i have a table like this:
name date time
tom | 2011-07-04 | 01:09:52
tom | 2011-07-04 | 01:09:52
mad | 2011-07-04 | 02:10:53
mad | 2009-06-03 | 00:01:01
i want oldest name first:
SELECT *
ORDER BY date ASC, time ASC
GROUP BY name
(->doesn't work!)
now it should give me first mad(has earlier date) then tom
but with GROUP BY name ORDER BY date ASC, time ASC gives me the newer mad first because it groups before it sorts!
again: the problem is that i can't sort by date and time before i group because GROUP BY must be before ORDER BY!
Another method:
SELECT *
FROM (
SELECT * FROM table_name
ORDER BY date ASC, time ASC
) AS sub
GROUP BY name
GROUP BY groups on the first matching result it hits. If that first matching hit happens to be the one you want then everything should work as expected.
I prefer this method as the subquery makes logical sense rather than peppering it with other conditions.
As I am not allowed to comment on user1908688's answer, here a hint for MariaDB users:
SELECT *
FROM (
SELECT *
ORDER BY date ASC, time ASC
LIMIT 18446744073709551615
) AS sub
GROUP BY sub.name
https://mariadb.com/kb/en/mariadb/why-is-order-by-in-a-from-subquery-ignored/
I think this is what you are seeking :
SELECT name, min(date)
FROM myTable
GROUP BY name
ORDER BY min(date)
For the time, you have to make a mysql date via STR_TO_DATE :
STR_TO_DATE(date + ' ' + time, '%Y-%m-%d %h:%i:%s')
So :
SELECT name, min(STR_TO_DATE(date + ' ' + time, '%Y-%m-%d %h:%i:%s'))
FROM myTable
GROUP BY name
ORDER BY min(STR_TO_DATE(date + ' ' + time, '%Y-%m-%d %h:%i:%s'))
This worked for me:
SELECT *
FROM your_table
WHERE id IN (
SELECT MAX(id)
FROM your_table
GROUP BY name
);
Use a subselect:
select name, date, time
from mytable main
where date + time = (select min(date + time) from mytable where name = main.mytable)
order by date + time;
If you wont sort by max date and group by name, you can do this query:
SELECT name,MAX(date) FROM table group by name ORDER BY name
where date may by some date or date time string. It`s response to you max value of date by each one name
Another way to solve this would be with a LEFT JOIN, which could be more efficient. I'll first start with an example that considers only the date field, as probably it is more common to store date + time in one datetime column, and I also want to keep the query simple so it's easier to understand.
So, with this particular example, if you want to show the oldest record based on the date column, and assuming that your table name is called people you can use the following query:
SELECT p.* FROM people p
LEFT JOIN people p2 ON p.name = p2.name AND p.date > p2.date
WHERE p2.date is NULL
GROUP BY p.name
What the LEFT JOIN does, is when the p.date column is at its minimum value, there will be no p2.date with a smaller value on the left join and therefore the corresponding p2.date will be NULL. So, by adding WHERE p2.date is NULL, we make sure to show only the records with the oldest date.
And similarly, if you want to show the newest record instead, you can just change the comparison operator in the LEFT JOIN:
SELECT p.* FROM people p
LEFT JOIN people p2 ON p.name = p2.name AND p.date < p2.date
WHERE p2.date is NULL
GROUP BY p.name
Now, for this particular example where date+time are separate columns, you would need to add them in some way if you want to query based on the datetime of two columns combined, for example:
SELECT p.* FROM people p
LEFT JOIN people p2 ON p.name = p2.name AND p.date + INTERVAL TIME_TO_SEC(p.time) SECOND > p2.date + INTERVAL TIME_TO_SEC(p2.time) SECOND
WHERE p2.date is NULL
GROUP BY p.name
You can read more about this (and also see some other ways to accomplish this) on the The Rows Holding the Group-wise Maximum of a Certain Column page.
I had a different variation on this question where I only had a single DATETIME field and needed a limit after a group by or distinct after sorting descending based on the datetime field, but this is what helped me:
select distinct (column) from
(select column from database.table
order by date_column DESC) as hist limit 10
In this instance with the split fields, if you can sort on a concat, then you might be able to get away with something like:
select name,date,time from
(select name from table order by concat(date,' ',time) ASC)
as sorted
Then if you wanted to limit you would simply add your limit statement to the end:
select name,date,time from
(select name from table order by concat(date,' ',time) ASC)
as sorted limit 10
In Oracle, This work for me
SELECT name, min(date), min(time)
FROM table_name
GROUP BY name
work for me mysql
select * from (SELECT number,max(date_added) as datea FROM sms_chat group by number) as sup order by datea desc
This is not the exact answer, but this might be helpful for the people looking to solve some problem with the approach of ordering row before group by in mysql.
I came to this thread, when I wanted to find the latest row(which is order by date desc but get the only one result for a particular column type, which is group by column name).
