解析 JSON 为对象 JAVA,无需 Root
我的服务 ALFRESCO REST 的响应是:
[
{
"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"
}
]
我想解析对象列表:
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;
}
我认为有两种解决方案:
- 如何将标签 Memberships 添加到 JSON 结果用于封装 整体。
- 如何将 JSON 结果直接解析到我的列表
谢谢
The response of my service ALFRESCO REST is:
[
{
"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"
}
]
And I want to parse to list of object:
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;
}
I think that there are two solutions:
- how to add the tag Memberships to JSON result for encapsulates
the whole. - how to parse JSON result directly to my list
Thanks
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正如 a-better-java-json-library 中的回答,我会使用 google-gson 库。
As answered in a-better-java-json-library I would use the google-gson library.
谢谢奥佐利。我的问题的答案是:
Thank you Ozoli. The answer to my question is:
您还可以使用 http://jsongen.byingtondesign.com/ 从 json 响应生成 java 代码,然后使用jackson 库 ( http://jackson.codehaus.org/ ) 将该响应数据绑定到您的对象):
You can also use http://jsongen.byingtondesign.com/ to generate java code from json response and then use jackson library ( http://jackson.codehaus.org/ ) to bind that response data to your object(s):
抱歉没有格式化代码
sorry for not formatting code