Msql last year data only - mysql

With mysql i am trying to fetch amount which is donated last year only,
for exampe: this is my sample table. here i would like to fech amount which is donated in
last year only. here most recent donation is done on oct-14, so i want data from oct -14 to oct - 13 and not sep -13
+--------+-------+------+---------+
| nid | amount | time | month |
+--------+-------+------+---------+
| X | 30 | |sep-13 |
+--------+-------+------+---------+
| X | 30 | |sep-13 |
+--------+-------+------+---------+
| A | 10 | |oct-13 |
+--------+-------+------+---------+
| A | 10 | |oct-13 |
+--------+-------+------+---------+
| B | 20 | |oct-14 |
+--------+-------+------+---------+
| C | 20 | |oct-14 |
+--------+-------+------+---------+

If your 'month' column is a date column, you can use EXTRACT function.
SELECT * FROM _table_ WHERE EXTRACT(year from `month`) = 13;

SELECT * FROM TABLE WHERE RIGHT(MONTH, 2) = '13'
IF THE DATE LAST YEAR IS A STATIC VALUE AND WON'T NEED TO BE ADJUSTED FOR A YEAR TO YEAR PROJECT, WHETHER IT'S VARCHAR OR DATE THIS VALUE WILL BE TREATED AS A STRING AND THE RIGHT 2 CHARACTERS WILL REMAIN THE SAME. -Sorry about the caps.
If you're looking for a dynamic approach...
SELECT * FROM TABLE WHERE STR_TO_DATE(MONTH, '%Y') = YEAR(CURDATE() - 1)
OR something like the above. Again, this is dependent on the data type that defines that field. But this should do the trick.

If you need to return rows based on maximum year in table then one of these queries can help:
If month is DATE type:
SELECT *
FROM sample_table
WHERE YEAR(month) = YEAR((SELECT MAX(month) FROM sample_table))
If month is character type:
SELECT *
FROM sample_table
WHERE RIGHT(month,2) = (SELECT MAX(RIGHT(month,2)) FROM sample_table)

Related

SQL count row equal to column

Here is my table:
+--------+---------------------+
| roomNo | date |
+--------+---------------------+
| 1 | 2017-05-17 16:05:00 |
| 1 | 2017-05-17 15:05:00 |
| 2 | 2019-05-20 12:30:00 |
| 2 | 2019-05-15 10:30:00 |
| 2 | 2019-05-14 08:00:00 |
+--------+---------------------+
I want to get the day where the room is used at least once and which day(s) had the most operations in it and how many times, in the current year. I don't know how to compare the dates.
The expected result would be something like :
+--------+------------+------------+
| roomNo | date | operations |
+--------+------------+------------+
| 2 | 2019-05-20 | 3 |
+--------+------------+------------+
We can use MySQL DATE function to lop off times from DATETIME and TIMESTAMP columns. Or we could use MySQL DATE_FORMAT function, to return just year, month day.
We can use an aggregate function like COUNT or SUM in a query with GROUP BY to get counts by room and day.
If "current year" means from Jan 1 thru Dec 31, we can use expression to derive date values of '2019-01-01' and '2020-01-01', and do a comparison of the date column to those values in the WHERE clause.
As a start, consider this:
SELECT t.roomno
, DATE(t.date) AS date_
, COUNT(*) AS cnt_
FROM mytable t
WHERE t.date >= DATE_FORMAT(NOW(),'%Y-01-01') + INTERVAL 0 YEAR
AND t.date < DATE_FORMAT(NOW(),'%Y-01-01') + INTERVAL 1 YEAR
GROUP
BY t.roomno
, DATE(t.date)
ORDER
BY t.roomno
, cnt_ DESC
If the goal is to just return one of the rooms that has the highest number of uses, we could use a LIMIT clause, and order by the highest count to lowest,
ORDER
BY cnt_ DESC
, t.roomno
LIMIT 1
If the results are more complex than that, we can omit the LIMIT clause, and use the result from that query as an inline view in an outer query.
With MySQL 8.0, we can use common table expression (CTE) and window/analytic functions, to get more elaborate results.

