I have made a view(joining four tables) like below:
ID | BookID | date | points |
1 | 11 | 2014-11-01 | 15 |
1 | 11 | 2015-01-01 | 16 |
1 | 11 | 2014-12-01 | 17 |
1 | 12 | 2014-02-11 | 18 |
1 | 12 | 2014-03-11 | 19 |
1 | 12 | 2014-04-11 | 15 |
1 | 13 | 2014-12-23 | 121 |
1 | 14 | 2014-01-15 | 113 |
1 | 14 | 2014-02-08 | 112 |
I want the result of this view as below
ID | BookID | Date | points |
1 | 11 | 2015-01-01 | 16 |
1 | 12 | 2014-04-11 | 15 |
1 | 13 | 2014-12-23 | 121 |
1 | 14 | 2014-02-08 | 112 |
It should be like Distincit Book ID with max date and showing as seprate points.
So far i have tried the group by with join and group by with date. But it is getting a bit over as i am unable to find a solution to this.
My Query is:
SELECT m1.* FROM viewPoints m1 LEFT JOIN viewPoints m2
ON (m1.BookID = m2.BookID AND m1.Date < m2.Date)
WHERE m1.ID= 1 and m2.Date IS NULL
ORDER BY m1.BookID
Any help! Thanks in Advance.
Maybe this is what you want?
select v.*
from viewPoints v
join (
select
BookID,
max(date) max_date
from viewPoints
where points is not null
group by BookID
) v2 on v.BookID = v2.BookID and v.date = v2.max_date
where v.points is not null
order by v.BookID
Sample SQL Fiddle
Sample output:
| ID | BOOKID | DATE | POINTS |
|----|--------|---------------------------------|--------|
| 1 | 11 | January, 01 2015 00:00:00+0000 | 16 |
| 1 | 12 | April, 11 2014 00:00:00+0000 | 15 |
| 1 | 13 | December, 23 2014 00:00:00+0000 | 121 |
| 1 | 14 | February, 08 2014 00:00:00+0000 | 112 |
SELECT *
FROM tablename
WHERE DATE
IN (
SELECT MAX( DATE )
FROM tablename
GROUP BY bookid
ORDER BY DATE DESC
)
ORDER BY DATE DESC
CREATE VIEW [BOOKLIST] AS
SELECT m1.* FROM viewPoints m1 LEFT JOIN viewPoints m2
ON (m1.BookID = m2.BookID AND m1.Date < m2.Date)
WHERE m1.ID= 1 and m2.Date IS NULL
ORDER BY m1.BookID
SELECT ID, DISTINCT BookID, Date, points FROM BOOKLIST
WHERE Date BETWEEN "start date" AND "end date"
Related
select * from new_joiner;
+------+--------------+
| id | date_of_join |
+------+--------------+
| 1 | 2020-01-10 |
| 2 | 2020-01-02 |
| 3 | 2020-01-05 |
| 4 | 2020-02-10 |
| 5 | 2020-02-11 |
| 6 | 2020-07-11 |
| 7 | 2020-07-11 |
| 8 | 2020-07-11 |
| 9 | 2020-07-11 |
| 10 | 2020-07-11 |
| 11 | 2020-05-01 |
| 12 | 2020-05-02 |
| 13 | 2020-05-03 |
| 14 | 2020-05-04 |
| 15 | 2020-05-05 |
| 16 | 2020-05-05 |
| 17 | 2020-05-06 |
+------+--------------+
select MONTHNAME(date_of_join) as MONTHNAME,
count(id) as JOINEE
from new_joiner
where MONTH(date_of_join)>=1
group by MONTH(date_of_join);
+-----------+--------+
| MONTHNAME | JOINEE |
+-----------+--------+
| January | 3 |
| February | 2 |
| May | 7 |
| July | 5 |
+-----------+--------+
I want a query that gives me the monthly data change compare to previous month.
For example: new joinee in Jan was 3, and in Feb it was 2, so compare to Jan in Feb month -1 joined, so the query should output me:
+-----------+-------------+
| MONTHNAME | JOINEE_DIFF |
+-----------+-------------+
| February | -1 |
| Mar | -2 |
| April | 0 |
| May | 7 |
| June | -7 |
| July | 5 |
| Aug | -5 |
| Sep | 0 |
| Oct | 0 |
| Nov | 0 |
| Dec | 0 |
+-----------+-------------+
Ignore Jan as it doesn't have a previous month and assume we have data only for a given year say 2020. Require data for all months from Feb to Dec.
