How to generate a lot of tables from SQL? - sql

I have the following tables:
Table names: US states - 50 tables with following format
columns: state, zip code, address
zip code is not unique
I want to do the following with one SQL query:
generate a list of unique zip codes ~40,000
create tables with names of the zip code ~40,000
in each table - only the records with corresponding zip code
columns: state, zip code, address
How to create such a SQL query to be most effective and cheapest way to create all these tables?

With current features you could do a couple "cheats":
Define a linear transform from zipcode to dates. For example, DATE_ADD('1980-01-01', INTERVAL zipcode DAYS)
Create a partitioned table - use that synthetic date as partition.
Write a query that inserts all your data into that table SELECT *, synthetic_date(zipcode) FROM ``tables*``.
Now you have a partitioned table, with each partition containing data only for each zip code! Cost is linear, just one scan of all your data.
And if you'd like, now you can copy each partition to a new table - at no cost (but be aware of daily quotas). See https://cloud.google.com/bigquery/docs/managing-partitioned-tables#copying_a_single_partition.

Related

Hive bucketing based on a string column value

I understand we can partition a hive table based on a column and apply a filter to insert records through static or dynamic partitioning.
But if I'll need to bucket based on a particular column value, how do I do that?
Let us say I have users from 500 different countries
username string
country string
If I need to create a hive table and bucket it based on country, is that possible? ideally one bucket per country.
Yes you need to cluster your data based on country. and you need to define the number of buckets based on the total number of countries.
for e.g. if there are data coming in from 500 countries, then the following should do:
create table (
name string,
country string
) clustered by (country) into 500 buckets;
But having said that, I would suggest you to go for dynamic partitioning for better query plan.

