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
Related
I have the following data in mysql:
date,value,id
2016-01-01,0,1
2016-01-04,1,2
2016-01-10,2,3
2016-01-25,1,4
2016-01-26,10,5
I have another table with just dates and ids that I want to insert in the first table using the following rule: set the value as the value at the last date before the date of the given id. That is, if I have to introduce the following elements:
date,id
2016-01-02,6
2016-01-03,7
2016-01-11,8
2016-01-28,9
2016-01-28,10
I want the final table to be:
date,value,id
2016-01-01,0,1
2016-01-04,1,2
2016-01-10,2,3
2016-01-25,1,4
2016-01-26,10,5
2016-01-02,0,6
2016-01-03,0,7
2016-01-11,28
2016-01-28,10,9
2016-01-28,10,10
Can you help me please?
You can get the "previous" value using a correlated subquery:
select x.date, x.id,
(select t.value
from t
where t.date <= x.date
order by t.date desc
limit 1
) as value
from (select '2016-01-02' as date, 6 as id union all
select '2016-01-03' as date, 7 as id union all
select '2016-01-11' as date, 8 as id union all
select '2016-01-28' as date, 9 as id union all
select '2016-01-28' as date, 10 as id
) x;
If you want to insert these into the table, simply put an insert clause before the select. If you want a result set with all rows, then use union all with the other table.
I have the following data inside a table:
id person_id item_id price
1 1 1 10
2 1 1 20
3 1 3 50
Now what I want to do is group by the item ID, select the id that has the highest value and take the price.
E.g. the sum would be: (20 + 50) and ignore the 10.
I am using the following:
SELECT SUM(`price`)
FROM
(SELECT id, person_id, item_id, price
FROM `table` tbl
INNER JOIN person p USING (person_id)
WHERE p.person_id = 1
ORDER BY id DESC) x
GROUP BY item_id
However, this query is still adding (10 + 20 + 50), which is obviously not what I need to have.
Any ideas to where I am going wrong?
Here is what you are trying to achieve. First you need grouping in a subquery and not in outer query. In outer query you need only sum:
SELECT SUM(`price`)
FROM
(SELECT MAX(price) as price
FROM `table` tbl
INNER JOIN person p USING (person_id)
WHERE p.person_id = 1
GROUP BY item_id) x
http://sqlfiddle.com/#!9/40803/5
SELECT SUM(t1.price)
FROM tbl t1
LEFT JOIN tbl t2
ON t1.person_id= t2.person_id
AND t1.item_id = t2.item_id
AND t1.id<t2.id
WHERE t1.person_id = 1
AND t2.id IS NULL;
I'm not sure if this is the only requirement you have. If so, try this.
SELECT SUM(price)
FROM
(SELECT MAX(price)
FROM table
WHERE person_id = 1
GROUP BY item_id)
First of all - you don't need the person table, because the other table already contains the person_id. So i removed it from the examples.
Your query returns a sum of prices for each item.
If you replace SELECT SUM(price) with SELECT item_id, SUM(price) you wil get
item_id SUM(`price`)
1 30
3 50
But that is not what you want. Neither is it what you wrote in the question " (10 + 20 + 50)".
Now replacing the first line with SELECT id, item_id, SUM(price) you will get one row for each item with the highest id.
id item_id price
2 1 20
3 3 50
This works because of the "undocumented feature" of MySQL, wich allows you to select columns that are not listed in the GROUP BY clause and get the first row from the subselect each group (each item in this case).
Now you only need to sum the price column in an additional outer select
SELECT SUM(price)
FROM (
SELECT id, item_id ,price
FROM (
SELECT id, person_id, item_id, price
FROM `table` tbl
WHERE tbl.person_id = 1
ORDER BY id DESC ) x
GROUP BY item_id
) y
However i do not recomend to use that "feature". While it still works on MySQL 5.6, you never know if that will work with newer versions. It already doesn't work on MariaDB.
Instead you can determite the MAX(id) for each item in an subselect, select only the rows with the determined ids and get the summed price of them.
SELECT SUM(`price`)
FROM `table` tbl
WHERE tbl.id IN (
SELECT MAX(tbl2.id)
FROM `table` tbl2
WHERE tbl2.person_id = 1
GROUP BY tbl2.item_id
)
Another solution (wich internaly does the same) is
SELECT SUM(`price`)
FROM `table` tbl
JOIN (
SELECT MAX(tbl2.id) as id
FROM `table` tbl2
WHERE tbl2.person_id = 1
GROUP BY tbl2.item_id
) x ON x.id = tbl.id
Alex's solution also works fine, if the groups (number of rows per person and item) are rather small.
