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;
Hi i am trying to get the count of records per day which i can do, but i also want the date to be show, for example,
Result
Date | Count
26/01/2015 20
25/01/2015 | 413
Here is an example of my data.
I would think this would work. Replace 'yourTable' with your table name
SELECT Date, COUNT(*) FROM yourTable GROUP BY Date;
Get the total count and group them by date.
SELECT `date`, COUNT(*) as Total
FROM `table`
GROUP BY `date`
ORDER BY `date`;
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
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 :)
What might be wrong with this query:
select count(customer_email) as num_prev
from _pj_cust_email_by_date
where order_date < '2011-02'
and customer_email is not null
group by customer_email having count(order_date) > 0;
Which returns row results such as:
1
2
3
2
1
5
4
When I'm trying to get a full count of how many customers in total purchased during the specified date range?
_pj_cust_email_by_date is a view that returns only email address and order date in YYYY-MM-DD format. I do not have access to use anything save for this view.
The GROUP BY is causing that.
It causes one result row to be returned per group, in this for each distinct value of customer_email.
If you want the total number of distinct email addresses, then you need to drop the GROUP BY clause and change the COUNT to COUNT(DISTINCT customer_email).
You need to subquery it further
select count(*) CustomerCount
from (
select count(customer_email) as num_prev
from _pj_cust_email_by_date
where order_date < '2011-02'
and customer_email is not null
group by customer_email having count(order_date) > 0;
) as innercount
That would normally be the approach, but since you're using having count(order_date) > 0, I think you only need
select count(distinct customer_email) as num_prev
from _pj_cust_email_by_date
where order_date < '2011-02' and customer_email is not null
Because the HAVING clause will never detail with empty order_dates which makes the HAVING clause a dud, actually.