Question

我有一个对象，我想生成一个JSON对象。目前我有一个被另一个对象（ResourceObject）包装的对象。此资源对象仅在顶级对象上具有公开变量的子集。

我使用此资源对象通过注释来生成json。这是为了保持原始对象更清洁，因为它在代码中的其他位置使用。但是，维护此过滤代码会增加复杂性，并且主要是样板代码。

理想情况下，我想传递对象和预期的架构来为我做过滤，但我没有看到类似的东西。我希望我的搜索条件不正确。

示例Class1

  SELECT a.id
       , c.name_first
       , c.name_last
       , a.created             AS app_created
       , a.edited
       , a.comments            AS custcomment
       , c.name_company
       , a.loan_amount
       , a.product_type
       , ap.packager           AS placed_with
       , a.introducer_id
       , a.status
       , c.contact_number
       , c.email
       , a.database
       , bd.legal_structure
       , ca.id                 AS address_id
       , ca.address_postcode
       , ca.resident_status 
  FROM cl_application a
  LEFT
  JOIN cl_application_business_details bd
    ON bd.id = a.id
  LEFT
  JOIN cl_customer c
    ON c.id = a.customer_id
   AND c.deleted IS NOT NULL
  LEFT
  JOIN cl_application_packager ap
    ON ap.application_id = a.id
   AND ap.status != 'Declined'
   AND ap.deleted = '0000-00-00 00:00:00'
  LEFT
  JOIN ( SELECT ca1.customer_id
              , MAX(ca1.created) AS latest_created
           FROM cl_customer_address ca1
          WHERE ca1.deleted = '0000-00-00 00:00:00'
          GROUP BY ca1.customer_id
       ) lc
    ON lc.customer_id = a.customer_id
  LEFT
  JOIN cl_customer_address ca
    ON ca.customer_id = lc.customer_id
   AND ca.created     = lc.latest_created
   AND ca.deleted     = '0000-00-00 00:00:00'
 WHERE a.deleted = '0000-00-00 00:00:00'
 GROUP BY a.created, a.id
 ORDER BY a.created, a.id

示例资源类

Class1 {
    private String name = "C1";
    private String version = "1.0";

    public String getName() {
        return name;
    }

    public String getVersion() {
        return version;
    }
}

示例架构

ResourceClass1 {
    private Class1 class1;

    ResourceClass1 (Class1 c1) {
        class1 = c1;
    }

    public String getName() {
        return class1.getName();
    }
}

现在，从Resource类中获取JSON会给我一些影响：

  {
  "$schema": "http://json-schema.org/draft-04/schema#",
  "title": "Class1",
  "type" : "object",
  "properties" : {
    "name" : {
      "type" : "string"
    },
  }
}

是否有一个包只通过传入schema和java对象来获得相同的json？

如何在给定模式和对象的情况下创建JSON对象

0 个答案: