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 have a very simple table which consists of the following columns:
id | customer_id | total | created_at
I was running this query to get the results per day for the last ten days:
SELECT SUM(total) AS total, DATE_FORMAT(created_at, "%d/%m/%Y") AS date
FROM table
WHERE created_at BETWEEN "2017-02-20" AND "2017-03-01"
GROUP BY created_at
ORDER BY created_at DESC
This works fine, but I've just noticed that there's an issue with imported rows being duplicated for some reason so I'd like to update the query to be able to handle the situation if it ever happens again, in other words select one row instead of all when the date and customer id are the same (the total is also identical).
If I add customer_id to the group by that seems to work but the trouble with that is then the query returns a result per day for each customer when I only want the overall total.
I've tried a couple of things but I haven't cracked it yet, I think it will be achievable using a sub query and/or an inner join, I have tried this so far but the figures are very wrong:
SELECT
created_at,
(
SELECT SUM(total)
FROM table test
WHERE test.created_at = table.created_at
AND test.customer_id = table.customer_id
GROUP BY customer_id, created_at
LIMIT 1
) AS total
FROM table
WHERE created_at BETWEEN "2017-02-20" AND "2017-03-01"
GROUP BY created_at
ORDER BY created_at DESC
It's also a large table so finding a performant way to do this is also important.
First, are you sure that created_at is a date and not a datetime? This makes a big difference.
You can do what you want using two levels of aggregation:
SELECT SUM(max_total) AS total, DATE_FORMAT(created_at, '%d/%m/%Y') AS date
FROM (SELECT t.customer_id, t.created_at, MAX(total) as max_total
FROM table t
WHERE t.created_at BETWEEN '2017-02-20' AND '2017-03-01'
GROUP BY t.customer_id, t.created_at
) t
GROUP BY created_at
ORDER BY created_at DESC;
I've a table cart which structure is following: id, item_id, session_id, date, num, so I need rows to be grouped by session_id column and sorted by date column in the same time, is it possible?
Yes it is possible.
... order by session_id, date
This will order firstly by session_id, and then by date. I believe that is what you want?
Try this:
SELECT * FROM YourTable
GROUP BY session_id ORDER BY date DESC;
Hope this helps!
You cant group by session_id and in the same time order by date.
You will get:
ORDER BY clause because it is not contained in either an aggregate function or the GROUP BY clause
what you can do is:
SELECT [session_id],[date]
From TableName
group by [session_id], [date]
order by [date] desc
I'm trying to execute a query like this:
SELECT MAX(counter), MIN(counter) ,
my_date IN (SELECT my_date FROM my_table WHERE counter = MAX(counter) ) AS max_Date ,
my_date IN (SELECT my_date FROM my_table WHERE counter = MIN(counter) ) AS min_Date
FROM my_table;
and it's giving me the "invalid use of group function" error. what I want to do is to find the date for the maximum counter and then find the date for the minimum counter. Any help!! really appreciate it .. thanks.
You're trying to use the result of aggregate functions (max()/min()) on a row-by-row basis, but those results are not available until the DB has scanned the entire table.
e.g. it's a chicken and egg problem. You need to count chickens, but the eggs that will produce the chickens haven't even been layed yet.
That's why there's HAVING clauses, which allow you to use the results of aggregate functions to do filtering.
Try this for the subqueries:
SELECT my_date FROM my_table HAVING counter = MIN(counter)
^^^^^^
You can get the dates where the largest and smallest counter values appear using a trick with group_concat() and substring_index():
SELECT MAX(counter), MIN(counter) ,
substring_index(group_concat(my_date order by counter desc), ',', 1) as max_date,
substring_index(group_concat(my_date order by counter), ',', 1) as min_date
FROM my_table;
Note: You probably want to format the date first to your liking.
You can also do this with a join.
The problem with your query is:
where counter = min(counter)
You can't include aggregation functions in the where clause, because both are referring to the table in the subquery. You could possibly do this using aliaes, but why bother? There are other ways to write the query.
You need a subselect to get the max and min counters and then join back against the table a couple of times to get the other values from those rows.
SELECT MaxCounter, MinCounter, a.my_date, b.my_date
FROM (SELECT MAX(counter) AS MaxCounter, MIN(counter) AS MinCounter
FROM my_table) Sub1
INNER JOIN my_table a ON Sub1.MaxCounter
INNER JOIN my_table b ON Sub1.MinCounter
Note that this does assume that counter is unique!
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 :)