2016-08-21 68 views
0

我有JSON格式的数据,其中嵌套数组。下面是一个例子:将嵌套数组加载到bigquery中

"data": {"events": [[1, 1271, 518, 945], [1, 1287, 495, 963],... 

子阵列的长度可以是3或4,并且所述第一数量的数据类型(有大约30种不同的)。有没有办法将这些数据加载到bigQuery而不转换成字典的“记录”?

感谢, 亚龙

- 编辑 -

this问题,即有一个解决办法,但有一个固定长度的子阵,所以不适用我想..

+0

目前尚不清楚预期的最终表 - 举例说明! –

+0

重要的是,您可以使用投票下方发布的答案左侧的勾号标记接受的答案。请参阅http://meta.stackexchange。com/questions/5234/how-does-accepting-an-answer-work#5235为什么它很重要。答案投票也很重要。表决有用的答案。还有更多......当某人回答你的问题时,你可以查看该怎么做 - http://stackoverflow.com/help/someone-answers。 –

回答

0

无法直接加载数组数组;您需要使用记录来包装数组的内层。标准SQL的参考涉及到这一点(尽管在语言本身方面,没有加载数据):https://cloud.google.com/bigquery/sql-reference/arrays#building-arrays-of-arrays

+0

感谢您的回答。我试图避免必须对数据进行转换才能将其转化为记录/结构。有没有办法上传这个嵌套数组作为字符串/文本blob? – WeaselFox

+0

这个过去的问题可能会有所帮助:http://stackoverflow.com/questions/37660579/bigquery-create-column-of-json-datatype你可以将它加载为一个字符串,然后使用BigQuery的JSON函数来提取你想要的部分作为查询的一部分。 –

1

这可能是错误的方向,因为它是不完全清楚什么是你的最终目标,但让我尽量帮你
不知怎的,我觉得你的目标表有望成为类似下面

type metric1 metric2 metric3 
    1  1271  518  945 
    1  1287  495  963 

所以,我的建议是让你的两个步骤

步骤1 - 只需一个字段加载数据作为CSV - 假设表theTable与现场data

       data 
{"data": {"events": [[1, 1271, 518, 945], [1, 1287, 495, 963]]}} 
{"data": {"events": [[2, 111, 222, 333], [3, 444, 555, 666], [4, 777, 888, 999]]}} 

第二步 - 过程theTable产生预期的架构(请参阅对答案的顶部)和保存到决赛桌。您可以使用下面的查询该

SELECT 
    NTH(1, SPLIT(y)) AS type, 
    NTH(2, SPLIT(y)) AS metric1, 
    NTH(3, SPLIT(y)) AS metric2, 
    NTH(4, SPLIT(y)) AS metric3, 
FROM (
    SELECT 
    REPLACE(REPLACE(COALESCE(y0, y1, y2, y3, y4, y5, y6), '[', ''), ']', '') AS y 
    FROM (
    SELECT 
     IF(k=0, JSON_EXTRACT(data, '$.data.events[0]'), NULL) AS y0, 
     IF(k=1, JSON_EXTRACT(data, '$.data.events[1]'), NULL) AS y1, 
     IF(k=2, JSON_EXTRACT(data, '$.data.events[2]'), NULL) AS y2, 
     IF(k=3, JSON_EXTRACT(data, '$.data.events[3]'), NULL) AS y3, 
     IF(k=4, JSON_EXTRACT(data, '$.data.events[4]'), NULL) AS y4, 
     IF(k=5, JSON_EXTRACT(data, '$.data.events[5]'), NULL) AS y5, 
     IF(k=6, JSON_EXTRACT(data, '$.data.events[6]'), NULL) AS y6, 
    FROM theTable AS a 
    CROSS JOIN (
     SELECT k FROM (SELECT 0 AS k), (SELECT 1 AS k), (SELECT 2 AS k), 
     (SELECT 3 AS k), (SELECT 4 AS k), (SELECT 5 AS k), (SELECT 6 AS k) 
    ) AS b 
) 
    HAVING NOT y IS NULL 
) 

其结果将是

type metric1 metric2 metric3 
    1  1271  518  945 
    1  1287  495  963 
    2  111  222  333 
    3  444  555  666 
    4  777  888  999 

正如你所看到的 - 这个特定的查询多达7次阵列支持,但是你可以减少或改变增加此在三个地方

#1

REPLACE(REPLACE(COALESCE(y0, y1, y2, y3, y4, y5, y6), '[', ''), ']', '') AS y  

#2

0码
IF(k=0, JSON_EXTRACT(data, '$.data.events[0]'), NULL) AS y0, 
IF(k=1, JSON_EXTRACT(data, '$.data.events[1]'), NULL) AS y1, 
IF(k=2, JSON_EXTRACT(data, '$.data.events[2]'), NULL) AS y2, 
IF(k=3, JSON_EXTRACT(data, '$.data.events[3]'), NULL) AS y3, 
IF(k=4, JSON_EXTRACT(data, '$.data.events[4]'), NULL) AS y4, 
IF(k=5, JSON_EXTRACT(data, '$.data.events[5]'), NULL) AS y5, 
IF(k=6, JSON_EXTRACT(data, '$.data.events[6]'), NULL) AS y6, 

#3

SELECT k FROM (SELECT 0 AS k), (SELECT 1 AS k), (SELECT 2 AS k), 
(SELECT 3 AS k), (SELECT 4 AS k), (SELECT 5 AS k), (SELECT 6 AS k) 

最后,测试只是转换逻辑,W/O负载的实际数据 - 您可以使用下面的脚本

SELECT 
    NTH(1, SPLIT(y)) AS type, 
    NTH(2, SPLIT(y)) AS metric1, 
    NTH(3, SPLIT(y)) AS metric2, 
    NTH(4, SPLIT(y)) AS metric3, 
FROM (
    SELECT 
    REPLACE(REPLACE(COALESCE(y0, y1, y2, y3, y4, y5, y6), '[', ''), ']', '') AS y 
    FROM (
    SELECT 
     IF(k=0, JSON_EXTRACT(data, '$.data.events[0]'), NULL) AS y0, 
     IF(k=1, JSON_EXTRACT(data, '$.data.events[1]'), NULL) AS y1, 
     IF(k=2, JSON_EXTRACT(data, '$.data.events[2]'), NULL) AS y2, 
     IF(k=3, JSON_EXTRACT(data, '$.data.events[3]'), NULL) AS y3, 
     IF(k=4, JSON_EXTRACT(data, '$.data.events[4]'), NULL) AS y4, 
     IF(k=5, JSON_EXTRACT(data, '$.data.events[5]'), NULL) AS y5, 
     IF(k=6, JSON_EXTRACT(data, '$.data.events[6]'), NULL) AS y6, 
    FROM (
    SELECT data FROM 
     (SELECT '{"data": {"events": [[1, 1271, 518, 945], [1, 1287, 495, 963]]}}' AS data), 
     (SELECT '{"data": {"events": [[2, 111, 222, 333], [3, 444, 555, 666], [4, 777, 888, 999]]}}' AS data) 
    ) AS a 
    CROSS JOIN (
     SELECT k FROM (SELECT 0 AS k), (SELECT 1 AS k), (SELECT 2 AS k), 
     (SELECT 3 AS k), (SELECT 4 AS k), (SELECT 5 AS k), (SELECT 6 AS k) 
    ) AS b 
) 
    HAVING NOT y IS NULL 
) 

希望这是有帮助的!

+0

Upvote梦幻般的逻辑和努力 – BigDaddy