MYSQL - Finding row delta using date rows that include holidays

I have a table that includes a field with dates (call it date) and a field with a cumulative running total (call it X) | call it table SAMPLE.
***My data in field DATE does not include weekends and holidays.
I can find the delta in the numbers from day to day by simply subtracting any chosen value in "X" and subtracting that from the row above.
Here's my current query:
select
date,
a.X - b.X as 'Daily Total'
from SAMPLE as a
left join SAMPLE as b
on b.date = if(weekday(a.date) = 0 , a.date - interval 3 day, a.date- interval 1 day);
The problem is that the above values work until I hit dates with holidays. If Monday is a holiday, then the values return null because a.date - interval 1 day will not exist. What's the best way to go about solving the holidays issue?
the below are the current results:
+------------+---------------+
| date | X |
+------------+---------------+
| 2018-03-26 | -40105.00 |
| 2018-03-27 | 28470.00 |
| 2018-03-28 | 5265.00 |
| 2018-03-29 | -23010.00 |
| 2018-04-02 | NULL |
| 2018-04-03 | -24830.00 |
| 2018-04-04 | -21970.00 |
| 2018-04-05 | -9620.00 |
| 2018-04-06 | 36465.00 |
Thanks in advance!!
I will sort the table by date then assign a sequence or series of numbers from 1 to n. I will then subtract the value of current row from the previous row except the first row. For first row, i will copy the value X.
select rnk2.`date`,
case when rnk1.r1=1 and rnk2.r2=1 then rnk1.X else rnk2.X-rnk1.X end as 'Daily Total'
from (
select `date`,X,#r1:=#r1+1 as r1
from samples, (select #r1:=0) a
order by `date` ) rnk1
inner join
(select `date`,X,#r2:=#r2+1 as r2
from samples, (select #r2:=0) b
order by `date`) rnk2
on (rnk1.r1=1 and rnk2.r2=1) or (rnk1.r1+1=rnk2.r2)
order by rnk2.`date`

Obtain a list of month and records from the current date(month) in two different years

I have such a set of records in a table in a mysql database;
Date | Number_of_leaves
10th-December-2015 | 10 leaves
6th-August-2015 | 10 leaves
15th-September-2015 | 14 leaves
15th-January-2016: | 100 leaves
7th-November-2015: | 4 leaves
9th-October -2015: | 200 leaves
How can i return a list months and their records for just the past 4 months from Jan-2016 backwards? In other words, i need a result for the past 4 months including the current one like this:
January 2016 | 100 leaves
December 2015 | 10 leaves
November 2015 | 4 leaves
October 2015 | 200 leaves
The above is the kind of result which shows the month and the corresponding year with the number of leaves collected in that month and corresponding year
Schema
create table xyz
( id int auto_increment primary key,
theDate date not null,
leaves int not null
);
-- truncate table xyz;
insert xyz(theDate,leaves) values
('2016-04-10',444510),
('2016-02-10',55510),
('2015-12-10',10),
('2015-08-06',10),
('2015-09-15',14),
('2016-01-15',100),
('2015-11-07',4),
('2015-10-09',200);
Query 1
select month(theDate) as m,
year(theDate) as y,
sum(leaves) as leaves
from xyz
where theDate<='2016-02-01'
group by month(theDate),year(theDate)
order by theDate desc
limit 4;
or
Query 2
select concat(monthname(theDate),' ',year(theDate)) as 'Month/Year',
sum(leaves) as leaves
from xyz
where theDate<='2016-02-01'
group by month(theDate),year(theDate)
order by theDate desc
limit 4;
+---------------+--------+
| Month/Year | leaves |
+---------------+--------+
| January 2016 | 100 |
| December 2015 | 10 |
| November 2015 | 4 |
| October 2015 | 200 |
+---------------+--------+
op is your table name first use str_to_date for convert string to date format .we use if because your date format is different
select * FROM (
SELECT *,
IFNULL(
IFNULL(
str_to_date(Date,'%D-%b-%Y'),str_to_date(Date,'%d-%M-%Y')) ,
str_to_date(Date,'%D-%M-%Y')
)
f_date
FROM `op`
order by number_of_leaves DESC,f_date ASC
) tab
group by month(tab.f_date) LIMIT 5

