I have two tables, linked by an ID column. Each table has a date column, formatted as 8 numbers ie. 20230102.
My goal is simple: find the earliest date for each ID in both tables, and determine if these dates match.
Historical_Period
ID
StartDate
1
18020101
1
19000217
CountryName
ID
StartDate
1
15161129
1
20020301
Here is what I have tried:
Select
*
from Historical_Period t1
inner join CountryName on CountryName.ID=t1.ID
where CountryName.StartDate <> (Select min(Historical_Period.StartDate) from Historical_Period) group by t1.StartDate;
The result is a mess. I neither get the earliest dates for each ID, and some of them match...
select HID, CID, Hdate, Cdate, case when Hdate <> Cdate then '!Matched' else
'Matched' end from
(select ID as HID, MIN(StartDate) as Hdate from Historical_Period group by
ID) a,
(select ID as CID, MIN(StartDate) as Cdate from CountryName group by ID) b
where a.HID = b.CID
Related
having a list of people like:
name date_of_birth
john 1987-09-08
maria 1987-09-08
samuel 1987-09-09
claire 1987-09-10
jane 1987-09-10
rose 1987-09-12
...
How can I get a result view using SQL of how many people are born up to that date, like the output for that table should be:
date count
1987-09-08 2
1987-09-09 3
1987-09-10 5
1987-09-11 5
1987-09-12 6
...
Thanks!
Here is another way, in addition to Gordon's answer. It uses joins:
SELECT
t1.date_of_birth,
COUNT(*) AS count
FROM (SELECT DISTINCT date_of_birth FROM yourTable) t1
INNER JOIN yourTable t2
ON t1.date_of_birth >= t2.date_of_birth
GROUP BY
t1.date_of_birth;
Note: I left out a step. Apparently you also want to report missing dates. If so, then you may replace what I aliased as t1 with a calendar table. For the sake of demonstration, you can inline all the dates:
SELECT
t1.date_of_birth,
COUNT(*) AS count
FROM
(
SELECT '1987-09-08' AS date_of_birth UNION ALL
SELECT '1987-09-09' UNION ALL
SELECT '1987-09-10' UNION ALL
SELECT '1987-09-11' UNION ALL
SELECT '1987-09-12'
) t1
LEFT JOIN yourTable t2
ON t1.date_of_birth >= t2.date_of_birth
GROUP BY
t1.date_of_birth;
Demo
In practice, your calendar table would be a bona fide table which just contains all the dates you want to appear in your result set.
One method is a correlated subquery:
select dob.date_of_birth,
(select count(*) from t where t.date_of_birth <= dob.date_of_birth) as running_count
from (select distinct date_of_birth from t) dob;
This is not particularly efficient. If your data has any size, variables are better (or window functions if you are using MySQL 8.0):
select date_of_birth,
(#x := #x + cnt) as running_count
from (select date_of_birth, count(*) as cnt
from t
group by date_of_birth
order by date_of_birth
) dob cross join
(select #x := 0) params;
Use subquery with correlation approach :
select date_of_birth, (select count(*)
from table
where date_of_birth <= t.date_of_birth
) as count
from table t
group by date_of_birth;
Hello i am having two different table with same field created_date (datetime)
now i want records which counts daywise records with joining table i have done for individual counting as below query :
SELECT DATE(created_date), COUNT(*) FROM table1 GROUP BY DAY(created_date)
SELECT DATE(created_date), COUNT(*) FROM table2 GROUP BY DAY(created_date)
and i am getting results for individuals something like this:
RESULT I NEED :
DATE(created_date) count(table1) count(table2)
2016-12-01 10 3
2016-12-02 1 0
2016-12-05 1 0
2016-11-29 1 0
2016-11-30 4 1
Now i just want to join these result WITH INDIVIDUAL VIEW COUNT ACCORDING TO TABLE can anyone please help me out with this profile....
First take a UNION between your two tables, then use conditional aggregation to determine the counts for each of the two tables. Note that I introduce a field called table_name to keep track of data from each of the two tables.
