to pojo usually we do this @POST Call<User> getDataFromServer(@Url String url, @Body HashMap<String,Object> hashMap); ApiCalls api = retrofit.create(ApiCalls.class); Call<User> call = api.getDataFromServer(StringConstants.URL,hashMap); call.enqueue(new Callback<User>() { //Response and failure callbacks } where User is my Pojo class. But for every other request i need to make different pojo and write same code for retrofit call.I want to make a single method

IntelliJ IDEA 10 generate entity (POJO) from DB model

﹥>﹥吖頭↗ 提交于 2019-11-26 18:53:30
问题 How can I generate entity (POJO) from database model using IntelliJ IDEA 10. I create "Data source" in IntelliJ but I have not any idea how can I generate the POJO. 回答1: UPDATE: In IntelliJ 16 this feature in now implemented. The steps to do it are: 1. Database view context menu 2. Scripted Extensions 3. Generate POJOs You can read more here: Feature request: allow "generate classes from database schema" for plain-JDBC developers First you need tell intelliJ that you are using Hibernate (I

Jackson JSON Deserialization with Root Element

瘦欲@ 提交于 2019-11-26 17:29:52
I am having a question with Jackson that I think should be simple to solve, but it is killing me. Let's say I have a java POJO class that looks like this (assume Getters and Setters for me): class User { private String name; private Integer age; } And I want to deserialize JSON that looks like this into a User object: { "user": { "name":"Sam Smith", "age":1 } } Jackson is giving me issues because the User is not the first-level object in the JSON. I could obviously make a UserWrapper class that has a single User object and then deserialize using that but I know there must be a more elegant

Spring Data JPA map the native query result to Non-Entity POJO

帅比萌擦擦* 提交于 2019-11-26 12:51:16
I have a Spring Data repository method with a native query @Query(value = "SELECT g.*, gm.* FROM group g LEFT JOIN group_members gm ON g.group_id = gm.group_id and gm.user_id = :userId WHERE g.group_id = :groupId", nativeQuery = true) GroupDetails getGroupDetails(@Param("userId") Integer userId, @Param("groupId") Integer groupId); and I'd like to map the result to Non-Entity POJO GroupDetails . Is it possible and if so, could you please provide an example ? Assuming GroupDetails as in orid's answer have you tried JPA 2.1 @ConstructorResult ? @SqlResultSetMapping( name="groupDetailsMapping",

Date format Mapping to JSON Jackson

落花浮王杯 提交于 2019-11-26 12:02:54
I have a Date format coming from API like this: "start_time": "2015-10-1 3:00 PM GMT+1:00" Which is YYYY-DD-MM HH:MM am/pm GMT timestamp. I am mapping this value to a Date variable in POJO. Obviously, its showing conversion error. I would like to know 2 things: What is the formatting I need to use to carry out conversion with Jackson? Is Date a good field type for this? In general, is there a way to process the variables before they get mapped to Object members by Jackson? Something like, changing the format, calculations, etc. pb2q What is the formatting I need to use to carry out conversion

How to create a POJO?

别说谁变了你拦得住时间么 提交于 2019-11-26 11:13:47
Recently I've started hearing about "POJOs" (Plain Old Java Objects). I googled it, but still don't understand the concept well. Can anyone give me a clear description of a POJO? Consider a class "Person" with variables "id, name, address, salary" -- how would I create a POJO for this scenario? Is the code below a POJO? public class Person { //variables People people = new People(); private int id; private String name; private String address; private int salary; public int getId() { return id; } public String getName() { return name; } public String getAddress() { return address; } public int

How can I iterate over a map of <String, POJO>?

狂风中的少年 提交于 2019-11-26 10:04:41
问题 I\'ve got a Map<String, Person> (actually I\'m using a more complex POJO but simplifying it for the sake of my question) Person looks like : class Person { String name; Integer age; //accessors } How can I iterate through this map, printing out the key, then the person name, then the person age such as : System.out.println(String.format(\"Key : %s Name : %s Age : %s\", a, b, c)); A being the key from Map< String , Person> B being the name from Person.getName() C being the age from Person