You have used group by in main query, but it is on subquery like
SELECT id, person_id, item_id, SUM(`price`) FROM ( SELECT MAX(price) FROM `table` tbl WHERE p.person_id = 1 GROUP BY item_id ) AS x
Let's say I have two columns: id and date.
I want to give it an id and it'll find all the duplicates of the value date of the column id.
Example:
id |date
1 |2013-09-16
2 |2013-09-16
3 |2013-09-23
4 |2013-09-23
I want to give it id 1 (without giving anything about date) and it'll give me a table of 2 columns listing the duplicates of id 1's date
Thanks in advance!
select * from your_table
where `date` in
(
select `date`
from your_table
where id = 1
)
or if you like to use a join
select t.*
from your_table t
inner join
(
select `date`
from your_table
where id = 1
) x on x.date = t.date
I have a table :
ID | time
1 | 300
1 | 100
1 | 200
2 | 200
2 | 500
I want to get 2nd row for every ID
I know that I can get 1st row as
select ID,time from T group by ID;
But I don't know about how to get 2nd row for every ID.
I know about limit and offset clause in mysql, but can't figure out how to use them here.
How can I do it ?
EDIT : Actually, time is not ordered. I forgot to specify that. I have made an edit in the table.
i have just an idee how to make it but i couldnt fix it , maybe you can fix it. any suggest is appreciated to correct my query
first this to select the first row of each id.
SELECT min(id) id
FROM TableName t2
group by id
then select the min(id) which are not in the first query to select to min(id) (which is second row)
like that
SELECT min(id) id ,time
FROM TableName
WHERE id NOT IN (
SELECT min(id) id
FROM TableName
GROUP BY id
)
GROUP BY id
** as i said its just suggest . it returns me 0 values.if u fix it let me edit my post to be helpful
here a demo
SELECT ID, MAX(time) time
FROM
(
select ID, Time
from TableName a
where
(
select count(*)
from TableName as f
where f.ID = a.ID and f.time <= a.time
) <= 2
) s
GROUP BY ID
SQLFiddle Demo
SELECT x.*
FROM test x
JOIN test y
ON y.id = x.id
AND y.time >= x.time
GROUP
BY id,time
HAVING COUNT(*) = n;
Note that any entries with less than n results will be omitted
You cannot do this with the tables that you have. You could make a valiant attempt with:
select id, time
from (select id, time
from t
group by t
) t
where not exists (select 1 from t t2 where t2.id = t.id and t2.time = t.time)
group by id
That is, attempt to filter out the first row.
The reason this is not possible is because tables are inherently unordered, so there is not real definition of "second" in your tables. This gives the SQL engine the opportunity to rearrange the rows as it sees fit during processing -- which can result in great performance gains.
Even the construct that you are using:
select id, time
from t
group by id
is not guaranteed to return time from the first row. This is a (mis)feature of MySQL called Hidden Columns. It is really only intended for the case where all the values are the same. I will admit that in practice it seems to get the value from the first row, but you cannot guarantee that.
Probably your best solution is to select the data into a new table that has an auto-incrementing column:
create table newtable (
autoid int auto_increment,
id int,
time int
);
insert into newtable(id, time)
select id, time from t;
In practice, this will probably keep the same order as the original table, and you can then use the autoid to get the second row. I want to emphasize, though, the "in practice". There is no guarantee that the values are in the correct order, but they probably will be.
I have a table tbl_patient and I want to fetch last 2 visit of each patient in order to compare whether patient condition is improving or degrading.
tbl_patient
id | patient_ID | visit_ID | patient_result
1 | 1 | 1 | 5
2 | 2 | 1 | 6
3 | 2 | 3 | 7
4 | 1 | 2 | 3
5 | 2 | 3 | 2
6 | 1 | 3 | 9
I tried the query below to fetch the last visit of each patient as,
SELECT MAX(id), patient_result FROM `tbl_patient` GROUP BY `patient_ID`
Now i want to fetch the 2nd last visit of each patient with query but it give me error
(#1242 - Subquery returns more than 1 row)
SELECT id, patient_result FROM `tbl_patient` WHERE id <(SELECT MAX(id) FROM `tbl_patient` GROUP BY `patient_ID`) GROUP BY `patient_ID`
Where I'm wrong
select p1.patient_id, p2.maxid id1, max(p1.id) id2
from tbl_patient p1
join (select patient_id, max(id) maxid
from tbl_patient
group by patient_id) p2
on p1.patient_id = p2.patient_id and p1.id < p2.maxid
group by p1.patient_id
id11 is the ID of the last visit, id2 is the ID of the 2nd to last visit.