SELECT t.created_date,
SUM(CASE WHEN t.table_name = 'one' THEN 1 ELSE 0 END) AS count_table_one,
SUM(CASE WHEN t.table_name = 'two' THEN 1 ELSE 0 END) AS count_table_two
FROM
(
SELECT DATE(created_date) AS created_date, 'one' AS table_name
FROM table1
UNION ALL
SELECT DATE(created_date), 'two'
FROM table2
) t
GROUP BY t.created_date
I used DATE consistently everywhere to make the query correct.
Try This:
SELECT created_date, sum(countTable1) countTable1,
sum(countTable2) countTable2
FROM (
SELECT DATE(created_date) created_date, COUNT(*) countTable1, NULL countTable2
FROM table1 GROUP BY DAY(created_date)
UNION ALL
SELECT DATE(created_date) created_date, NULL, COUNT(*) countTable2
FROM table2 GROUP BY DAY(created_date)) t GROUP BY t.created_date
You have a problem in your queries, you are grouping by DAY(date) and showing 'date' so the result will be first date with day(date), yet repeating it to avoid misunderstanding :)
select IFNULL(A.cd, B.cd), A.cnt, B.cnt from
(SELECT DAY(created_date) d, DATE(created_date) cd, COUNT(*) cnt
FROM table1 GROUP BY DAY(created_date)) as A
LEFT JOIN
(SELECT DAY(created_date) d, DATE(created_date) cd , COUNT(*) cnt
FROM table2 GROUP BY DAY(created_date)) B ON B.d = A.d
Its not too hard just use union if no need to allow duplicate row else use union all for all(means allow duplicate as well).
SELECT DATE(created_date), COUNT(*) FROM table1 GROUP BY DAY(created_date)
UNION
SELECT DATE(created_date), COUNT(*) FROM table2 GROUP BY DAY(created_date)
SELECT T.create_date,ISNULL(T.count,0)AS Counttable1,ISNULL(X.count,0)AS Counttable2 FROM(SELECT DATE(created_date) AS create_date,COUNT(*) as count FROM table1 GROUP BY DAY(created_date)) AS T LEFTJOIN(SELECT DATE(created_date) AS create_date, COUNT(*) as count FROM table2 GROUP BY DAY(created_date))AS X ON T.create_date=X.create_date
You actually need a SQL UNION. JOIN natuarually eliminate counts becuase the maytch fields. I.e. if you had 2016-12-01 in both table1 andtable2 then a JOIN on created_date would give you a count of 1 instead of a count of 2.
SELECT DATE(total.created_date), COUNT(*)
FROM (
SELECT created_date FROM table1
UNION ALL
SELECT created_date FROM table2) as total
GROUP BY total.created_date
HERE you simply union the two tables since they have a matching column name. Then you get back every date from both tables. That is in the inner query. The outer query then does the counting.
Hope that makes sense.
Good day.
I seem to be struggling with what seems like a simple problem.
I have a table that has a value connected to a date (Monthly) for a finite number of ID's
ie. Table1
ID | Date ---| Value
01 | 2015-01 | val1
01 | 2015-02 | val2
02 | 2015-01 | val1
02 | 2015-03 | val2
So ID: 02 does not have a value for date 2015-02.
I would like to return all ID's and Dates that do not have a value.
Date range is: select distinct date from Table1
I can't seem to think outside the realms of selecting and joining on the same table.
I need to include the ID in my select to I can somehow select the ID and Date range that exists for that ID and compare to the entire date range, to get all the dates for each ID that isn't in the "entire" date range.
Please advise.
