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gson 復雜json數據

江奕云1年前6瀏覽0評論

Gson是一個優秀的Java庫,可以幫助我們快速方便地處理JSON數據。當然,對于復雜的JSON數據,Gson同樣能夠勝任。

舉個例子,假設我們的JSON數據如下:

{
"name": "張三",
"age": 25,
"address": {
"city": "北京",
"street": "朝陽區沙河",
"zipcode": "100025"
},
"phoneNumbers": [
"13012345678",
"13987654321"
]
}

我們可以通過以下代碼把JSON字符串轉為Java對象:

Gson gson = new Gson();
String json = "{
\"name\": \"張三\",
\"age\": 25,
\"address\": {
\"city\": \"北京\",
\"street\": \"朝陽區沙河\",
\"zipcode\": \"100025\"
},
\"phoneNumbers\": [
\"13012345678\",
\"13987654321\"
]
}";
Person person = gson.fromJson(json, Person.class);

其中Person是一個Java類,對應著JSON數據中的這個對象。例如:

class Person {
String name;
int age;
Address address;
ListphoneNumbers;
}
class Address {
String city;
String street;
String zipcode;
}

在轉為Java對象之后,我們就可以愉快地使用它們了。例如,我們可以打印出這個人的名字:

System.out.println(person.name); //輸出:張三

同樣,我們也可以把一個Java對象轉為JSON字符串:

Person person = new Person();
person.name = "李四";
person.age = 30;
Address address = new Address();
address.city = "上海";
address.street = "浦東新區陸家嘴";
address.zipcode = "200120";
person.address = address;
ListphoneNumbers = new ArrayList<>();
phoneNumbers.add("13100000000");
phoneNumbers.add("13911111111");
person.phoneNumbers = phoneNumbers;
String json = gson.toJson(person);
System.out.println(json);
//輸出:{"name":"李四","age":30,"address":{"city":"上海","street":"浦東新區陸家嘴","zipcode":"200120"},"phoneNumbers":["13100000000","13911111111"]}

通過Gson,我們可以快速方便地處理復雜的JSON數據。希望這篇文章對大家有所幫助。