Your first query doesn't get the last visits, since it gives results 5 and 6 instead of 2 and 9.
You can try this query:
SELECT patient_ID,visit_ID,patient_result
FROM tbl_patient
where id in (
select max(id)
from tbl_patient
GROUP BY patient_ID)
union
SELECT patient_ID,visit_ID,patient_result
FROM tbl_patient
where id in (
select max(id)
from tbl_patient
where id not in (
select max(id)
from tbl_patient
GROUP BY patient_ID)
GROUP BY patient_ID)
order by 1,2
SELECT id, patient_result FROM `tbl_patient` t1
JOIN (SELECT MAX(id) as max, patient_ID FROM `tbl_patient` GROUP BY `patient_ID`) t2
ON t1.patient_ID = t2.patient_ID
WHERE id <max GROUP BY t1.`patient_ID`
There are a couple of approaches to getting the specified resultset returned in a single SQL statement.
Unfortunately, most of those approaches yield rather unwieldy statements.
The more elegant looking statements tend to come with poor (or unbearable) performance when dealing with large sets. And the statements that tend to have better performance are more un-elegant looking.
Three of the most common approaches make use of:
correlated subquery
inequality join (nearly a Cartesian product)
two passes over the data
Here's an approach that uses two passes over the data, using MySQL user variables, which basically emulates the analytic RANK() OVER(PARTITION ...) function available in other DBMS:
SELECT t.id
, t.patient_id
, t.visit_id
, t.patient_result
FROM (
SELECT p.id
, p.patient_id
, p.visit_id
, p.patient_result
, #rn := if(#prev_patient_id = patient_id, #rn + 1, 1) AS rn
, #prev_patient_id := patient_id AS prev_patient_id
FROM tbl_patients p
JOIN (SELECT #rn := 0, #prev_patient_id := NULL) i
ORDER BY p.patient_id DESC, p.id DESC
) t
WHERE t.rn <= 2
Note that this involves an inline view, which means there's going to be a pass over all the data in the table to create a "derived tabled". Then, the outer query will run against the derived table. So, this is essentially two passes over the data.
This query can be tweaked a bit to improve performance, by eliminating the duplicated value of the patient_id column returned by the inline view. But I show it as above, so we can better understand what is happening.
This approach can be rather expensive on large sets, but is generally MUCH more efficient than some of the other approaches.
Note also that this query will return a row for a patient_id if there is only one id value exists for that patient; it does not restrict the return to just those patients that have at least two rows.
It's also possible to get an equivalent resultset with a correlated subquery:
SELECT t.id
, t.patient_id
, t.visit_id
, t.patient_result
FROM tbl_patients t
WHERE ( SELECT COUNT(1) AS cnt
FROM tbl_patients p
WHERE p.patient_id = t.patient_id
AND p.id >= t.id
) <= 2
ORDER BY t.patient_id ASC, t.id ASC
Note that this is making use of a "dependent subquery", which basically means that for each row returned from t, MySQL is effectively running another query against the database. So, this will tend to be very expensive (in terms of elapsed time) on large sets.
As another approach, if there are relatively few id values for each patient, you might be able to get by with an inequality join:
SELECT t.id
, t.patient_id
, t.visit_id
, t.patient_result
FROM tbl_patients t
LEFT
JOIN tbl_patients p
ON p.patient_id = t.patient_id
AND t.id < p.id
GROUP
BY t.id
, t.patient_id
, t.visit_id
, t.patient_result
HAVING COUNT(1) <= 2
Note that this will create a nearly Cartesian product for each patient. For a limited number of id values for each patient, this won't be too bad. But if a patient has hundreds of id values, the intermediate result can be huge, on the order of (O)n**2.
Try this..
SELECT id, patient_result FROM tbl_patient AS tp WHERE id < ((SELECT MAX(id) FROM tbl_patient AS tp_max WHERE tp_max.patient_ID = tp.patient_ID) - 1) GROUP BY patient_ID
Why not use simply...
GROUP BY `patient_ID` DESC LIMIT 2
... and do the rest in the next step?