Thank you
Not very clear about your last two sentences. But you can play with the following query with different #max_days and #min_date:
-- DROP TABLE table1;
CREATE TABLE table1(ID int not null, `date` date not null, value varchar(64) not null);
INSERT table1(ID,`date`,value)
VALUES (1,'2015-01-01','v1'),(1,'2015-01-02','v2'),(2,'2015-01-01','v1'),(2,'2015-01-03','v2'),(4,'2015-01-01','v1'),(4,'2015-01-04','v2');
SELECT * FROM table1;
SET #day=0;
SET #max_days=5;
SET #min_date='2015-01-01';
SELECT i.ID,d.`date`
FROM (SELECT DISTINCT ID FROM table1) i
CROSS JOIN (
SELECT TIMESTAMPADD(DAY,#day,#min_date) AS `date`,#day:=#day+1 AS day_num
FROM table1 WHERE #day<#max_days) d
LEFT JOIN table1 t
ON t.ID=i.ID
AND t.`date`=d.`date`
WHERE t.`date` IS NULL
ORDER BY i.ID,d.`date`;
I now understand your requirement of dates being taken from the table; you want to find any gaps in the date ranges for each id.
This does what you need, but can probably be improved. Explanation below and you can view a working example.
DROP TABLE IF EXISTS Table1;
DROP TABLE IF EXISTS Year_Month_Calendar;
CREATE TABLE Table1 (
id INTEGER
,date CHAR(7)
,value CHAR(4)
);
INSERT INTO Table1
VALUES
(1,'2015-01','val1')
,(1,'2015-02','val2')
,(2,'2015-01','val1')
,(2,'2015-03','val1');
CREATE TABLE Year_Month_Calendar (
date CHAR(10)
);
INSERT INTO Year_Month_Calendar
VALUES
('2015-01')
,('2015-02')
,('2015-03');
SELECT ID_Year_Month.id, ID_Year_Month.date, Table1.id, Table1.date
FROM (
SELECT Distinct_ID.id, Year_Month_Calendar.date
FROM Year_Month_Calendar
CROSS JOIN
( SELECT DISTINCT id FROM Table1 ) AS Distinct_ID
WHERE Year_Month_Calendar.date >= (SELECT MIN(date) FROM Table1 WHERE id=Distinct_ID.ID)
AND Year_Month_Calendar.date <= (SELECT MAX(date) FROM Table1 WHERE id=Distinct_ID.ID)
) AS ID_Year_Month
LEFT JOIN Table1
ON ID_Year_Month.id = Table1.id AND ID_Year_Month.date = Table1.date
-- WHERE Table1.id IS NULL
ORDER BY ID_Year_Month.id, ID_Year_Month.date
Explanation
You need a calendar table which contains all dates (year/months) to cover the data you are querying.
CREATE TABLE Year_Month_Calendar (
date CHAR(10)
);
INSERT INTO Year_Month_Calendar
VALUES
('2015-01')
,('2015-02')
,('2015-03');
The inner select creates a table with all dates between the min and max date for each id.
SELECT Distinct_ID.id, Year_Month_Calendar.date
FROM Year_Month_Calendar
CROSS JOIN
( SELECT DISTINCT id FROM Table1 ) AS Distinct_ID
WHERE Year_Month_Calendar.date >= (SELECT MIN(date) FROM Table1 WHERE id=Distinct_ID.ID)
AND Year_Month_Calendar.date <= (SELECT MAX(date) FROM Table1 WHERE id=Distinct_ID.ID)
This is then LEFT JOINED to the original table to find the missing rows.
If you only want to return the missing row (my query displays the whole table to show how it works), add a WHERE clause to restrict the output to those rows where an id and date is not returned from Table1
Original answer before comments
You can do this without a tally table, since you say
Date range is: select distinct date from Table1
I've slightly changed the field names to avoid reserved words in SQL.
SELECT id_table.ID, date_table.`year_month`, table1.val
FROM (SELECT DISTINCT ID FROM table1) AS id_table
CROSS JOIN
(SELECT DISTINCT `year_month` FROM table1) AS date_table
LEFT JOIN table1
ON table1.ID=id_table.ID AND table1.`year_month` = date_table.`year_month`
ORDER BY id_table.ID
I've not filtered the results, in order to show how the query is working. To return the rows where only where a date is missing, add WHERE table1.year_month IS NULL to the outer query.