One other approach to solve such problem is to make use of aggregation.
So, we can let the query run as usual, which sorted asc and introduce new field as max(doc) as latest_doc, which will give the latest date, with grouped by the same column.
Suppose, you want to find the data of a particular column now and max aggregation cannot be done.
In general, to finding the data of a particular column, you can make use of GROUP_CONCAT aggregator, with some unique separator which can't be present in that column, like GROUP_CONCAT(string SEPARATOR ' ') as new_column, and while you're accessing it, you can split/explode the new_column field.
Again, this might not sound to everyone. I did it, and liked it as well because I had written few functions and I couldn't run subqueries. I am working on codeigniter framework for php.
Not sure of the complexity as well, may be someone can put some light on that.
Regards :)
Related
I'm trying to do something I thought was relatively simple. Get the last entered value of a user.
I have a table which tracks all their entries called plan_activities_logs. I used this statement to get all the activity regarding a single user:
SELECT created_at as last_active, plan_id, plan_value
from plan_activity_logs where plan_id IN (select id from plans where tile_id = 30);
and it gives me back a table that looks like this:
but when I try to do something like this:
SELECT MAX(created_at) as last_active, plan_id, plan_value from plan_activity_logs where plan_id IN (select id from plans where tile_id = 30);
I get this:
while the date and id are correct the plan value is the wrong value. Any idea what I'm doing wrong?
Any idea what I'm doing wrong?
you are not doing anything wrong. According to the manual
If ONLY_FULL_GROUP_BY is disabled, a MySQL extension to the standard SQL use of GROUP BY permits the select list, HAVING
condition, or ORDER BY list to refer to nonaggregated columns even if
the columns are not functionally dependent on GROUP BY columns. This
causes MySQL to accept the preceding query. In this case, the server
is free to choose any value from each group, so unless they are the
same, the values chosen are nondeterministic, which is probably not
what you want.
In other words, this behavior is in accordance with the specification. The query returns a random non-deterministic value because you are using MySql nonstandard extension to the group by query.
This should work
SELECT created_at as last_active, plan_id, plan_value from plan_activity_logs where plan_id IN (select id from plans where tile_id = 30) order by created_at desc limit 1;
Try this query:
select #rn := 1;
select created_at as last_active,
plan_id,
plan_value
from (
select created_at as last_active,
plan_id,
plan_value,
#rn := #rn + 1 rn
from plan_activity_logs
where plan_id IN (select id from plans where tile_id = 30)
order by created_at desc
) a where rn = 1;
Hello there I have the following doubt I want to count how many times in a month I enter data.
My database is:
Date:
10/2010
10/2010
09/2010
08/2010
I have the following query.
SELECT DISTINCT (date)
FROM employee
WHERE date
IN (SELECT date
FROM employee
GROUP BY date
HAVING count( date ) >0)
ORDER BY date DESC;
This query gives me:
Date:
10/2017
8/2017
9/2017
But I want you to give me something like that.
Count | Date
2 | 10/2017
1 | 9/2017
1 | 10/2017
I hope I have explained my regards.
You're overcomplicating it; no subquery, or DISTINCT, needed.
SELECT `date`, count(*)
FROM `employee`
GROUP BY `date`
HAVING count(*) > 0
ORDER BY `date` DESC;
I am a little confused as to what reason you would have for the HAVING count() > 0 though; the only way something could have a zero count would mean it wasn't in the table (and therefore wouldn't show up anyway).
Other observations:
DISTINCT is not a function; enclosing the date in parenthesis in the SELECT clause has absolutely no effect. (Also, DISTINCT is almost never appropriate for a GROUPing query.)
COUNT(somefield) is the same as COUNT(1), COUNT(*). If you want the count of unique values you can do COUNT(DISTINCT somefield); but it wouldn't make sense to COUNT(DISTINCT groupingfield) as that would always result in 1.
The query you wrote is a bit complicated. Distinct and group by are doing the same thing for you here. When you do a group by count will automatically give you the count of grouped rows. Also you will have unique dates as well. Try this.
SELECT count(date), date
FROM employee
GROUP BY date
HAVING count( date ) >0
ORDER BY date DESC;
I need to get the earliest date from the following table based on column 'ItemNo'.
ItemNo PO_number Date
110913 PO-8048 9/15/2015
110913 PO-8036 9/30/2015
110652 PO-1011 10/19/2015
110652 PO-1011 10/10/2015
110009 PO-1016 7/1/2015
110009 PO-1087 6/20/2015
110888 PO-7171 4/1/2015
Your query result should be look like this.
ItemNo PO_number Date
110913 PO-8048 9/15/2015
110652 PO-1011 10/10/2015
110009 PO-1087 6/20/2015
110888 PO-7171 4/1/2015
Any help would be greatly appreciated.
