Roc*_*och 7 json javabeans gson
我想知道是否有一个网页/软件可以将Json feed对象"转换"为具有属性的Java对象.
例如 :
{
'firstName': 'John',
'lastName': 'Smith',
'address': {
'streetAddress': '21 2nd Street',
'city': 'New York'
}
}
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会成为:
class Person {
private String firstName;
private String lastName;
private Address address;
public String getFirstName() { return firstName; }
public String getLastName() { return lastName; }
public Address getAddress() { return address; }
public void setFirstName(String firstName) { this.firstName = firstName; }
public void setLastName(String lastName) { this.lastName = lastName; }
public void setAddress(Address address) { this.address = address; }
public String toString() {
return String.format("firstName: %s, lastName: %s, address: [%s]", firstName, lastName, address);
}
}
class Address {
private String streetAddress;
private String city;
public String getStreetAddress() { return streetAddress; }
public String getCity() { return city; }
public void setStreetAddress(String streetAddress) { this.streetAddress = streetAddress; }
public void setCity(String city) { this.city = city; }
public String toString() {
return String.format("streetAddress: %s, city: %s", streetAddress, city);
}
}
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我不是因为我很懒,但是我要解析的JSON有很多属性.
您可能想看看 Gson,Google 的 JSON Java 库。它具有将对象从 Java 序列化为 JSON 以及从 JSON 反序列化为 Java 对象的方法。
我从未将它用于反序列化,因此我无法提供太多详细信息,例如它将如何处理嵌套对象(地址),但我想它能够处理它。
API 文档:http://google-gson.googlecode.com/svn/tags/1.3/docs/javadocs/index.html 用户指南:https://sites.google.com/site/gson/gson-user -指导
祝你好运 !