将具有不同类型对象（字符串、数组）的 JSON 转换为 Map答案

【问题标题】：Convert JSON with different types of objects (strings, array) to Map将具有不同类型对象（字符串、数组）的 JSON 转换为 Map
【发布时间】：2015-10-26 15:50:29
【问题描述】：

我正在尝试使用砖房库“brickhouse.udf.json.FromJsonUDF”将 JSON 对象转换为 Hive 中的地图对象。

问题是，我的 json 对象包含不同类型的值：字符串和另一个数组的一个数组。

我的 json 看起来像这样：

'{"key1":"value1","key2":"value2","key3":"value3","key4":[["0","1","nnn"],["1","3","mmm"],["1","3","ggg"],["1","5","kkk"],["4","5","ppp"]]}'

我可以使用以下方法正确读取数组元素的数组 (key4)：

select from_json('{"key1":"value1","key2":"value2","key3":"value3","key4":[["0","1","nnn"],["1","3","mmm"],["1","3","ggg"],["1","5","kkk"],["4","5","ppp"]]}', 'map<string,array<array<string>>>') from my_table limit 1;

Which gives me:

{"key1":[],"key3":[],"key2":[],"key4":[["0","1","nnn"],["1","3","mmm"],["1","3","ggg"],["1","5","kkk"],["4","5","ppp"]]}

你可以看到除了key4之外的所有元素都是空的。

或者我可以阅读其他元素，但 key4 使用：

select from_json('{"key1":"value1","key2":"value2","key3":"value3","key4":[["0","1","nnn"],["1","3","mmm"],["1","3","ggg"],["1","5","kkk"],["4","5","ppp"]]}', 'map<string,string>') from my_table limit 1;

Which gives me:

{"key1":"value1","key3":"value3","key2":"value2","key4":null}

但是如何将所有元素正确转换为结果映射对象上的键值对呢？

已编辑：

我的实际数据是一个由两个组件组成的数组，它们是 json 对象：

[{"key1":"value1", "key2":"value2"}{"key3":"value3","key4":"value4","key5":"value5","key6":[["0","1","nnn"],["1","3","mmm"],["1","3","ggg"],["1","5","kkk"],["4","5","ppp"]]}]

是否可以创建一个 struct 对象，其中包含两个 json 对象作为两个映射对象，以便我可以访问第一个或第二个结构元素，然后使用键选择相应映射对象的值？

例如：假设我想要的最终结果被称为struct_result，我将从第一个组件访问value1，例如：

struct_result.t1["key1"]

这会给我“value1”。

这个库可以实现吗？

【问题讨论】：

标签： arrays json hive brickhouse

【解决方案1】：

这可以使用命名结构来完成。您需要创建一个 named_struct，并独立指定每个键的类型。

例如

select_from_json('{"key1":"value1","key2":"value2","key3":"value3","key4":[["0","1","nnn"],["1","3","mmm"],["1","3","ggg"],["1","5","kkk"],["4","5","ppp"]]}',
    named_struct("key1","", "key2", "", "key3", ""
        "key4", array(array("")))
from my_table;

这将使用“named_struct”UDF 创建一个模板对象，或者您可以使用等效的字符串类型定义。

【讨论】：