Java 为 Kotlin 创建 POJO 类
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Create POJO Class for Kotlin
提问by Pratik Butani
I want to create POJO class for Kotlin, as we know that www.jsonschema2pojo.orgconverts JSON to POJO so we can use it with gson.
我想为 Kotlin 创建 POJO 类，因为我们知道www.jsonschema2pojo.org将 JSON 转换为 POJO，因此我们可以将它与 gson 一起使用。
Anyone know how to create Gson POJO for Kotlin QUICKLY?
有人知道如何快速为 Kotlin 创建 Gson POJO吗？
Edited:
编辑：
I know its use Data classes, but is there any simplest way to create it?
我知道它使用 Data 类，但是有没有最简单的方法来创建它？
采纳答案by Pratik Butani
Yes, I got solution
是的，我得到了解决方案
for Example:
例如：
{
"foo": "string",
"bar": "integer",
"baz": "boolean"
}
My POJO Class Created using http://www.jsonschema2pojo.org/
我的 POJO 类使用http://www.jsonschema2pojo.org/创建
Example.java
例子.java
public class Example {
@SerializedName("foo")
@Expose
private String foo;
@SerializedName("bar")
@Expose
private String bar;
@SerializedName("baz")
@Expose
private String baz;
public String getFoo() {
return foo;
}
public void setFoo(String foo) {
this.foo = foo;
}
public String getBar() {
return bar;
}
public void setBar(String bar) {
this.bar = bar;
}
public String getBaz() {
return baz;
}
public void setBaz(String baz) {
this.baz = baz;
}
}
Converted KotlinClass using Code -> Convert Java File to Kotlin Fileor CTRL + ALT + SHIFT + K
使用或转换Kotlin类Code -> Convert Java File to Kotlin FileCTRL + ALT + SHIFT + K
Example.kt
例子.kt
class Example {
@SerializedName("foo")
@Expose
var foo: String? = null
@SerializedName("bar")
@Expose
var bar: String? = null
@SerializedName("baz")
@Expose
var baz: String? = null
}
Thank you all.
谢谢你们。
回答by PRATEEK BHARDWAJ
data class VideoGame(val name: String, val publisher: String, var reviewScore: Int)
//Constructor
val game: VideoGame = VideoGame("Gears of War", "Epic Games", 8)
print(game.name) // "Gears of War"
print(game.publisher) // "Epic Games"
print(game.reviewScore) // 8
game.reviewScore = 7
回答by MatPag
A feature request about Kotlin support to auto generate data classes have been filled herein jsonschema2pojo github repository. Currently, there is no jsonschema2kotlin web utility available.
关于自动科特林支持某种功能的请求产生的数据类已经充满这里的jsonschema2pojo GitHub的仓库。目前，没有可用的 jsonschema2kotlin Web 实用程序。
If you don't have any problem installing a new plugin on Android Studio, follow the accepted answer, otherwise the best you can do is to use jsonschema2pojoto convert JSON to Java POJO and the use the Android Studio 3.0+ feature that converts a Java file to a Kotlin one.
如果您在 Android Studio 上安装新插件没有任何问题，请按照已接受的答案进行操作，否则您能做的最好的事情就是使用jsonschema2pojo将 JSON 转换为 Java POJO，并使用 Android Studio 3.0+ 功能转换 Java文件到 Kotlin 文件。
回答by Nabin Khatiwada
If I got your question, you might be searching some plugin for converting to POJO. So
RoboPOJOGeneratormay help you. You can use a plugin from File>Setting>Plugin>Browse Repositoriesand search for RoboPOJOGenerator.
To use this plugin you first need to create a separate package like "data", right-click the package and you will see Generate POJO from JSON. Also, you need to include gsonlibrary in gradlebecause this plugin will automatically generate annotation of gsonlike @SerializedName, etc.
如果我收到您的问题，您可能正在搜索一些插件以转换为 POJO。所以
RoboPOJOGenerator可以帮到你。您可以使用插件 fromFile>Setting>Plugin>Browse Repositories并搜索RoboPOJOGenerator. 要使用此插件，您首先需要创建一个单独的包，如“data”，右键单击该包，您将看到 Generate POJO from JSON. 此外，您需要在其中包含gson库，gradle因为此插件会自动生成gsonlike@SerializedName等注释。
回答by u5167396
I think this should be the Plugin what you want
我认为这应该是你想要的插件
回答by Kishan Solanki
data class ModelUser(val imagePath: String,val userName: String)
Unbelievable Right! Its as simple as that. Just use datakeyword before classto create Data class in Kotlin.
难以置信的对！就这么简单。只需使用data关键字 beforeclass在 Kotlin 中创建 Data 类。
Data class provides you with everything, getters, setters, hashCode, toString and equals functions. So all you have to do is create an instance and start using the functions.
Data 类为您提供了一切，getter、setter、hashCode、toString 和 equals 函数。所以你所要做的就是创建一个实例并开始使用这些函数。
回答by Sunil
Try this
尝试这个
This is the simple way
这是简单的方法
- Right click on the package name and select New->Kotlin File/Class
- Name the name, In my case, I am naming this as Model you can whatever you like and click on ok
and Paste this code, This is you POJO/Model class
class Model { var uid: String? = null var name: String? = null }
- 右键单击包名并选择New->Kotlin File/Class
- 命名名称，就我而言，我将其命名为模型，您可以随心所欲，然后单击确定
并粘贴此代码，这是您的 POJO/Model 类
class Model { var uid: String? = null var name: String? = null }
How to use this
如何使用这个
val model=Model()
model.name="Sunil"
Log.e("Model after",model.name)
回答by Manish Gowardipe
In vs-code there is a plugin named Paste JSON as Code. it supports many languages. Paste Json as code
在 vs-code 中有一个名为 Paste JSON as Code 的插件。它支持多种语言。 将 Json 粘贴为代码