Couple of different ways you could approach this, one reasonable approach would be something like:
with min_rec as
(
select t.ItemNo, t.PO_number, t.Date, row_number() over(partition by t.ItemNo order by t.Date asc) as rn
from your_table t
)
select m.ItemNo, m.PO_number, m.Date
from min_rec m
where m.rn = 1;
Leveraging a CROSS APPLY would be another approach that would work as well, though in this particular case it likely wouldn't be a better performing approach (though as always, it depends):
select distinct c.ItemNo, c.PO_number, c.Date
from your_table t
cross apply (
select top 1 i.ItemNo, i.PO_number, i.Date
from your_table i
where i.ItemNo = t.ItemNo
order by i.Date asc) c;
And naturally, you could simply use a self-joining subquery (I'll skip the example on that one).
I have the following query:
SELECT t.ID, t.caseID, time
FROM tbl_test t
INNER JOIN (
SELECT ID, MAX( TIME )
FROM tbl_test
WHERE TIME <=1353143351
GROUP BY caseID
ORDER BY caseID DESC -- ERROR HERE!
) s
USING (ID)
It seems that I only get the correct result if I use the ORDER BY in the inner join. Why is that? I am using the ID for the join, so the order should take no effekt.
If I remove the order by, I get too old entries from the database.
ID is the primary key, the caseID is a kind of object with multiple entries with different timestamps.
This query is ambiguous:
SELECT ID, MAX( TIME )
FROM tbl_test
WHERE TIME <=1353143351
GROUP BY caseID
It's ambiguous because it does not guarantee that it returns the ID of the row where the MAX(TIME) occurs. It returns the MAX(TIME) for each distinct value of caseID, but the value of other columns (like ID) is chosen arbitrarily from members of the group.
In practice, MySQL chooses the row that it finds first in the group as it scans rows in storage order.
Example:
caseID ID time
1 10 15:00
1 12 18:00
1 14 13:00
The max time is 18:00, which is the row with ID 12. But the query will return ID 10, simply because it's the first one in the group. If you were to reverse the order with ORDER BY, it would return ID 14. Still not the row where the max time is found, but it's from the other end of the group of rows.
Your query works with ORDER BY caseID DESC because, by coincidence, your Time values increase with the increasing ID.
This sort of query is actually an error in standard SQL and most other brands of SQL database. MySQL permits it, trusting that you know how to form an unambiguous query.
The fix is to use columns in the select-list only if they are unambiguous, that is, if they are in the GROUP BY clause, then each group is guaranteed to have only one distinct value:
SELECT caseID, MAX( TIME )
FROM tbl_test
WHERE TIME <=1353143351
GROUP BY caseID
SELECT t.ID, t.caseID, time
FROM tbl_test t
INNER JOIN (
SELECT caseID, MAX( TIME ) maxtime
FROM tbl_test
WHERE TIME <=1353143351
GROUP BY caseID
) s
ON t.caseID = s.caseID and t.time = s.maxtime
You are seeing that issue because you are getting the MAX(TIME) per caseID, but since you are grouping by caseID and NOT ID, you are getting an arbitrary ID. That happens because when you use an aggregate function, like MAX, you must, for every non-grouped field in the select specify how you want to aggregate it. That means, if it's in the SELECT and NOT in the GROUP BY, you have to tell MySQL how to aggregate. If you don't then you get a RANDOM row (well, not random per se, but it's not going to be in an order that you necessarily expect).
The reason ORDER BY is working for you, is that it kind of tricks the query optimizer into sorting the results before grouping, which just so happens to produce the result you want, but be warned, that will not always be the case.
What you want is the ID that has the MAX(TIME) given a caseID. Which means your INNER join needs to connect by caseID (not ID) and time (which will give you 1 row per each 1 row in the outer table).
Barmar beat me to the actual query, but that's the way you want to go.
I've a table like (I am omitting unnecessary columns)
id:int | name:string | ts:DateTime
There are multiple entries. now What I want in my resultset is
date:Date | entries:int
e.g. how many entries were made on all dates. actually I gonna make a chart of it.
What SQL Query I need to use for this ? I can create a view of it
This should do the job:
SELECT
DATE(ts) AS date,
COUNT(*) AS entries
FROM table
GROUP BY date
You can try
SELECT ts AS `date`, COUNT(id) AS entries
FROM your_table
GROUP BY ts
If you just need date part, then try
SELECT DATE(ts) AS `date`, COUNT(id) AS entries
FROM your_table
GROUP BY DATE(ts)
To strip off the time part of your timestamps, use DATE(). Sort by the date, ascending to get them in the right order. Note that this will omit dates for which there are no entries. If you need to fill those in, it becomes somewhat more complicated.
SELECT
DATE(ts) AS date,
COUNT(*) AS entries
FROM tbl
GROUP BY DATE(ts)
ORDER BY DATE(ts) ASC