Understanding Hive table creation notation

I have come across Hive tables which I need to convert to Redshift/MySql equivalent.
I am having trouble understanding Hive query structure and would appreciate some help:
CREATE TABLE IF NOT EXISTS table_1 (
id BIGINT,
price DOUBLE,
asset string
)
PARTITIONED BY (
pt STRING
);
ALTER TABLE table_1 DROP IF EXISTS PARTITION (pt== '${yyyymmdd}');
INSERT OVERWRITE TABLE table_1 PARTITION (pt= '${yyyymmdd}')
select aa.id,aa.price,aa.symbol from
...
...
from
table_2 table
I am having trouble understanding the PARTITIONED BY clause. This, if I am understanding correctly, is different from MySQL table partitions, and is a Hive specific dynamic partition.
The partition does not define a column or a key, and partitions by the current date.
Does this mean that table_1 is partitioned by the date? Each day has a separate partition?
Then later on in the code there are notations similar to
inner join table_new table on table.pt = '${yyyymmdd}' and ...
In this context, does it mean only rows inserted on yyyymmdd are selected for the join?
Thank you.
Partition in Hive is a folder in HDFS by default with name key=value + metadata in the Hive metastore. You can alter partition location and create partition on top of any folder.
This PARTITIONED BY (pt STRING) defines partition column pt of type string, not date. Partition values are stored in the metadata. The pt column is not present in the table data files, it is only defined in PARTITIONED BY, all partition values are stored in the metadata. If you load partition dynamically, partition folder is being created with name pt='value'.
This sentence creates partition dynamically:
INSERT OVERWRITE TABLE table_1 PARTITION (pt)
select id, price, symbol
coln as pt --partition column should be the last one
from ...
And this sentence loads single STATIC partition:
INSERT OVERWRITE TABLE table_1 PARTITION (pt= '${yyyymmdd}')
select aa.id,aa.price,aa.symbol
from
No partition column is selected, partition value specified in the
PARTITION (pt= '${yyyymmdd}')
'${yyyymmdd}' here is a parameter with name yyyymmdd which is passed to the script using --hivevar like this:
hive --hivevar yyyymmdd=20200604 -f myscript.sql
You can pass ANY string as partition value in this case, though parameter name yyyymmdd suggests it's format.
BTW date format in hive is 'yyyy-MM-dd' Strings in 'yyyy-MM-dd' format can be implicitly converted to DATE.
I will try in one shot explain what is partitioning in Hive. First of all would be
WHEN TO USE TABLE PARTITIONING
Table partitioninig is good when:
Reading the entire dataset takes too long
Queries almost always filter on the partition columns
There are a reasonable number of different values for partition columns
Data generation of ETL process splits data by file or directory names
Partition column values are not in the data itself
Don't partition on columns with many unique values
Example: Partitioning customers by first name
CREATING PARTITIONED TABLES
To create a partitioned table, use the PARTITIONED BY clause in the CREATE TABLE statement.
The names and types of the partition columns must be specified
in the PARTITIONED BY clause, and only in the PARTITIONED BY clause.
They must not also appear in the list of all the other columns.
CREATE TABLE customers_by_country
(cust_id STRING, name STRING)
PARTITIONED BY (country STRING)
ROW FORMAT DELIMITED FIELDS TERMINATED BY '\t';
The example CREATE TABLE statement shown above creates the table customers_by_country,
which is partitioned by the STRING column named country.
Notice that the country column appears only in the PARTITIONED BY clause,
and not in the column list above it.
This example specifies only one partition column, but you can specify more than one by using
a comma-separated column list in the PARTITIONED BY clause.
Aside from these specific differences, this CREATE TABLE statement is the same
as the statement used to create an equivalent non-partitioned table.
Table partitioning is implemented in a way that is mostly transparent
to a user issuing queries with Hive.
A partition column is what’s known as a virtual column, because its values are not stored within the data files.
Following is the result of the DESCRIBE command on customers_by_country;
it displays the partition column country just as if it were a normal column within the table.
You can refer to partition columns in any of the usual clauses of a SELECT statement.
name type comment
cust_id string
name string
country string
You can load data in partitioned tables dynamically or statically
LOADING DATA WITH DYNAMIC PARTITION
One way to load data into a partitioned table is to use dynamic partitioning,
which automatically defines partitions when you load the data, using the values in the partition column.
(The other way is to manually define the partitions with Static Partitioning)
To use dynamic partitioning, you must load data using an INSERT statement.
In the INSERT statement, you must use the PARTITION clause to list the partition columns.
The data you are inserting must include values for the partition columns.
The partition columns must be the rightmost columns in the data you are inserting,
and they must be in the same order as they appear in the PARTITION clause.
INSERT OVERWRITE TABLE customers_by_country
PARTITION(country)
SELECT cust_id, name, country FROM customers;
The example shown above uses an INSERT … SELECT statement
to load data into the customers_by_country table with dynamic partitioning.
Notice that the partition column, country, is included
in the PARTITION clause and is specified last in the SELECT list.
When Hive executes this statement, it automatically creates partitions
for the country column and loads the data into these partitions based on the values in the country column.
The resulting data files in the partition subdirectories do not include values for the country column.
Since the country is known based on which subdirectory a data file is in,
it would be redundant to include country values in the data files as well.
Look at the contents of the customers_by_country directory.
It should now have one subdirectory for each value in the country column.
Look at the file in one of those directories.
Notice that the file contains the row for the customer from that country,
and no others; notice also that the country value is not included.
Note: Hive includes a safety feature that prevents users
from accidentally creating or overwriting a large number of partitions.
(See “Risks of Using Partitioning” for more about this.)
By default, Hive sets the property hive.exec.dynamic.partition.mode to strict.
This prevents you from using dynamic partitioning, though you can still use static partitions.
You can disable this safety feature in Hive by setting
the property hive.exec.dynamic.partition.mode to nonstrict:
SET hive.exec.dynamic.partition.mode=nonstrict;
Then you can use the INSERT statement to load the data dynamically.
Hive properties set in Beeline are for the current session only,
so the next time you start a Hive session this property will be set back to strict.