SQL Fiddle
You will need a tally table(s) or month/year tables. So you can then generate all of the potential combinations you want to test with. As far as exactly how to use it your example could use some expanding on such as last 12 months, last3 months, etc. but here is an example that might help you understand what you are looking for:
http://rextester.com/ZDQS5259
CREATE TABLE IF NOT EXISTS Tbl (
ID INTEGER
,Date VARCHAR(10)
,Value VARCHAR(10)
);
INSERT INTO Tbl VALUES
(1,'2015-01','val1')
,(1,'2015-02','val2')
,(2,'2015-01','val1')
,(2,'2015-03','val1');
SELECT yr.YearNumber, mn.MonthNumber, i.Id
FROM
(
SELECT 2016 as YearNumber
UNION SELECT 2015
) yr
CROSS JOIN (
SELECT 1 MonthNumber
UNION SELECT 2
UNION SELECT 3
UNION SELECT 4
UNION SELECT 5
UNION SELECT 6
UNION SELECT 7
UNION SELECT 8
UNION SELECT 9
UNION SELECT 10
UNION SELECT 11
UNION SELECT 12
) mn
CROSS JOIN (
SELECT DISTINCT ID
FROM
Tbl
) i
LEFT JOIN Tbl t
ON yr.YearNumber = CAST(LEFT(t.Date,4) as UNSIGNED)
AND mn.MonthNumber = CAST(RIGHT(t.Date,2) AS UNSIGNED)
AND i.ID = t.ID
WHERE
t.ID IS NULL
The basic idea to determine what you don't know is to generate all possible combinations of something could be. E.g. Year X Month X DISTINCT Id and then join back to figure out what is missing.
Probably not the prettiest but this should work.
select distinct c.ID, c.Date, d.Value
from (select a.ID, b.Date
from (select distinct ID from Table1) as a, (select distinct Date from Table1) as b) as c
left outer join Table1 d on (c.ID = d.ID and c.Date = d.Date)
where d.Value is NULL
I want to use 1 table to create a new table using 2 sets of queries.
To test out the code: http://sqlfiddle.com/#!9/02e3ff/5
Reference table:
Desired table:
They share the same order_id.
type = A, updated_at = pDate
type = B, updated_at = dDate
Query 1:
select t.order_id, t.updated_at as pDate, weekday(t.updated_at) from transactions t
where t.type = 'A' group by t.order_id
Query 2:
select t.order_id, max(t.updated_at) as dDate, weekday(max(t.updated_at)) from transactions t
where t.type= 'B'
group by t.order_id;
For type = A, I want to get the earliest updated_at date, while for type = B, I want to get the latest updated_at date.
Currently, I tried union but they give me 2 rows instead of the desired table.
How do I join or union these 2 queries to get the desired table?
Alternatively, is there a better method to do this? Thanks!
You can try something like this:
SELECT order_id, min(pDate) pDate, max(dDate) dDate FROM(
SELECT
order_id,
if(type='A',updated_at,null) pDate,
if(type='B',updated_at,null) dDate
FROM transactions
) as d
GROUP BY order_id
SQLFiddle
I have the following table:
ID Date FirstName Dept
1 1/2/12 James Act
1 2/5/12 Mike IT
2 5/6/12 Joe HR
2 7/6/12 Keith IT
What I need to do that for each ID, I need to get the max date.
I need to show ID, Date, FirstName, Dept for the record for each ID that has the Max Date.
So in this case for ID of 1, I would show 1 2/5/12 Mike IT
How do I do this in SQL Server T-SQL?
I know I need to do group by.
The table name is TblAct
You will use the MAX() function with a GROUP BY
select t1.id, t1.date, t1.fname, t1.dept
from tblAct t1
inner join
(
SELECT Max(Date) maxdate, ID
from TblAct
GROUP BY id
) t2
on t1.id = t2.id
and t1.date = t2.maxdate
See SQL Fiddle with Demo
You can do this with windows/ranking functions:
select ID, Date, FirstName, Dept
from (select t.*,
row_number() over (partition by id order by date desc) as seqnum
from t
) t
where seqnum = 1
This is ordering all the rows for each id by date, in reverse order. It then selects the first of them.
dont use group by :
select * from tblAct t1
where date=(select max(date) from tblAct where t1.id = id)
just enjoy.