Fetch Unit consumption date-wise

I am struggling in to get result from mysql in the following way. I have 10 records in mysql db table having date and unit fields. I need to get used units on every date.
Table structure as follows, adding today unit with past previous unit in every record:
Date Units
---------- ---------
10/10/2012 101
11/10/2012 111
12/10/2012 121
13/10/2012 140
14/10/2012 150
15/10/2012 155
16/10/2012 170
17/10/2012 180
18/10/2012 185
19/10/2012 200
Desired output will be :
Date Units
---------- ---------
10/10/2012 101
11/10/2012 10
12/10/2012 10
13/10/2012 19
14/10/2012 10
15/10/2012 5
16/10/2012 15
17/10/2012 10
18/10/2012 5
19/10/2012 15
Any help will be appreciated. Thanks
There's a couple of ways to get the resultset. If you can live with an extra column in the resultset, and the order of the columns, then something like this is a workable approach.
using user variables
SELECT d.Date
, IF(#prev_units IS NULL
,#diff := 0
,#diff := d.units - #prev_units
) AS `Units_used`
, #prev_units := d.units AS `Units`
FROM ( SELECT #prev_units := NULL ) i
JOIN (
SELECT t.Date, t.Units
FROM mytable t
ORDER BY t.Date, t.Units
) d
This returns the specified resultset, but it includes the Units column as well. It's possible to have that column filtered out, but it's more expensive, because of the way MySQL processes an inline view (MySQL calls it a "derived table")
To remove that extra column, you can wrap that in another query...
SELECT f.Date
, f.Units_used
FROM (
query from above goes here
) f
ORDER BY f.Date
but again, removing that column comes with the extra cost of materializing that result set a second time.
using a semi-join
If you are guaranteed to have a single row for each Date value, either stored as a DATE, or as a DATETIME with the timecomponent set to a constant, such as midnight, and no gaps in the Date value, and Date is defined as DATE or DATETIME datatype, then another query that will return the specifid result set:
SELECT t.Date
, t.Units - s.Units AS Units_Used
FROM mytable t
LEFT
JOIN mytable s
ON s.Date = t.Date + INTERVAL -1 DAY
ORDER BY t.Date
If there's a missing Date value (a gap) such that there is no matching previous row, then Units_used will have a NULL value.
using a correlated subquery
If you don't have a guarantee of no "missing dates", but you have a guarantee that there is no more than one row for a particular Date, then another approach (usually more expensive in terms of performance) is to use a correlated subquery:
SELECT t.Date
, ( t.Units - (SELECT s.Units
FROM mytable s
WHERE s.Date < t.Date
ORDER BY s.Date DESC
LIMIT 1)
) AS Units_used
FROM mytable t
ORDER BY t.Date, t.Units
spencer7593's solution will be faster, but you can also do something like this...
SELECT * FROM rolling;
+----+-------+
| id | units |
+----+-------+
| 1 | 101 |
| 2 | 111 |
| 3 | 121 |
| 4 | 140 |
| 5 | 150 |
| 6 | 155 |
| 7 | 170 |
| 8 | 180 |
| 9 | 185 |
| 10 | 200 |
+----+-------+
SELECT a.id,COALESCE(a.units - b.units,a.units) units
FROM
( SELECT x.*
, COUNT(*) rank
FROM rolling x
JOIN rolling y
ON y.id <= x.id
GROUP
BY x.id
) a
LEFT
JOIN
( SELECT x.*
, COUNT(*) rank
FROM rolling x
JOIN rolling y
ON y.id <= x.id
GROUP
BY x.id
) b
ON b.rank= a.rank -1;
+----+-------+
| id | units |
+----+-------+
| 1 | 101 |
| 2 | 10 |
| 3 | 10 |
| 4 | 19 |
| 5 | 10 |
| 6 | 5 |
| 7 | 15 |
| 8 | 10 |
| 9 | 5 |
| 10 | 15 |
+----+-------+
This should give the desired result. I don't know how your table is called so I named it "tbltest".
Naming a table date is generally a bad idea as it also refers to other things (functions, data types,...) so I renamed it "fdate". Using uppercase characters in field names or tablenames is also a bad idea as it makes your statements less database independent (some databases are case sensitive and some are not).
SELECT
A.fdate,
A.units - coalesce(B.units, 0) AS units
FROM
tbltest A left join tbltest B ON A.fdate = B.fdate + INTERVAL 1 DAY

