如何在 bigquery 中 delete/update 嵌套数据
How to delete/update nested data in bigquery
有没有办法在 bigquery 中 delete/update 嵌套字段?
假设我有这个数据
wives.age wives.name name
21 angel adam
20 kale
21 victoria rossi
20 jessica
或 json:
{"name":"adam","wives":[{"name":"angel","age":21},{"name":"kale","age":20}]}
{"name":"rossi","wives":[{"name":"victoria","age":21},{"name":"jessica","age":20}]}
从上面的数据可以看出。
亚当有 2 个妻子,分别是天使和凯尔。如何:
- 删除羽衣甘蓝记录。
- 将杰西卡更新为德西卡
我试图 google 这个,但找不到。我也试过 unnest 等,但没有运气。
我们之所以要这样做是因为我们将数组插入到错误的记录中,并希望remove/update array data with some condition
以下适用于 BigQuery 标准 SQL
#standardSQL
WITH updates AS (
SELECT 'rossi' name, 'jessica' oldname, 'dessica' newname UNION ALL
SELECT 'rossi' name, 'victoria' oldname, 'polly' newname UNION ALL
SELECT 'adam' name, 'angel' oldname, 'jen' newname
), divorces AS (
SELECT 'adam' name, 'kale' wifename UNION ALL
SELECT 'adam' name, 'milly' wifename UNION ALL
SELECT 'rossi' name, 'linda' wifename
)
SELECT t.name,
ARRAY(
SELECT AS STRUCT
age,
CASE
WHEN NOT oldname IS NULL THEN newname
ELSE name
END name
FROM UNNEST(wives)
LEFT JOIN UNNEST(updates) ON t.name = u.name AND name = oldname
LEFT JOIN UNNEST(divorces) AS wifename ON t.name = d.name AND name = wifename
WHERE wifename IS NULL
) waves
FROM `project.dataset.table` t
LEFT JOIN (
SELECT name, ARRAY_AGG(STRUCT(oldname, newname)) updates
FROM updates GROUP BY name
) u ON t.name = u.name
LEFT JOIN (
SELECT name, ARRAY_AGG(wifename) divorces
FROM divorces GROUP BY name
) d ON t.name = d.name
您可以使用下面的虚拟数据测试/玩上面的内容
#standardSQL
WITH `project.dataset.table` AS (
SELECT 'adam' name, [STRUCT<age INT64, name STRING>(21, 'angel'), (20, 'kale'), (22, 'milly')] wives UNION ALL
SELECT 'rossi', [STRUCT<age INT64, name STRING>(21, 'victoria'), (20, 'jessica'), (23, 'linda')]
), updates AS (
SELECT 'rossi' name, 'jessica' oldname, 'dessica' newname UNION ALL
SELECT 'rossi' name, 'victoria' oldname, 'polly' newname UNION ALL
SELECT 'adam' name, 'angel' oldname, 'jen' newname
), divorces AS (
SELECT 'adam' name, 'kale' wifename UNION ALL
SELECT 'adam' name, 'milly' wifename UNION ALL
SELECT 'rossi' name, 'linda' wifename
)
SELECT t.name,
ARRAY(
SELECT AS STRUCT
age,
CASE
WHEN NOT oldname IS NULL THEN newname
ELSE name
END name
FROM UNNEST(wives)
LEFT JOIN UNNEST(updates) ON t.name = u.name AND name = oldname
LEFT JOIN UNNEST(divorces) AS wifename ON t.name = d.name AND name = wifename
WHERE wifename IS NULL
) waves
FROM `project.dataset.table` t
LEFT JOIN (
SELECT name, ARRAY_AGG(STRUCT(oldname, newname)) updates
FROM updates GROUP BY name
) u ON t.name = u.name
LEFT JOIN (
SELECT name, ARRAY_AGG(wifename) divorces
FROM divorces GROUP BY name
) d ON t.name = d.name
结果符合预期
name waves.age waves.name
adam 21 jen
rossi 21 polly
20 dessica
希望您能够将以上内容应用到您的实际案例中:o)
有没有办法在 bigquery 中 delete/update 嵌套字段?