Assuming you have data for every month, you can use lag():
select MONTHNAME(date_of_join) as MONTHNAME,
count(id) as JOINEE,
(count(*) - lag(count(*)) over (order by min(date_of_join)) as diff
from new_joiner
where MONTH(date_of_join) >= 1
group by MONTH(date_of_join);
Note that using months without years if fraught with peril. Also, the month() of any well-formed date should be larger than 1.
All this suggests a query more like:
select *
from (select MONTHNAME(date_of_join) as MONTHNAME,
count(id) as JOINEE,
(count(*) - lag(count(*)) over (order by min(date_of_join)) as diff,
min(date_of_join) as min_date_of_join
from new_joiner
where date_of_join >= '2020-01-01' and date_of_join < '2021-01-01'
group by MONTH(date_of_join)
) t
where diff is not null
order by min_date_of_join;
Use a correlated subquery to get the number of joinees of previous month and subtract it:
SELECT
t.monthname,
joinee - (SELECT COUNT(*) FROM new_joiner WHERE MONTH(date_of_join) = t.month - 1) JOINEE_DIFF
FROM (
SELECT MONTH(date_of_join) month, MONTHNAME(date_of_join) monthname,
COUNT(id) joinee
FROM new_joiner
GROUP BY month, monthname
) t
WHERE t.month > 1;
I am trying to show invoices for every single day, so for that purpose I used group by on created date and sum on subtotal. This is how I done it :
SELECT
`main_table`.*,
SUM(subtotal) AS `total_sales`
FROM
`sales_invoice` AS `main_table`
GROUP BY
DATE_FORMAT(created_at, "%m-%y")
Its working, but I also want to get the Invoice # from and Invoice # to for every date. Is it possible to do it with single query ?
EDIT :
Table Structure :
------------------------------------------------
| id | inoice_no | created_at | subtotal
| 1 | 34 | 2015-03-17 05:55:27 | 5
| 2 | 35 | 2015-03-17 12:35:00 | 7
| 3 | 36 | 2015-03-20 01:40:00 | 3
| 4 | 37 | 2015-03-20 07:05:13 | 6
| 5 | 38 | 2015-03-20 10:25:23 | 1
| 6 | 39 | 2015-03-24 12:00:00 | 6
------------------------------------------------
Output
---------------------------------------------------------------
| id | inoice_no | created_at | subtotal | total_sales
| 2 | 35 | 2015-03-17 12:35:00 | 7 | 12
| 5 | 38 | 2015-03-20 10:25:23 | 1 | 10
| 6 | 39 | 2015-03-24 12:00:00 | 6 | 6
-----------------------------------------------------------------
What I Expect
---------------------------------------------------------------
| id | inoice_no | created_at | subtotal | total_sales | in_from | in_to
| 2 | 35 | 2015-03-17 12:35:00 | 7 | 12 | 34 | 35
| 5 | 38 | 2015-03-20 10:25:23 | 1 | 10 | 36 | 38
| 6 | 39 | 2015-03-24 12:00:00 | 6 | 6 | 39 | 39
-----------------------------------------------------------------
If your invoice number is INTEGER then below query will give you the result what you want:
SELECT DATE_FORMAT(A.created_at, "%m-%y") AS InvoiceDate,
MIN(A.invoiveNo) AS FromInvoiceNo,
MAX(A.invoiveNo) AS ToInvoiceNo,
SUM(A.subtotal) AS total_sales
FROM sales_invoice AS A
GROUP BY InvoiceDate;
I guess salesid is primaryid in sales_invoice table.
