scd2 table implementation in google bigquery
I am trying to create a SCD (slowly changing dimension) type 2 table in BigQuery without using any DML's
test schema :
id | date | name | valid_from | valid_to | flag
I need to capture the name changes for a specific id.
Thanks,
google-bigquery scd2
edited Nov 26 '18 at 6:30
TeeKea
3,22851932
asked Nov 25 '18 at 19:21
dineshachantadineshachanta
211
add a comment |
I am trying to create a SCD (slowly changing dimension) type 2 table in BigQuery without using any DML's
test schema :
id | date | name | valid_from | valid_to | flag
I need to capture the name changes for a specific id.
Thanks,
google-bigquery scd2
edited Nov 26 '18 at 6:30
TeeKea
3,22851932
asked Nov 25 '18 at 19:21
dineshachantadineshachanta
211
add a comment |
I am trying to create a SCD (slowly changing dimension) type 2 table in BigQuery without using any DML's
test schema :
id | date | name | valid_from | valid_to | flag
I need to capture the name changes for a specific id.
Thanks,
google-bigquery scd2
edited Nov 26 '18 at 6:30
TeeKea
3,22851932
asked Nov 25 '18 at 19:21
dineshachantadineshachanta
211
I am trying to create a SCD (slowly changing dimension) type 2 table in BigQuery without using any DML's
test schema :
id | date | name | valid_from | valid_to | flag
I need to capture the name changes for a specific id.
Thanks,
google-bigquery scd2
google-bigquery scd2
edited Nov 26 '18 at 6:30
TeeKea
3,22851932
asked Nov 25 '18 at 19:21
dineshachantadineshachanta
211
edited Nov 26 '18 at 6:30
TeeKea
3,22851932
asked Nov 25 '18 at 19:21
dineshachantadineshachanta
211
edited Nov 26 '18 at 6:30
TeeKea
3,22851932
edited Nov 26 '18 at 6:30
TeeKea
3,22851932
edited Nov 26 '18 at 6:30
TeeKea
3,22851932
3,22851932
asked Nov 25 '18 at 19:21
dineshachantadineshachanta
211
asked Nov 25 '18 at 19:21
dineshachantadineshachanta
211
asked Nov 25 '18 at 19:21
dineshachantadineshachanta
211
211
add a comment |
add a comment |
1 Answer
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You can create artificially date partitioned tables to achieve this. This way, each date's data is unique and can be accessed by `select distinct column from `dataset.table$yyyymmdd;` or for all the history, just do select distinct column from dataset.table;. Also, you can always overwrite/append to a given date's partition without harming others.
answered Nov 26 '18 at 0:52
khankhan
2,12493053
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can create artificially date partitioned tables to achieve this. This way, each date's data is unique and can be accessed by `select distinct column from `dataset.table$yyyymmdd;` or for all the history, just do select distinct column from dataset.table;. Also, you can always overwrite/append to a given date's partition without harming others.
answered Nov 26 '18 at 0:52
khankhan
2,12493053
add a comment |
You can create artificially date partitioned tables to achieve this. This way, each date's data is unique and can be accessed by `select distinct column from `dataset.table$yyyymmdd;` or for all the history, just do select distinct column from dataset.table;. Also, you can always overwrite/append to a given date's partition without harming others.
answered Nov 26 '18 at 0:52
khankhan
2,12493053
add a comment |
You can create artificially date partitioned tables to achieve this. This way, each date's data is unique and can be accessed by `select distinct column from `dataset.table$yyyymmdd;` or for all the history, just do select distinct column from dataset.table;. Also, you can always overwrite/append to a given date's partition without harming others.
answered Nov 26 '18 at 0:52
khankhan
2,12493053
You can create artificially date partitioned tables to achieve this. This way, each date's data is unique and can be accessed by `select distinct column from `dataset.table$yyyymmdd;` or for all the history, just do select distinct column from dataset.table;. Also, you can always overwrite/append to a given date's partition without harming others.
answered Nov 26 '18 at 0:52
khankhan
2,12493053
answered Nov 26 '18 at 0:52
khankhan
2,12493053
answered Nov 26 '18 at 0:52
khankhan
2,12493053
answered Nov 26 '18 at 0:52
khankhan
2,12493053
2,12493053
add a comment |
add a comment |
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