假设我有这个数据
wives.age wives.name name
21 angel adam
20 kale
21 victoria rossi
20 jessica
或 json:
{"name":"adam","wives":[{"name":"angel","age":21},{"name":"kale","age":20}]}
{"name":"rossi","wives":[{"name":"victoria","age":21},{"name":"jessica","age":20}]}
从上面的数据可以看出。 亚当有 2 个妻子,分别是天使和凯尔。如何:
- 删除羽衣甘蓝记录。
- 将杰西卡更新为德西卡
我试图 google 这个,但找不到。我也试过 unnest 等,但没有运气。
我们之所以要这样做是因为我们将数组插入到错误的记录中,并希望remove/update array data with some condition
以下适用于 BigQuery 标准 SQL
#standardSQL
WITH updates AS (
SELECT 'rossi' name, 'jessica' oldname, 'dessica' newname UNION ALL
SELECT 'rossi' name, 'victoria' oldname, 'polly' newname UNION ALL
SELECT 'adam' name, 'angel' oldname, 'jen' newname
), divorces AS (
SELECT 'adam' name, 'kale' wifename UNION ALL
SELECT 'adam' name, 'milly' wifename UNION ALL
SELECT 'rossi' name, 'linda' wifename
)
SELECT t.name,
ARRAY(
SELECT AS STRUCT
age,
CASE
WHEN NOT oldname IS NULL THEN newname
ELSE name
END name
FROM UNNEST(wives)
LEFT JOIN UNNEST(updates) ON t.name = u.name AND name = oldname
LEFT JOIN UNNEST(divorces) AS wifename ON t.name = d.name AND name = wifename
WHERE wifename IS NULL
) waves
FROM `project.dataset.table` t
LEFT JOIN (
SELECT name, ARRAY_AGG(STRUCT(oldname, newname)) updates
FROM updates GROUP BY name
) u ON t.name = u.name
LEFT JOIN (
SELECT name, ARRAY_AGG(wifename) divorces
FROM divorces GROUP BY name
) d ON t.name = d.name
您可以使用下面的虚拟数据测试/玩上面的内容
#standardSQL
WITH `project.dataset.table` AS (
SELECT 'adam' name, [STRUCT<age INT64, name STRING>(21, 'angel'), (20, 'kale'), (22, 'milly')] wives UNION ALL
SELECT 'rossi', [STRUCT<age INT64, name STRING>(21, 'victoria'), (20, 'jessica'), (23, 'linda')]
), updates AS (
SELECT 'rossi' name, 'jessica' oldname, 'dessica' newname UNION ALL
SELECT 'rossi' name, 'victoria' oldname, 'polly' newname UNION ALL
SELECT 'adam' name, 'angel' oldname, 'jen' newname
), divorces AS (
SELECT 'adam' name, 'kale' wifename UNION ALL
SELECT 'adam' name, 'milly' wifename UNION ALL
SELECT 'rossi' name, 'linda' wifename
)
SELECT t.name,
ARRAY(
SELECT AS STRUCT
age,
CASE
WHEN NOT oldname IS NULL THEN newname
ELSE name
END name
FROM UNNEST(wives)
LEFT JOIN UNNEST(updates) ON t.name = u.name AND name = oldname
LEFT JOIN UNNEST(divorces) AS wifename ON t.name = d.name AND name = wifename
WHERE wifename IS NULL
) waves
FROM `project.dataset.table` t
LEFT JOIN (
SELECT name, ARRAY_AGG(STRUCT(oldname, newname)) updates
FROM updates GROUP BY name
) u ON t.name = u.name
LEFT JOIN (
SELECT name, ARRAY_AGG(wifename) divorces
FROM divorces GROUP BY name
) d ON t.name = d.name
结果符合预期
name waves.age waves.name
adam 21 jen
rossi 21 polly
20 dessica
希望您能够将以上内容应用到您的实际案例中:o)