select * from(
SELECT
`main_table`.*,
SUM(subtotal) AS `total_sales`
FROM
`sales_invoice` AS `main_table`
GROUP BY
DATE_FORMAT(created_at, "%m-%y")
order by main_table.salesid limit 1
union all
SELECT
`main_table`.*,
SUM(subtotal) AS `total_sales`
FROM
`sales_invoice` AS `main_table`
GROUP BY
DATE_FORMAT(created_at, "%m-%y")
order by main_table.salesid desc limit 1
)a
Let's say I have a table "calendar"
+------------+
| day_date |
+------------+
| 2015-01-01 |
| 2015-01-02 |
| 2015-01-03 |
| .......... |
| 2015-07-14 |
| 2015-07-15 |
+------------+
With this query I can select the WEEK (that I need)
SELECT WEEK(day_date,1) AS NUM_WEEK,
YEAR(day_date) AS YEAR,
STR_TO_DATE(CONCAT(YEAR(day_date),WEEK(day_date,1),' Monday'), '%X%V %W') AS date_start
FROM calendar
GROUP BY NUM_WEEK
And this is the result:
+----------+------+------------+
| NUM_WEEK | YEAR | date_start |
+----------+------+------------+
| 29 | 2015 | 2015-07-20 |
| 30 | 2015 | 2015-07-27 |
| 31 | 2015 | 2015-08-03 |
| 32 | 2015 | 2015-08-10 |
| 33 | 2015 | 2015-08-17 |
| 34 | 2015 | 2015-08-24 |
| 35 | 2015 | 2015-08-31 |
| 36 | 2015 | 2015-09-07 |
| 37 | 2015 | 2015-09-14 |
| 38 | 2015 | 2015-09-21 |
| 39 | 2015 | 2015-09-28 |
| 40 | 2015 | 2015-10-05 |
| 41 | 2015 | 2015-10-12 |
| 42 | 2015 | 2015-10-19 |
| 43 | 2015 | 2015-10-26 |
+----------+------+------------+
Now I have another table:
+----+------------+--------+---------------------+
| id | id_account | amount | date_transaction |
+----+------------+--------+---------------------+
| 1 | 283 | 150 | 2015-06-21 15:50:47 |
| 2 | 283 | 47.74 | 2015-07-23 15:55:44 |
| 3 | 281 | 21.55 | 2015-08-24 12:27:11 |
| 4 | 283 | 11.22 | 2015-08-25 10:00:54 |
+----+------------+--------+---------------------+
They are gaps in date.
With a similar query:
SELECT WEEK(date_transaction,1) AS NUM_WEEK,
YEAR(date_transaction) AS YEAR,
STR_TO_DATE(CONCAT(YEAR(date_transaction),WEEK(date_transaction,1),' Monday'), '%X%V %W')
AS date_start,
transaction.id_account,
SUM(amount) as total FROM transaction
INNER JOIN account ON account.id_account = transaction.id_account
WHERE amount > 0 AND transaction.id_account
IN ( SELECT id_account FROM account WHERE id_customer = 12 )
GROUP BY id_account, WEEK(date_transaction,1)
I obtain this result (probably data are not accurate, referring to previous tables, just to explain).
+----------+------+------------+-----------+----------+
| NUM_WEEK | YEAR | date_start | idAccount | total |
+----------+------+------------+-----------+----------+
| 29 | 2015 | 2015-07-20 | 281 | 22377.00 |
| 30 | 2015 | 2015-07-27 | 281 | 11550.00 |
| 32 | 2015 | 2015-08-04 | 281 | 4500.00 |
| 30 | 2015 | 2015-07-27 | 283 | 1500 |
+----------+------+------------+-----------+----------+
What I would, RIGHT (or LEFT) JOINING the two tables?
The min (and max) WEEK, so I can... (see 2)
Fill the gaps with missing WEEKS with NULL VALUES.
E.g., in a more complicated resultset:
+----------+------+------------+-----------+----------+
| NUM_WEEK | YEAR | date_start | idAccount | total |
+----------+------+------------+-----------+----------+
| 29 | 2015 | 2015-07-20 | 281 | 22377.00 |
| 30 | 2015 | 2015-07-27 | 281 | 11550.00 |
| 31 | 2015 | 2015-07-02 | 281 | NULL |
| 32 | 2015 | 2015-08-09 | 281 | 4500.00 |
| 29 | 2015 | 2015-08-09 | 283 | NULL |
| 30 | 2015 | 2015-07-16 | 283 | 1500 |
| 31 | 2015 | 2015-07-16 | 283 | NULL |
| 32 | 2015 | 2015-07-16 | 283 | NULL |
+----------+------+------------+-----------+----------+
Note, for example, that id=283 now has NULL at WEEK 29, 31 and 32, for example, like id=281 has NULL in WEEK 31.