Counting appointments for each day using MYSQL

I'm in trouble with a mysql statement counting appointments for one day within a given time period. I've got a calendar table including starting and finishing column (type = DateTime). The following statement should count all appointments for November including overall appointments:
SELECT
COUNT('APPOINTMENTS') AS Count,
DATE(c.StartingDate) AS Datum
FROM t_calendar c
WHERE
c.GUID = 'blalblabla' AND
((DATE(c.StartingDate) <= DATE('2012-11-01 00:00:00')) AND (DATE(c.EndingDate) >= DATE('2012-11-30 23:59:59'))) OR
((DATE(c.StartingDate) >= DATE('2012-11-01 00:00:00')) AND (DATE(c.EndingDate) <= DATE('2012-11-30 23:59:59')))
GROUP BY DATE(c.StartingDate)
HAVING Count > 1
But how to include appointments that starts before a StartingDate and ends on the StartingDate?
e.g.
StartingDate = 2012-11-14 17:00:00, EndingDate = 2012-11-15 08:00:00
StartingDate = 2012-11-15 09:00:00, EndingDate = 2012-11-15 10:00:00
StartingDate = 2012-11-15 11:00:00, EndingDate = 2012-11-15 12:00:00
My statement returns a count of 2 for 15th of November. But that's wrong because the first appointment is missing. How to include these appointments? What I am missing, UNION SELECT, JOIN, sub selection?
A possible solution?
SELECT
c1.GUID, COUNT('APPOINTMENTS') + COUNT(DISTINCT c2.ANYFIELD) AS Count,
DATE(c1.StartingDate) AS Datum,
COUNT(DISTINCT c2.ANYFIELD)
FROM
t_calendar c1
LEFT JOIN
t_calendar c2
ON
c2.ResourceGUID = c1.ResourceGUID AND
(DATE(c2.EndingDate) = DATE(c1.StartingDate)) AND
(DATE(c2.StartingDate) < DATE(c1.StartingDate))
WHERE
((DATE(c1.StartingDate) <= DATE('2012-11-01 00:00:00')) AND (DATE(c1.EndingDate) >= DATE('2012-11-30 23:59:59'))) OR
((DATE(c1.StartingDate) >= DATE('2012-11-01 00:00:00')) AND (DATE(c1.EndingDate) <= DATE('2012-11-30 23:59:59')))
GROUP BY
c1.ResourceGUID,
DATE(c1.StartingDate)
First: Consolidate range checking
First of all your two range where conditions can be replaced by a single one. And it also seems that you're only counting appointments that either completely overlap target date range or are completely contained within. Partially overlapping ones aren't included. Hence your question about appointments that end right on the range starting date.
To make where clause easily understandable I'll simplify it by using:
two variables to define target range:
rangeStart (in your case 1st Nov 2012)
rangeEnd (I'll rather assume to 1st Dec 2012 00:00:00.00000)
won't be converting datetime to dates only (using date function) the way that you did, but you can easily do that.
With these in mind your where clause can be greatly simplified and covers all appointments for given range:
...
where (c.StartingDate < rangeEnd) and (c.EndingDate >= rangeStart)
...
This will search for all appointments that fall in target range and will cover all these appointment cases:
start end
target range |==============|
partial front |---------|
partial back |---------|
total overlap |---------------------|
total containment |-----|
Partial front/back may also barely touch your target range (what you've been after).
Second: Resolving the problem
Why you're missing the first record? Simply because of your having clause that only collects those groups that have more than 1 appointment starting on a given day: 15th Nov has two, but 14th has only one and is therefore excluded because Count = 1 and is not > 1.
To answer your second question what am I missing is: you're not missing anything, actually you have too much in your statement and needs to simplified.
Try this statement instead that should return exactly what you're after:
select count(c.GUID) as Count,
date(c.StartingDate) as Datum
from t_calendar c
where (c.GUID = 'blabla') and
