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Google Gson的使用方法,实现Json结构的相互转换
在Java开发中,有时需要保存一个数据结构成字符串,可能你会考虑用Json,但是当Json字符串转换成Java对象时,转换成的是JsonObject,并不是你想要的Class类型的对象,操作起来就很不是愉悦,下面说的就可以解决了这种问题。
首先,需要把Google的Gson的Jar包导入到项目中,这个导入包的简单步骤就不展示了,Gson的下载链接:http://download.csdn.net/detail/qxs965266509/8367275
现在,我先自定义一个Class类
public class Student { public int id; public String nickName; public int age; public ArrayList<String> books; public HashMap<String, String> booksMap; }
案例一,案例二,案例三都是把Java的Class对象使用Gson转换成Json的字符串
案例一:
仅包含基本数据类型的数据结构
Gson gson = new Gson(); Student student = new Student(); student.id = 1; student.nickName = "乔晓松"; student.age = 22; student.email = "965266509@qq.com"; Log.e("MainActivity", gson.toJson(student));
输出结果是 :
{"email":"965266509@qq.com","nickName":"乔晓松","id":1,"age":22}
案例二:
除了基本数据类型还包含了List集合
Gson gson = new Gson(); Student student = new Student(); student.id = 1; student.nickName = "乔晓松"; student.age = 22; student.email = "965266509@qq.com"; ArrayList<String> books = new ArrayList<String>(); books.add("数学"); books.add("语文"); books.add("英语"); books.add("物理"); books.add("化学"); books.add("生物"); student.books = books; Log.e("MainActivity", gson.toJson(student));输出结果是 :
{"books":["数学","语文","英语","物理","化学","生物"],"email":"965266509@qq.com","nickName":"乔晓松","id":1,"age":22}
案例三:
除了基本数据类型还包含了List和Map集合
Gson gson = new Gson(); Student student = new Student(); student.id = 1; student.nickName = "乔晓松"; student.age = 22; student.email = "965266509@qq.com"; ArrayList<String> books = new ArrayList<String>(); books.add("数学"); books.add("语文"); books.add("英语"); books.add("物理"); books.add("化学"); books.add("生物"); student.books = books; HashMap<String, String> booksMap = new HashMap<String, String>(); booksMap.put("1", "数学"); booksMap.put("2", "语文"); booksMap.put("3", "英语"); booksMap.put("4", "物理"); booksMap.put("5", "化学"); booksMap.put("6", "生物"); student.booksMap = booksMap; Log.e("MainActivity", gson.toJson(student));
输出结果是 :
{"books":["数学","语文","英语","物理","化学","生物"],"booksMap":{"3":"英语","2":"语文","1":"数学","6":"生物","5":"化学","4":"物理"},"email":"965266509@qq.com","nickName":"乔晓松","id":1,"age":22}
案例四:
把案例三输出的字符串使用Gson转换成Student对象
Gson gson = new Gson(); Student student = new Student(); student.id = 1; student.nickName = "乔晓松"; student.age = 22; student.email = "965266509@qq.com"; ArrayList<String> books = new ArrayList<String>(); books.add("数学"); books.add("语文"); books.add("英语"); books.add("物理"); books.add("化学"); books.add("生物"); student.books = books; HashMap<String, String> booksMap = new HashMap<String, String>(); booksMap.put("1", "数学"); booksMap.put("2", "语文"); booksMap.put("3", "英语"); booksMap.put("4", "物理"); booksMap.put("5", "化学"); booksMap.put("6", "生物"); student.booksMap = booksMap; String result = gson.toJson(student); Student studentG = gson.fromJson(result, Student.class); Log.e("MainActivity", "id:" + studentG.id); Log.e("MainActivity", "nickName:" + studentG.nickName); Log.e("MainActivity", "age:" + studentG.age); Log.e("MainActivity", "email:" + studentG.email); Log.e("MainActivity", "books size:" + studentG.books.size()); Log.e("MainActivity", "booksMap size:" + studentG.booksMap.size());输出结果是 :
id:1 nickName:乔晓松 age:22 email:965266509@qq.com books size:6 booksMap size:6
通过这4个案例我解决你一定就把Gson的基本用法学会了,当然我们的需求可能需要把List或者Map等集合的泛型换成我们自定义个class,这也是可以解决的,请看案例
案例五:
泛型的使用
public HashMap<String,Book> booksMap; public class Book{ public int id; public String name; }
把booksMap转换成字符串和上面的案例是一样的,但是booksMap的Json字符串换成booksMap的实例对象就有点不同了,因为booksMap有自定义的泛型
HashMap<String, Book> booksMap = gson.fromJson(result, new TypeToken<HashMap<String, Book>>() { }.getType());
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源代码下载链接:http://download.csdn.net/detail/qxs965266509/8367689
Google Gson的使用方法,实现Json结构的相互转换