hive sql 处理多层 json 数组

1. 背景

json 字符串值数据示例：

json 复制代码

{
    "score": 1,
    "submitTime": 1712491933,
    "answerFlag": 1,
    "groupId": 1755547960,
    "answers": [
        {
            "value": "[1, 2, 3]",
            "ids": [
                4,
                5,
                6
            ],
            "isPic": 0,
            "duration": 22314,
            "status": 1,
            "tid": 1
        },
        {
            "value": "aabbcc",
            "lessons": [
                44,
                55,
                66
            ],
            "isPic": 0,
            "duration": 22314,
            "status": 2,
            "tid": 2
        }
    ],
    "questionType": 65
}

现在这个 json 字符串形式的字段值在 hive 表的某个字段中，我需要获取到 "answers" 这个 json 数组，然后将其按照数组长度，列转行到多行数据，每行数据一个子 json ，并且从中获取到每个子 json 的 "tid" 和 "status" 值，理想情况下，我需要这行数据处理完之后，结果如下表所示。

tid	status
1	1
2	2

2. 常见方案

通过 ge_json_object 函数，先获取到 "answers" 对应的 json 数组字符串，然后通过正则替换掉 [ 和 ] 符号，之后将 },{ 符号替换为 }我是分隔符{，最后将 我是分隔符 作为 split 函数的分隔符号，将字符串分割，再通过 lateral view explode() 语法，将数组放到多行。多行数据，都是处理好的 json 对象，之后通过 get_json_object 函数后取需要的字段值即可。具体代码示例如下。

获取 "answers" 对应的 json 数组。

sql 复制代码

with table1 as (
    select '{"score":1,"submitTime":1712491933,"answerFlag":1,"groupId":1755547960,"answers":[{"value":"[1, 2, 3]","ids":[4,5,6],"isPic":0,"duration":22314,"status":1,"tid":1},{"value":"aabbcc","lessons":[44,55,66],"isPic":0,"duration":22314,"status":2,"tid":2}],"questionType":65}' as col
)

select get_json_object(col, '$.answers') as answers
from table1

结果如下所示，正常获取到 "answers" 下的 json 数组，结果为字符串。

answers
${"value":"\[1, 2, 3$ ","ids": $4,5,6$ ,"isPic":0,"duration":22314,"status":1,"tid":1},{"value":"aabbcc","lessons": $44,55,66$ ,"isPic":0,"duration":22314,"status":2,"tid":2}]

将最外层的 [ 和 ] 符号去掉
sql 复制代码
```
with table1 as (
    select '{"score":1,"submitTime":1712491933,"answerFlag":1,"groupId":1755547960,"answers":[{"value":"[1, 2, 3]","ids":[4,5,6],"isPic":0,"duration":22314,"status":1,"tid":1},{"value":"aabbcc","lessons":[44,55,66],"isPic":0,"duration":22314,"status":2,"tid":2}],"questionType":65}' as col
)

select regexp_replace(answers, '\\[|\\]', '') as answers
from (
    select get_json_object(col, '$.answers') as answers
    from table1
    ) as a
```
结果如下所示。

answers

{"value":"1, 2, 3","ids":4,5,6,"isPic":0,"duration":22314,"status":1,"tid":1},{"value":"aabbcc","lessons":44,55,66,"isPic":0,"duration":22314,"status":2,"tid":2}

由于第一步处理的结果最外层是 json 数组，左右有 [ 和 ] 符号，但是由于内层还有子 json 数组，这种直接全局替换的方式，会将内层子 json 数组的 [、] 符号也一并去掉，可以查看下面的结果，"ids":4,5,6,，就是因为全局替换，造成整个 json 结构被破坏，之后将无法使用 get_json_object() 函数来获取想要的 key 对应的值了。

answers
{"value":"1, 2, 3","ids":4,5,6,"isPic":0,"duration":22314,"status":1,"tid":1},{"value":"aabbcc","lessons":44,55,66,"isPic":0,"duration":22314,"status":2,"tid":2}

由此可见，这种方式只适合于 "answers" 下的 json 数组内的每个 json 对象中都只包含 json 对象才行，不能再包含 json 数组，否则就会造成处理错误，拿不到想要的数据。

3. 推荐方案

3.1 具体步骤

获取 "answers" 对应的 json 数组。

sql 复制代码

with table1 as (
    select '{"score":1,"submitTime":1712491933,"answerFlag":1,"groupId":1755547960,"answers":[{"value":"[1, 2, 3]","ids":[4,5,6],"isPic":0,"duration":22314,"status":1,"tid":1},{"value":"aabbcc","lessons":[44,55,66],"isPic":0,"duration":22314,"status":2,"tid":2}],"questionType":65}' as col
)

select get_json_object(col, '$.answers') as answers
from table1

结果如下所示，正常获取到 "answers" 下的 json 数组，结果为字符串。

answers
${"value":"\[1, 2, 3$ ","ids": $4,5,6$ ,"isPic":0,"duration":22314,"status":1,"tid":1},{"value":"aabbcc","lessons": $44,55,66$ ,"isPic":0,"duration":22314,"status":2,"tid":2}]

