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 = "[e-mail]";Log.e("MainActivity", gson.toJson(student));

输出结果是 ：

{"email":"[e-mail]","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 = "[e-mail]";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":"[e-mail]","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 = "[e-mail]";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":"[e-mail]","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 = "[e-mail]";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:1nickName:乔晓松age:22email:[e-mail] size:6booksMap 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