But you or your system administrator can configure properties permanently, if necessary.
When you run some SELECT queries on the partitioned table, if the table is big enough you can note significant difference in the time it takes to run.
Notice that you will not query the table any differently than you would query the customers table.
LOADING DATA WITH STATIC PARTITIONING
One way to load data into a partitioned table is to use static partitioning,
in which you manually define the different partitions.
With static partitioning, you create a partition manually, using an ALTER TABLE … ADD PARTITION statement,
and then load the data into the partition.
For example, this ALTER TABLE statement creates the partition for Pakistan (pk):
ALTER TABLE customers_by_country
ADD PARTITION (country='pk');
Notice how the partition column name, which is country, and the specific value that defines this partition,
which is pk, are both specified in the ADD PARTITION clause.
This creates a partition directory named country=pk inside the customers_by_country table directory.
After the partition for Pakistan is created, you can add data into the partition using an INSERT … SELECT statement:
INSERT OVERWRITE TABLE customers_by_country
PARTITION(country='pk')
SELECT cust_id, name FROM customers WHERE country='pk'
Notice how in the PARTITION clause, the partition column name, which is country,
and the specific value, which is pk, are both specified, just like in the ADD PARTITION command used to create the partition.
Also notice that in the SELECT statement, the partition column is not included in the SELECT list.
Finally, notice that the WHERE clause in the SELECT statement selects only customers from Pakistan.
With static partitioning, you need to repeat these two steps for each partition:
first create the partition, then add data.
You can actually use any method to load the data; you need not use an INSERT statement.
You could instead use hdfs dfs commands or a LOAD DATA INPATH command.
But however you load the data, it’s your responsibility to ensure that data is stored in the correct partition subdirectories.
For example, data for customers in Pakistan must be stored in the Pakistan partition subdirectory,
and data for customers in other countries must be stored in those countries’ partition subdirectories.
Static partitioning is most useful when the data being loaded
into the table is already divided into files based on the partition column,
or when the data grows in a manner that coincides with the partition column:
For example, suppose your company opens a new store in a different country,
like New Zealand ('nz'), and you're given a file of data for new customers, all from that country.
You could easily add a new partition and load that file into it.
RISKS OF USING PARTITIONING
A major risk when using partitioning is creating partitions that lead you into the small files problem.
When this happens, partitioning a table will actually worsen query performance
(the opposite of the goal when using partitioning) because it causes too many small files to be created.
This is more likely when using dynamic partitioning, but it could still
happen with static partitioning—for example if you added a new partition to a sales table
on a daily basis containing the sales from the previous day,
and each day’s data is not particularly big.
When choosing your partitions, you want to strike a happy balance between too many partitions
(causing the small files problem) and too few partitions (providing performance little benefit).
The partition column or columns should have a reasonable number of values
for the partitions—but what you should consider reasonable is difficult to quantify.
Using dynamic partitioning is particularly dangerous because if you're not careful,
it's easy to partition on a column with too many distinct values.
Imagine a use case where you are often looking for data that falls within
a time frame that you would specify in your query.
You might think that it's a good idea to partition on a column that pertains to time.
But a TIMESTAMP column could have the time to the nanosecond, so every row could have a unique value;
that would be a terrible choice for a partition column! Even to the minute or hour could create
far too many partitions, depending on the nature of your data;
partitioning by larger time units like day, month, or even year might be a better choice.
As another example, consider an employees table.
This has five columns: empl_id, first_name, last_name, salary, and office_id.
Before reading on, think for a moment, which of these might be reasonable for partitioning
The column empl_id is a unique identifier.
If that were your partition column, you would have a separate partition for each employee,
and each would have exactly one row.
In addition, it's not likely you'll be doing a lot of queries looking for a particular value,
or even a particular range of values. This is a poor choice.
The column first_name will not have one per employee, but there will likely be many columns that have only one row.
This is also true for last_name.
Also, like empl_id, it's not likely you'll need filter queries based on these columns. These are also poor choices.
The column salary also will have many divisions
(and even more so if your salaries go to the cent rather than to the dollar as our sample table does).
While it may be that you'll sometimes want to query on salary ranges,
it's not likely you'll want to use individual salaries.
So salary is a poor choice.
A more limited salary_grades specification, like the ones in the salary_grades table,
might be reasonable if your use case involves looking at the data by salary grade frequently.
The office_id column identifies the office where an employee works.
This will have a much smaller number of unique values, even if you have a large company with offices in many cities.
It's imaginable that your use case might be to frequently filter
your employee data based on office location, too. So this would be a good choice.
You also can use multiple columns and create nested partitions.
For example, a dataset of customers might include country and state_or_province columns.
You can partition by country and then partition those further by state_or_province, so customers from Ontario,
Canada would be in the country=ca/state_or_province=on/ partition directory.
This can be extremely helpful for large amounts of data that you want to access either by country or by state or province.
However, using multiple columns increases the danger of creating too many partitions, so you must take extra care when doing so.
The risk of creating too many partitions is why Hive includes the property
hive.exec.dynamic.partition.mode, set to strict by default, which must be reset to nonstrict before you can create a partition.
Rather than automatically and mechanically resetting that property when you're about to load data dynamically,
take it as an opportunity to think about the partitioning columns
and maybe check the number of unique values you would get when you load the data.
And that's all.

