I have a table with different values in BigQuery and I want to select the min or the max of these values.
You can use min/max with group by to get min or max date per ID group
SELECT ID, MIN(DATE) AS MIN_DATE, MAX(DATE) AS MAX_DATE
FROM TABLE
GROUP BY ID
Related
I have the following query that allows me to aggregate the number of unique sellers/buyers for every single day from the Flipside API:
SELECT
date_trunc('day', block_timestamp) AS date,
COUNT(DISTINCT(seller_address)) AS unique_sellers,
COUNT(DISTINCT(buyer_address)) AS unique_buyers
FROM ethereum.core.ez_nft_sales
GROUP BY date
Now, I've been trying a lot of different things, but I can't for the life of me figure out how it would be possible to get the number of unique active addresses on a given day as I would need to somehow merge the sellers and buyers and then count the unique addresses. I would greatly appreciate any kind of help. Thanks in advance!
This is how I managed to solve the issue by using a separate query for the unique_active and merging them:
WITH
other_values AS (
SELECT
date_trunc('day', block_timestamp) AS date,
COUNT(DISTINCT seller_address) AS unique_sellers,
COUNT(DISTINCT buyer_address) AS unique_buyers
FROM ethereum.core.ez_nft_sales
GROUP BY date
),
unique_addresses AS (
SELECT
date,
COUNT(*) as unique_active
FROM (
SELECT
date_trunc('day', block_timestamp) as date,
seller_address as address
FROM ethereum.core.ez_nft_sales
GROUP BY date, seller_address
UNION
SELECT
date_trunc('day', block_timestamp) as date,
buyer_address as address
FROM ethereum.core.ez_nft_sales
GROUP BY date, buyer_address
)
GROUP BY date
)
SELECT * FROM other_values
LEFT JOIN unique_addresses
ON other_values.date = unique_addresses.date
ORDER BY other_values.date DESC
I have a table of accident date I want to calculate the maximum between the difference of date i and date i + 1 which are in the same column. when we declare an accident date, I want to find the record of days without accidents.
You can use lag(). Assuming a table structure like mytable(dt), where dt is of a date-like datatype, you would do:
select max(diff)
from (select dt - lag(dt) over(order by dt) diff from mytable) t
This is a sample data file
Data Contains unique IDs with different latitudes and longitudes on multiple timestamps.I would like to select the rows of latest 30 days of coordinates for each unique ID.Please help me on how to run the query .This date is in Hive table
Regards,
Akshay
According to your example above (where no current year dates for id=2,3), you can numbering date for each id (order by date descending) using window function ROW_NUMBER(). Then just get latest 30 values:
--get all values for each id where num<=30 (get last 30 days for each day)
select * from
(
--numbering each date for each id order by descending
select *, row_number()over(partition by ID order by DATE desc)num from Table
)X
where num<=30
If you need to get only unique dates (without consider time) for each id, then can try this query:
select * from
(
--numbering date for each id
select *, row_number()over(partition by ID order by new_date desc)num
from
(
-- move duplicate using distinct
select distinct ID,cast(DATE as date)new_date from Table
)X
)Y
where num<=30
In Oracle this will be:
SELECT * FROM TEST_DATE1
WHERE DATEUPDT > SYSDATE - 30;
select * from MyTable
where
[Date]>=dateadd(d, -30, getdate());
To group by ID and perform aggregation, something like this
select ID,
count(*) row_count,
max(Latitude) max_lat,
max(Longitude) max_long
from MyTable
where
[Date]>=dateadd(d, -30, getdate())
group by ID;
I have a table with 3 columns: id, date and amount, but I would like to get accumulated SUM for each date (Last column).
Do you have an easy solution how to add this column?
I am trying with this:
SELECT date, sum(amount) as accumulated
FROM table group by date
WHERE max(date);
Should I user OVER() for this?
Use a window function to the total for each day:
SELECT date,
amount,
sum(amount) over (partition by date) as accumulated
FROM the_table;
However this will only work, if your dates all have the same time part (in Oracle a DATE column also contains a time). To make sure you ignore the time part, use trunc() to make sure all time parts are normalized to 00:00:00
SELECT date,
amount,
sum(amount) over (partition by trunc(date)) as accumulated
FROM the_table;
Use This:
SELECT T.ID, T.DATE, T.AMOUNT, (SELECT SUM(S.AMOUNT) FROM TABLE S WHERE S.DATE=T.DATE) ACCUMULATED
from
table T
This will give you the records from the table with a sum for all records for the date.
I have a table in which minimum and maximum temperature is stored per order no and date. I want to pick the minimum temperature and maximum temperature for each day. This should be done using SQL script.
You have to use group by clause, and aggregate functions min, max as below:
select date, min(temperature), max(temperature)
from table
group by date
It will work if your date have only year, month and day (01/11/2012).
In oracle:
SELECT TO_CHAR(DATE_VAL,'DD-MM-YYYY'), MAX(temperature),
MIN(temperature) FROM table_name group by TO_CHAR(DATE_VAL,'DD-MM-YYYY');
In MySQl:
SELECT DATE_FORMAT(DATE_VAL, '%d-%m-%Y'), max(temperature),
min(temperature) from table_name group by DATE_FORMAT(DATE_VAL, '%d-%m-%Y');