(c.StartingDate < str_to_date('2012-12-01', '%Y-%m-%d') and
(c.EndingDate >= str_to_date('2012-11-01', '%Y-%m-%d'))
group by date(c.StartingDate)
I used str_to_date function to make string to date conversion more safe.
I'm not really sure why you included having in your statement, because it's not really needed. Unless your actual statement is more complex and you only included part that's most relevant. In that case you'll likely have to change it to:
having Count > 0
Getting appointment count per day in any given date range
There are likely other ways as well but the most common way would be using a numbers or ?calendar* table that gives you the ability to break a range into individual points - days. They you have to join your appointments to this numbers table and provide results.
I've created a SQLFiddle that does the trick. Here's what it does...
Suppose you have numbers table Num with numbers from 0 to x. And appointments table Cal with your records. Following script created these two tables and populates some data. Numbers are only up to 100 which is enough for 3 months worth of data.
-- appointments
create table Cal (
Id int not null auto_increment primary key,
StartDate datetime not null,
EndDate datetime not null
);
-- create appointments
insert Cal (StartDate, EndDate)
values
('2012-10-15 08:00:00', '2012-10-20 16:00:00'),
('2012-10-25 08:00:00', '2012-11-01 03:00:00'),
('2012-11-01 12:00:00', '2012-11-01 15:00:00'),
('2012-11-15 10:00:00', '2012-11-16 10:00:00'),
('2012-11-20 08:00:00', '2012-11-30 08:00:00'),
('2012-11-30 22:00:00', '2012-12-05 00:00:00'),
('2012-12-01 05:00:00', '2012-12-10 12:00:00');
-- numbers table
create table Nums (
Id int not null primary key
);
-- add 100 numbers
insert into Nums
select a.a + (10 * b.a)
from (select 0 as a union all
select 1 union all
select 2 union all
select 3 union all
select 4 union all
select 5 union all
select 6 union all
select 7 union all
select 8 union all
select 9) as a,
(select 0 as a union all
select 1 union all
select 2 union all
select 3 union all
select 4 union all
select 5 union all
select 6 union all
select 7 union all
select 8 union all
select 9) as b
Now what you have to do now is
Select a range of days which you do by selecting numbers from Num table and convert them to dates.
Then join your appointments to those dates so that those appointments that fall on particular day are joined to that particular day
Then just group all these appointments per each day and get results
Here's the code that does this:
-- just in case so comparisons don't trip over
set names 'latin1' collate latin1_general_ci;
-- start and end target date range
set #s := str_to_date('2012-11-01', '%Y-%m-%d');
set #e := str_to_date('2012-12-01', '%Y-%m-%d');
-- get appointment count per day within target range of days
select adddate(#s, n.Id) as Day, count(c.Id) as Appointments
from Nums n
left join Cal c
on ((date(c.StartDate) <= adddate(#s, n.Id)) and (date(c.EndDate) >= adddate(#s, n.Id)))
where adddate(#s, n.Id) < #e
group by Day;
And this is the result of this rather simple select statement:
| DAY | APPOINTMENTS |
-----------------------------
| 2012-11-01 | 2 |
| 2012-11-02 | 0 |
| 2012-11-03 | 0 |
| 2012-11-04 | 0 |
| 2012-11-05 | 0 |
| 2012-11-06 | 0 |
| 2012-11-07 | 0 |
| 2012-11-08 | 0 |
| 2012-11-09 | 0 |
| 2012-11-10 | 0 |
| 2012-11-11 | 0 |
| 2012-11-12 | 0 |
| 2012-11-13 | 0 |
| 2012-11-14 | 0 |
| 2012-11-15 | 1 |
| 2012-11-16 | 1 |
| 2012-11-17 | 0 |
| 2012-11-18 | 0 |
| 2012-11-19 | 0 |
| 2012-11-20 | 1 |
| 2012-11-21 | 1 |
| 2012-11-22 | 1 |
| 2012-11-23 | 1 |
| 2012-11-24 | 1 |
| 2012-11-25 | 1 |
| 2012-11-26 | 1 |
| 2012-11-27 | 1 |
| 2012-11-28 | 1 |
| 2012-11-29 | 1 |
| 2012-11-30 | 2 |