Extract rows which haven't been extracted already using SQL Server stored procedure

I have a table Customers. I'm trying to design a way which will extract data from Customers table daily and create a CSV of this data. I want to pick only those records which haven't been extracted yet. How can I keep track of whether it has been extracted or not? I cannot alter the Customers table to add a flag.
So far I'm planning to use a Stage table which will have this flag. So I'm writing a stored procedure to get the data from the Customers table and have the flag set to 0 for each of these records. And use SSIS to create the CSV after pulling this data from stage table and once the records have been extracted into CSV update the staging table with flag=1 for those records.
What is a good design for this problem?
Customer table:
CustomerID | Name | RecordCreated | RecordUpdated
Create another table tblExportedEmpID with a column CustomerID. Add the customer id of each customer extracted from the Customer table into that new table. And to extract the customer from the Customer table which are not extracted yet, you can use this query :
select * from customer where customerid not in(select customerid from tblExportedEmpID)
You have RecordCreated and RecordUpdated. Why even bother with a separate record-per table if you have that information?
You'll need to create a table or equivalent "saved until next run" data area. The first thing you have your script do is grab the current time, and whatever was stored in that data area. Then, have your statement query everything:
SELECT <list of columns and transformation>
FROM Customers
WHERE recordCreated >= :lastRunTime AND recordCreated < :currentRunTime
(or recordUpdated, if you need to re-extract if the customer's name changes)
Note that you want the exclusive upper-bound (<) to cover the case where your stored timestamp has less resolution than the mechanism getting the timestamp.
For the last step, store off your run start - whatever the script grabbed for "current time" - into the "saved until next run" data area.

MS Access - create JOIN table to store value for every combination

I am working on an MS Access 2013 database. I have two tables:
Customers (28 records)
Chemicals (34 records)
I need to create a table for usage rates for each customer for each chemical.
The rates will be entered manually (at user's request). I am trying to determine how to create a new table where the customer-chemical fields will combine to be primary key.
The resulting table should have 28x34=952 unique records.
The goal is to then have a form wherein the user can select the customer, then the chemical, and edit the rate.
For any table/query creation I am comfortable using either the Access interface or SQL.
I will advise to create a new table containing 4 columns. The first column will be an 'id' it is going to be your primary key (auto-increment if you want), second column is the customer, then the chemical, and finally the rating. Then if you format your query to select 'rating' where customer='customer name' and chemical='chemical name', you should get the desired result you want.
Thank you for the reply. Did a little more wrestling with it and used the following SQL to create the table:
SELECT customers.customer, chemicals.chemical
INTO UsageRates
FROM Chemicals, Customers
Then adding a blank 'rate' field to the table.

How to take look up values from a look up table using SSIS

I have a scenario as described below need to create a SSIS Package for that.
I have 3 COLUMNS in source table which needs to be entered in destination table.
But all these columns has to be looked up in the look up table of destination database and then enter their ID's in the destination column.
For example
Source table has 3 columns with values
idnum static type timedimension geography modified date
1 price daydate france 8/12/2015
2 RetailpRICE WEEK ITALY 9/12/2014
I want a package which looks up the column values with the matchin ID and populates in the destination table...
I know we can use the LOOKUP transform to update the data for one single column in destination table what about the other columns which I need to insert along with the lookup insertion.
How can I achieve this ? Also is there a way to pull only the recent data from the source table using modified date column values
Use a different lookup for each lookup table that you need to reference to get the Ids. So if each of your columns that you want IDs for gets its ID from a different table, then you need to use three lookups, one after the other, until you have all three IDs.