将JSON解析为不带根的JAVA对象

我的户外休息服务的回应是：

[
{
"role": "SiteManager",
"authority":
{
    "authorityType": "USER",
    "fullName": "admin",
    "userName": "admin",
    "firstName": "Administrator",
    "lastName": "",
    "url": "\/alfresco\/service\/api\/people\/admin"
},
"url": "\/alfresco\/service\/api\/sites\/test3\/memberships\/admin"
}
,
{
"role": "SiteConsumer",
"authority":
{
    "authorityType": "GROUP",
    "shortName": "jamalgg",
    "fullName": "GROUP_jamalgg",
    "displayName": "jamalgg",
    "url": "\/alfresco\/service\/api\/groups\/jamalgg"
},
"url": "\/alfresco\/service\/api\/sites\/test3\/memberships\/GROUP_jamalgg"
}
,
{
"role": "SiteManager",
"authority":
{
    "authorityType": "GROUP",
    "shortName": "ALFRESCO_ADMINISTRATORS",
    "fullName": "GROUP_ALFRESCO_ADMINISTRATORS",
    "displayName": "ALFRESCO_ADMINISTRATORS",
    "url": "\/alfresco\/service\/api\/groups\/ALFRESCO_ADMINISTRATORS"
},
"url": "\/alfresco\/service\/api\/sites\/test3\/memberships\/GROUP_ALFRESCO_ADMINISTRATORS"
}
]

List<Memberships > listMemberships;

public class Memberships {
private String role;
private List<Authority> listAuthority ;
private String url;
}
public class Authority {
private String  authorityType;
private String  shortName;
private String  fullName;
private String  displayName;
private String  url;
}

我想解析到对象列表：

[
{
"role": "SiteManager",
"authority":
{
    "authorityType": "USER",
    "fullName": "admin",
    "userName": "admin",
    "firstName": "Administrator",
    "lastName": "",
    "url": "\/alfresco\/service\/api\/people\/admin"
},
"url": "\/alfresco\/service\/api\/sites\/test3\/memberships\/admin"
}
,
{
"role": "SiteConsumer",
"authority":
{
    "authorityType": "GROUP",
    "shortName": "jamalgg",
    "fullName": "GROUP_jamalgg",
    "displayName": "jamalgg",
    "url": "\/alfresco\/service\/api\/groups\/jamalgg"
},
"url": "\/alfresco\/service\/api\/sites\/test3\/memberships\/GROUP_jamalgg"
}
,
{
"role": "SiteManager",
"authority":
{
    "authorityType": "GROUP",
    "shortName": "ALFRESCO_ADMINISTRATORS",
    "fullName": "GROUP_ALFRESCO_ADMINISTRATORS",
    "displayName": "ALFRESCO_ADMINISTRATORS",
    "url": "\/alfresco\/service\/api\/groups\/ALFRESCO_ADMINISTRATORS"
},
"url": "\/alfresco\/service\/api\/sites\/test3\/memberships\/GROUP_ALFRESCO_ADMINISTRATORS"
}
]

List<Memberships > listMemberships;

public class Memberships {
private String role;
private List<Authority> listAuthority ;
private String url;
}
public class Authority {
private String  authorityType;
private String  shortName;
private String  fullName;
private String  displayName;
private String  url;
}

列出列表成员资格；
公共班级成员资格{
私有字符串角色；
私人名单管理局；
私有字符串url；
}
公共阶级权威{
私有字符串authorityType；
私有字符串短名称；
私有字符串全名；
私有字符串显示名；
私有字符串url；
}

我认为有两种解决办法：

如何将标签成员身份添加到封装的JSON结果中 全部

如何将JSON结果直接解析到我的列表

感谢

如中所述，我将使用库。

如中所述，我将使用库。

您还可以使用从json响应生成java代码，然后使用jackson库（）将响应数据绑定到您的对象：

您还可以使用从json响应生成java代码，然后使用jackson库（）将该响应数据绑定到您的对象：

---

谢谢你，小奥。我的问题的答案是：

Type targetType = new TypeToken<Collection<Memberships>>() {}.getType();
List<Memberships> list = (List<Memberships>)  new Gson().fromJson(renduJson,targetType);

Type targetType=new-TypeToken（）{}.getType（）；
List List=（List）new Gson（）.fromJson（renduJson，targetType）；
谢谢你，小奥。我的问题的答案是：

Type targetType = new TypeToken<Collection<Memberships>>() {}.getType();
List<Memberships> list = (List<Memberships>)  new Gson().fromJson(renduJson,targetType);

Type targetType=new-TypeToken（）{}.getType（）；
List List=（List）new Gson（）.fromJson（renduJson，targetType）；
很抱歉没有格式化代码

Type targetType = new TypeToken<Collection<Memberships>>() {}.getType();
List<Memberships> list = (List<Memberships>)new Gson().fromJson(rendu,targetType);

Type targetType=new-TypeToken（）{}.getType（）；
List List=（List）new Gson（）.fromJson（rendu，targetType）；
很抱歉没有格式化代码

Type targetType = new TypeToken<Collection<Memberships>>() {}.getType();
List<Memberships> list = (List<Memberships>)new Gson().fromJson(rendu,targetType);

Type targetType=new-TypeToken（）{}.getType（）；
List List=（List）new Gson（）.fromJson（rendu，targetType）；
这里甚至有一个例子：这里甚至有一个例子：