I prepared also SQLFiddle here: http://sqlfiddle.com/#!9/a8fdc/3
Thank you very much.
I take a look on your question and i came up with this solution. Here is how your query could look like:
SELECT t1.NUM_WEEK, t1.`YEAR`, t1.date_start, t1.id_account, t2.total
FROM (SELECT c.NUM_WEEK, c.`YEAR`, c.date_start, a.id_account
FROM (SELECT WEEK(day_date,1) AS NUM_WEEK,
YEAR(day_date) AS `YEAR`,
STR_TO_DATE(CONCAT(YEAR(day_date),WEEK(day_date,1),' Monday'), '%X%V %W') AS date_start,
(SELECT GROUP_CONCAT(id_account) FROM account WHERE id_customer=12) AS accounts_id
FROM calendar
GROUP BY NUM_WEEK) c
INNER JOIN account a
ON FIND_IN_SET(a.id_account, c.accounts_id)
ORDER BY a.id_account, c.NUM_WEEK) t1
LEFT JOIN
(SELECT WEEK(t.date_transaction,1) AS NUM_WEEK,
YEAR(t.date_transaction) AS `YEAR`,
STR_TO_DATE(CONCAT(YEAR(t.date_transaction),WEEK(t.date_transaction,1),' Monday'), '%X%V %W') AS date_start,
t.id_account, SUM(t.amount) AS total
FROM `transaction` t
INNER JOIN account a
ON a.id_account = t.id_account
WHERE t.amount > 0 AND
t.id_account IN (SELECT id_account FROM account WHERE id_customer = 12)
GROUP BY id_account, WEEK(date_transaction,1)) t2
ON t1.NUM_WEEK = t2.NUM_WEEK AND t1.YEAR = t2.YEAR AND t1.id_account = t2.id_account;
Here is SQL Fiddle for that so you can check up result. Hope that is what are you looking for.
Little explanation:
First think i done is that I little modified your first query where you extract data from table calendar and add there one new column called accounts_id. That query now look's like this:
SELECT WEEK(day_date,1) AS NUM_WEEK,
YEAR(day_date) AS `YEAR`,
STR_TO_DATE(CONCAT(YEAR(day_date),WEEK(day_date,1),' Monday'), '%X%V %W') AS date_start,
(SELECT GROUP_CONCAT(id_account) FROM account WHERE id_customer=12) AS accounts_id
FROM calendar
GROUP BY NUM_WEEK
Please pay attention on this line in SELECT statement
(SELECT GROUP_CONCAT(id_account) FROM account WHERE id_customer=12) AS accounts_id
Note that when you select for specific customer you need to change customer ID in this line too!!!
Here is Fiddle so you can check result that this query produce.
This is necessary because we need to connect each week with each account to get desired result.
Next step is to extend previous query so we could separate accounts_id column (look result of previous query) so we could get row for each value in that column. Extended query look like this:
SELECT c.NUM_WEEK, c.`YEAR`, c.date_start, a.id_account
FROM (SELECT WEEK(day_date,1) AS NUM_WEEK,
YEAR(day_date) AS `YEAR`,
STR_TO_DATE(CONCAT(YEAR(day_date),WEEK(day_date,1),' Monday'), '%X%V %W') AS date_start,
(SELECT GROUP_CONCAT(id_account) FROM account WHERE id_customer=12) AS accounts_id
FROM calendar
GROUP BY NUM_WEEK) c
INNER JOIN account a
ON FIND_IN_SET(a.id_account, c.accounts_id)
ORDER BY a.id_account, c.NUM_WEEK
and output you can see in this Fiddle
After that all we need to do is to make left join between this query and query you already wrote in your question (last query).
There might be a better solution or even this one maybe can be improved a little, but I don't have much time now to deal with that and this is the first think that cross my mind...
GL!
P. S. pay attention when you use reserved word in MySQL like YEAR, TRANSACTION etc for column name (as column_name).. that can cause you a treble if have to use them in name of column or table use backquote () to mark them (asyear`)...
I'm trying to select the most recent rows for every unique userid where pid = 50 and active = 1. I haven't been able to figure it out.