通过字符串截取的方式，将第一步的结果最前面和最后面的 [、] 符号去掉

sql 复制代码

with table1 as (
    select '{"score":1,"submitTime":1712491933,"answerFlag":1,"groupId":1755547960,"answers":[{"value":"[1, 2, 3]","ids":[4,5,6],"isPic":0,"duration":22314,"status":1,"tid":1},{"value":"aabbcc","lessons":[44,55,66],"isPic":0,"duration":22314,"status":2,"tid":2}],"questionType":65}' as col
)

select substring(answers, 2, length(answers) - 2) as answers
from (
    select get_json_object(col, '$.answers') as answers
    from table1
    ) as a

结果如下所示，最前面和最后面的 [、] 符号已经被去掉。

answers
{"value":" $1, 2, 3$ ","ids": $4,5,6$ ,"isPic":0,"duration":22314,"status":1,"tid":1},{"value":"aabbcc","lessons": $44,55,66$ ,"isPic":0,"duration":22314,"status":2,"tid":2}

最重要的一步 ：通过正则替换，只要匹配到 [.*] 内容，就直接替换为数字 0。
sql 复制代码
```
with table1 as (
    select '{"score":1,"submitTime":1712491933,"answerFlag":1,"groupId":1755547960,"answers":[{"value":"[1, 2, 3]","ids":[4,5,6],"isPic":0,"duration":22314,"status":1,"tid":1},{"value":"aabbcc","lessons":[44,55,66],"isPic":0,"duration":22314,"status":2,"tid":2}],"questionType":65}' as col
)

select substring(answers, 1, length(answers) - 3) as answers
from (
    select get_json_object(col, '$.answers') as answers
    from table1
    ) as a
```
结果如下所示，字符串中所有 [.*] 的部分，都已经被替换为数字 0

answers

{"value":"0","ids":0,"isPic":0,"duration":22314,"status":1,"tid":1},{"value":"aabbcc","lessons":0,"isPic":0,"duration":22314,"status":2,"tid":2}

将 json 数字替换为数字 0，是为了兼容 "[.*]" 和 [.*] 两种情况，不管 json 数组是不是被英文双引号包围，替换为数字 0，都是没问题的，都可以保证 json 的格式不被破坏。

answers
{"value":"0","ids":0,"isPic":0,"duration":22314,"status":1,"tid":1},{"value":"aabbcc","lessons":0,"isPic":0,"duration":22314,"status":2,"tid":2}

由于已经去掉了所有的子 json 数组，之后就可以按照传统的方式，将 },{ 替换为 }我是分隔符号{。

sql 复制代码

with table1 as (
    select '{"score":1,"submitTime":1712491933,"answerFlag":1,"groupId":1755547960,"answers":[{"value":"[1, 2, 3]","ids":[4,5,6],"duration":22314,"status":1,"tid":1},{"value":"aabbcc","lessons":[44,55,66],"isPic":0,"duration":22314,"status":2,"tid":2}],"questionType":65}' as col
)

select regexp_replace(a.answers, '\\}\\,\\{', '\\}我是分隔符号{') as answers
from (
    select regexp_replace(answers, '\\[.*\\]', '0') as answers
    from (
        select substring(answers, 2, length(answers) - 2) as answers
        from (
            select get_json_object(col, '$.answers') as answers
            from table1
            ) as a
        ) as a
    ) as a

answers
{"value":"0","ids":0,"isPic":0,"duration":22314,"status":1,"tid":1}我是分隔符号{"value":"aabbcc","lessons":0,"isPic":0,"duration":22314,"status":2,"tid":2}

通过 我是分隔符号 切换字符串，再通过 lateral view explode() 语法将 json 数组展开，最后通过 get_json_object() 函数，获取需要的值即可。

sql 复制代码

with table1 as (
    select '{"score":1,"submitTime":1712491933,"answerFlag":1,"groupId":1755547960,"answers":[{"value":"[1, 2, 3]","ids":[4,5,6],"duration":22314,"status":1,"tid":1},{"value":"aabbcc","lessons":[44,55,66],"isPic":0,"duration":22314,"status":2,"tid":2}],"questionType":65}' as col
)

select get_json_object(b.answer, '$.tid') as tid, get_json_object(b.answer, '$.status') as status
from (
    select regexp_replace(a.answers, '\\}\\,\\{', '\\}我是分隔符号{') as answers
    from (
        select regexp_replace(answers, '\\[.*\\]', '0') as answers
        from (
            select substring(answers, 2, length(answers) - 2) as answers
            from (
                select get_json_object(col, '$.answers') as answers
                from table1
                ) as a
            ) as a
        ) as a
    ) as a
lateral view explode(split(regexp_replace(a.answers, '\\}\\,\\{', '\\}我是分隔符号{'), '我是分隔符号')) b as answer

结果如下，可以看到，数据处理结果符合预期

tid	status
1	1
2	2

3.2 注意事项

上面的步骤 3，只适用于你想要展开最外层的这个 json 数组，并且完全不需要内部嵌套的子 json 数组才行，否则将内部的子 json 数组全部替换为数字 0 之后，你就获取不到子 json 数组数据了。

如果还想要内部的子 json 数组，单纯的 sql 应该是实现不了的，需要去自定义 udf，然后通过 java 代码一层一层解析了。