需要使用 Hive 数据构建 JSON 格式

Question

您能帮我解决以下要求吗？我有一个配置单元表数据并使用它来构建下面的 JSON。

------------------------
Table : global_product
-------------------------
------------------------------------------------------------------------
Columns             Datatypes
------------------------------------------------------------------------
AAA                 varchar(10)
BBB                 varchar(10)
GEOGRAPHIC          array<struct<CCC:double,DDD:double>>    

select 
AAA, 
BBB, 
GEOGRAPHIC 
from global_product;

o/p:

 ind, 
 hyd, 
 [{"CCC":4.7,"DDD":7.6},{"CCC":5.1,"DDD":7.5}, {"CCC":5.7,"DDD":9.5}]

---

需要使用上列数据框出JSON格式

"GLOBAL":
{
"PRODUCTS":
        "{
        "ITEMS":{"AAA":"ind","BBB":"hyd"},
        "GEOGRAPHIC":[
                        {"CCC":45.78,"DDD":87.56},
                        {"CCC":45.73,"DDD":97.59},
                        {"CCC":45.73,"DDD":97.59}
                    ]
                }"
        }

Answer 1

如果你对复杂类型没问题

struct<global: struct<products: struct<items: struct<AAA:double, BBB:double>,
                                      geographic: array< struct<ccc: double, ddd: double> >
                                      >
                   > 
      > 

然后你可以使用 named_struct() 构造。在这种情况下，所有属性名称都将小写。

演示

with your_data as (--This is initial data example
select 'ind' as AAA, 'hyd' as BBB, 
array(named_struct("CCC",4.7,"DDD",7.6),named_struct("CCC",5.1,"DDD",7.5), named_struct("CCC",5.7,"DDD",9.5)) as GEOGRAPHIC
)

select named_struct('GLOBAL', named_struct('PRODUCTS',named_struct('ITEMS', named_struct('AAA', AAA, 'BBB', BBB), 
                                            'GEOGRAPHIC', GEOGRAPHIC) 
                                          )  
                   )
     from your_data;

结果（打印不漂亮）：

{"global":{"products":{"items":{"aaa":"ind","bbb":"hyd"},"geographic":[{"ccc":4.7,"ddd":7.6},{"ccc":5.1,"ddd":7.5},{"ccc":5.7,"ddd":9.5}]}}} 

如果你对这样复杂的类型不满意，并且想要拥有像你一样的属性不小写的字符串，那么爆炸数组并连接所有内容

set hive.execution.engine=mr;
with your_data as (--This is initial data example, use your table instead of this example
select 'ind' as AAA, 'hyd' as BBB, 
array(named_struct("CCC",4.7,"DDD",7.6),named_struct("CCC",5.1,"DDD",7.5), named_struct("CCC",5.7,"DDD",9.5)) as GEOGRAPHIC
)

select    concat('"GLOBAL":\n{\n"PRODUCTS":\n          "{\n    "ITEMS":{"AAA":"',AAA,'","BBB":"',BBB,'"},', '\n',
          concat('     "GEOGRAPHIC": [\n         ',concat_ws(',',collect_list(concat('{"CCC": "',e.CCC, '", "DDD": "',DDD,'"}'))),
                 '\n          ]\n     }"\n }') 
          )
     from your_data --replace with your table name
          lateral view outer inline(GEOGRAPHIC) e as CCC, DDD
          group by AAA, BBB;

结果：

"GLOBAL":
{
"PRODUCTS":
          "{
    "ITEMS":{"AAA":"ind","BBB":"hyd"},
     "GEOGRAPHIC": [
         {"CCC": "4.7", "DDD": "7.6"},{"CCC": "5.1", "DDD": "7.5"},{"CCC": "5.7", "DDD": "9.5"}
          ]
     }"
 }

只需在 FROM 中使用您的表格而不是 CTE，它就可以处理您的数据。我添加了换行符 \n 和空格以产生类似的格式。