Here is a sample table
+-----+----------+-------+-----------------------+---------+
| id | userid | pid | start_date | active |
+-----+----------+-------+-----------------------+---------+
| 1 | 4 | 50 | 2015-05-15 12:00:00 | 1 |
| 2 | 4 | 50 | 2015-05-16 12:00:00 | 1 |
| 3 | 4 | 50 | 2015-05-17 12:00:00 | 0 |
| 4 | 4 | 51 | 2015-06-29 12:00:00 | 1 |
| 5 | 4 | 51 | 2015-06-30 12:00:00 | 1 |
| 6 | 5 | 50 | 2015-07-05 12:00:00 | 1 |
| 7 | 5 | 50 | 2015-07-06 12:00:00 | 1 |
| 8 | 5 | 51 | 2015-07-08 12:00:00 | 1 |
+-----+----------+-------+-----------------------+---------+
Desired Result
+-----+----------+-------+-----------------------+---------+
| id | userid | pid | start_date | active |
+-----+----------+-------+-----------------------+---------+
| 2 | 4 | 50 | 2015-05-16 12:00:00 | 1 |
| 7 | 5 | 50 | 2015-07-06 12:00:00 | 1 |
+-----+----------+-------+-----------------------+---------+
I've tried a bunch of things and this is the closest I got but unfortunately it is not quit there.
SELECT *
FROM mytable t1
WHERE
(
SELECT COUNT(*)
FROM mytable t2
WHERE
t1.userid = t2.userid
AND t1.start_date < t2.start_date
) < 1
AND pid = 50
AND active = 1
ORDER BY start_date DESC
plan
get last record grouping by userid where pid is 50 and is active
inner join to mytable to get the record info associated with last
query
select
my.*
from
(
select userid, pid, active, max(start_date) as lst
from mytable
where pid = 50
and active = 1
group by userid, pid, active
) maxd
inner join mytable my
on maxd.userid = my.userid
and maxd.pid = my.pid
and maxd.active = my.active
and maxd.lst = my.start_date
;
output
+----+--------+-----+------------------------+--------+
| id | userid | pid | start_date | active |
+----+--------+-----+------------------------+--------+
| 2 | 4 | 50 | May, 16 2015 12:00:00 | 1 |
| 7 | 5 | 50 | July, 06 2015 12:00:00 | 1 |
+----+--------+-----+------------------------+--------+
sqlfiddle
notes
as suggested by #Strawberry, updated to join also on pid and active. this will avoid the possibility of a record which is not active or not pid 50 but has exact same date also being rendered.
I have this tables SQL Fiddle
items table:
+----+----------+
| id | name |
+----+----------+
| 1 | Facebook |
| 2 | Twitter |
| 3 | Amazon |
+----+----------+
prices table:
+----+-----------+---------+-----------------------------+
| id | buy | item_id | created_at |
+----+-----------+---------+-----------------------------+
| 1 | 43000 | 1 | June, 18 2014 17:31:04+0000 |
| 2 | 44000 | 1 | June, 19 2014 17:31:04+0000 |
| 3 | 30000 | 2 | June, 20 2014 17:31:04+0000 |
| 4 | 33000 | 2 | June, 21 2014 17:31:04+0000 |
| 5 | 20000 | 3 | June, 22 2014 17:31:04+0000 |
| 6 | 21000 | 3 | June, 23 2014 17:31:04+0000 |
+----+-----------+---------+-----------------------------+
I want to get last prices per item and one before last price's buy field based on a price date
Desired output:
+----+---------+-----------------+---------+
| id | buy | last_before_buy | item_id |
+----+---------+-----------------+---------+
| 10 | 45000 | 43000 | 3 |
| 7 | 33000 | 31000 | 2 |
| 4 | 23000 | 23000 | 1 |
+----+---------+-----------------+---------+
Here's another way to do it:
select a.id, a.buy, b.buy last_before_buy, a.item_id
from (select * from prices WHERE (created_at <= NOW() - INTERVAL 5 DAY) order by id desc) a
join (select * from prices order by id desc) b on a.item_id = b.item_id and a.id > b.id
group by a.item_id;
fiddle
You can do this with the substring_index()/group_concat() trick:
select max(id) as id,
substring_index(group_concat(buy order by created_at desc), ',', 1) as buy,
substring_index(substring_index(group_concat(buy order by created_at desc), ',', 2), ',', -1) as lastbuy,
item_id
from prices p
group by item_id;