1. 背景
json 字符串值数据示例:
json
{
"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。
sqlwith 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, '$.answers') as answers from table1
结果如下所示,正常获取到 "answers" 下的 json 数组,结果为字符串。
answers [{"value":"[1, 2, 3]","ids":[4,5,6],"isPic":0,"duration":22314,"status":1,"tid":1},{"value":"aabbcc","lessons":[44,55,66],"isPic":0,"duration":22314,"status":2,"tid":2}] -
将最外层的
[
和]
符号去掉sqlwith table1 as ( select '{"score":1,"submitTime":1712491933,"answerFlag":1,"groupId":1755547960,"answers":[{"value":"[1, 2, 3]","ids":[4,5,6],"isPic":0,"duration":22314,"status":1,"tid":1},{"value":"aabbcc","lessons":[44,55,66],"isPic":0,"duration":22314,"status":2,"tid":2}],"questionType":65}' as col ) select regexp_replace(answers, '\\[|\\]', '') as answers from ( select get_json_object(col, '$.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。
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。
sqlwith 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, '$.answers') as answers from table1
结果如下所示,正常获取到 "answers" 下的 json 数组,结果为字符串。
answers [{"value":"[1, 2, 3]","ids":[4,5,6],"isPic":0,"duration":22314,"status":1,"tid":1},{"value":"aabbcc","lessons":[44,55,66],"isPic":0,"duration":22314,"status":2,"tid":2}] -
通过字符串截取的方式,将第一步的结果最前面和最后面的
[
、]
符号去掉sqlwith table1 as ( select '{"score":1,"submitTime":1712491933,"answerFlag":1,"groupId":1755547960,"answers":[{"value":"[1, 2, 3]","ids":[4,5,6],"isPic":0,"duration":22314,"status":1,"tid":1},{"value":"aabbcc","lessons":[44,55,66],"isPic":0,"duration":22314,"status":2,"tid":2}],"questionType":65}' as col ) select substring(answers, 2, length(answers) - 2) as answers from ( select get_json_object(col, '$.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。sqlwith 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, '$.answers') as answers from table1 ) as a
结果如下所示,字符串中所有
[.*]
的部分,都已经被替换为数字 0answers {"value":"0","ids":0,"isPic":0,"duration":22314,"status":1,"tid":1},{"value":"aabbcc","lessons":0,"isPic":0,"duration":22314,"status":2,"tid":2} 将 json 数字替换为数字 0,是为了兼容
"[.*]"
和[.*]
两种情况,不管 json 数组是不是被英文双引号包围,替换为数字 0,都是没问题的,都可以保证 json 的格式不被破坏。 -
由于已经去掉了所有的子 json 数组,之后就可以按照传统的方式,将
},{
替换为}我是分隔符号{
。sqlwith table1 as ( select '{"score":1,"submitTime":1712491933,"answerFlag":1,"groupId":1755547960,"answers":[{"value":"[1, 2, 3]","ids":[4,5,6],"duration":22314,"status":1,"tid":1},{"value":"aabbcc","lessons":[44,55,66],"isPic":0,"duration":22314,"status":2,"tid":2}],"questionType":65}' as col ) select regexp_replace(a.answers, '\\}\\,\\{', '\\}我是分隔符号{') as answers from ( select regexp_replace(answers, '\\[.*\\]', '0') as answers from ( select substring(answers, 2, length(answers) - 2) as answers from ( select get_json_object(col, '$.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。sqlwith 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。