[英]Extract multiple values from an array in JSON in BigQuery
I have a JSON in my database table like the following one, and let's say the column that contains a JSON file called favorites.我的数据库表中有一个 JSON,如下所示,假设该列包含一个名为 favorites 的 JSON 文件。
{
"info": {
"music": [{"Year":2021,"Name":"Stay","Singer":"Justin Bieber"},
{"Year":2015,"Name":"Love Yourself","Singer":"Justin Bieber"},
{"Year":2003,"Name":"Crazy In Love","Singer":"Beyonce"}
],
"movie": [{"Year":2018,"Name":"Green Book","Director":"Peter Farrelly"},
{"Year":2007,"Name":"Lust, Caution","Director":"Ang Lee"}
]
}
}
I wanted to select all values from tags and my expected table would be like as following:
-----------------------------------------------------------------------------
| Name | Singer |
----------------------------------------------------------------------------
| Stay,Love Yourself,Crazy In Love | Justin Bieber,Justin Bieber,Beyonce|
-----------------------------------------------------------------------------
I already know how to get the first value in an array with JSON_QUERY(json_col,'$.info.music[0].Name'), but I would like to extracted all the names or singers into one single column, and some arrays may have multiple items.我已经知道如何使用 JSON_QUERY(json_col,'$.info.music[0].Name') 获取数组中的第一个值,但我想将所有名称或歌手提取到一个列中,还有一些 arrays可能有多个项目。 Doe anyone have any suggestions?
有人有什么建议吗?
Consider below approach考虑以下方法
select
array(select json_extract_scalar(x, '$.Name') from unnest(json_extract_array(json_col, '$.info.music') || json_extract_array(json_col, '$.info.movie')) x) Name,
array(select json_extract_scalar(x, '$.Singer') from unnest(json_extract_array(json_col, '$.info.music')) x) Singer
from data
if applied to sample data in your question - output is如果应用于您问题中的样本数据 - output 是
I just realized - you wanted comma separated list - so consider below then我刚刚意识到 - 你想要逗号分隔列表 - 那么请考虑下面
select
(select string_agg(json_extract_scalar(x, '$.Name')) from unnest(json_extract_array(json_col, '$.info.music') || json_extract_array(json_col, '$.info.movie')) x) Name,
(select string_agg(json_extract_scalar(x, '$.Singer')) from unnest(json_extract_array(json_col, '$.info.music')) x) Singer
from data
with output与 output
Another sol.另一个溶胶。 One can use ARRAY_TO_STRING if you don't want aggregation.
如果您不想聚合,可以使用 ARRAY_TO_STRING。
with data as
(
select
"""
{
"info": {
"music": [{"Year":2021,"Name":"Stay","Singer":"Justin Bieber"},
{"Year":2015,"Name":"Love Yourself","Singer":"Justin Bieber"},
{"Year":2003,"Name":"Crazy In Love","Singer":"Beyonce"}
],
"movie": [{"Year":2018,"Name":"Green Book","Director":"Peter Farrelly"},
{"Year":2007,"Name":"Lust, Caution","Director":"Ang Lee"}
]
}
}
""" as _json
)
select array_to_string(
array(
select json_extract_scalar(x,"$.Name")
from data,
unnest(json_extract_array(_json,"$.info.music")) as x
),","
) as Name, array_to_string(
array(
select json_extract_scalar(x,"$.Singer")
from data,
unnest(json_extract_array(_json,"$.info.music")) as x
),","
) as Singer
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