So I have this data set (down below) and I'm simply trying to gather all data based on records in field 1 that have a count of more than 30 (meaning a distinct brand that has 30+ record entries) that's it lol!
I've been trying a lot of different distinct, count esc type of queries but I'm falling short. Any help is appreciated :)
Data Set
By using GROUP BY and HAVING you can achieve this. To select more columns remember to add them to the GROUP BY clause as well.
SELECT Mens_Brand FROM your_table
WHERE Mens_Brand IN (SELECT Mens_Brand
FROM your_table
GROUP BY Mens_Brand
HAVING COUNT(Mens_Brand)>=30)
You can simply use a window function (requires mysql 8 or mariadb 10.2) for this:
select Mens_Brand, Mens_Price, Shoe_Condition, Currency, PK
from (
select Mens_Brand, Mens_Price, Shoe_Condition, Currency, PK, count(1) over (partition by Mens_Brand) brand_count
from your_table
) counted where brand_count >= 30
I have a table about how often an event occurs, and I want to see first and last occurrence of every event. Assume I have three columns, event_id, event_time and event_date. I know queries like this are wrong, and give false result:
SELECT
event_id,
MIN(event_time),
MAX(event_time),
MIN(event_date),
MAX(event_date)
FROM events_table
WHERE 1
GROUP BY event_id
Is there any way I could do this without merging two columns?
Please be more precisely regarding column types (date,time,varchar..) and if you would look for occurence of the event by day or ever.
You can just concatenate these two columns and min,max it.
Following assumes event_time like '08:03:21' and event_date('2014-01-30') occurence ever:
SELECT id, event_id, min(concat(event_date,' ',event_time)),max(concat(event_date,' ',event_time))
from t_test
group by event_id
If you would like to check the first,last occurence per day, just group by day and remove the date from concat().
You forgot FROM
Here is another approach. Tested on my MySQL
SELECT a1.event_id,
Min(Concat(a2.event_date, ' ', a2.event_time)),
Max(Concat(a2.event_date, ' ', a2.event_time))
FROM event_log a1
INNER JOIN event_log a2
ON a2.event_id = a1.event_id
GROUP BY a1.event_id
I have a users table, and an appointments table. For any given day, I would like a query that selects
1) the user_id the appointment is scheduled with
2) the number of appointments for that user for the specified day.
It seems I can one or the other, but I'm unsure of how to do it with one query. For instance, I can do:
SELECT user_id FROM appt_tbl WHERE DATE(appt_date_time) = '2012-10-14'
group by user_id;
Which will give me the users that have an appointment that day, but how can I add to this query another column that will give me how many appointments each user has? Assuming I need some kind of subquery, but I'm unsure of how to structure that.
SQL uses the notion of "aggregate functions" to get you this information. You can use them with any aggregating query (i.e. it has "group by" in it).
SELECT user_id, count(*) as num_apts ...
Try adding COUNT(*) to your query:
SELECT user_id, COUNT(*) FROM appt_tbl WHERE DATE(appt_date_time) = '2012-10-14'
group by user_id;
I'm very sorry if the question seems too basic.
I've surfed entire Internet and StackOverflow for a finished solution, and did not find anything that I can understand, and can't write it myself, so have to ask it here.
I have a MySQL database.
It has a table named "posted".
It has 8 columns.
I need to output this result:
SELECT DISTINCT link FROM posted WHERE ad='$key' ORDER BY day, month
But I need not only the "link" column, but also other columns for this row.
Like for every row returned with this query I also need to know its "id" in the table, "day" and "month" values etc.
Please tell me what should I read to make it, or how to make it.
Please keep it as simple as possible, as I'm not an expert in MySQL.
Edit:
I tried this:
SELECT DISTINCT link,id,day,month FROM posted WHERE ad='$key' ORDER BY day, month
It doesn't work. It returns too many rows. Say there are 10 rows with same links, but different day/month/id. This script will return all 10, and I want only the first one (for this link).
The problem comes from instinctively believing that DISTINCT is a local pre-modifier for a column.
Hence, you "should" be able to type
XXbadXX SELECT col1, DISTINCT col2 FROM mytable XXbadXX
and have it return unique values for col2. Sadly, no. DISTINCT is actually a global post-modifier for SELECT, that is, as opposed to SELECT ALL (returning all answers) it is SELECT DISTINCT (returning all unique answers). So a single DISTINCT acts on ALL the columns that you give it.
This makes it real hard to use DISTINCT on a single column, while getting the other columns, without doing major extremely ugly backflips.
The correct answer is to use a GROUP BY on the columns that you want to have unique answers: SELECT col1, col2 FROM mytable GROUP BY col2 will give you arbitrary unique col2 rows, with their col1 data as well.
I tried this:
SELECT DISTINCT link,id,day,month FROM posted
WHERE ad='$key' ORDER BY day, month
It doesn't work. It returns too many rows. Say there are 10 rows with
same links, but different day/month/id. This script will return all
10, and I want only the first one (for this link).
What you're asking doesn't make sense.
Either you want the distinct value of all of link, id, day, month, or you need to find a criterion to choose which of the values of id, day, month you want to use, if you just want at most one distinct value of link.
Otherwise, what you're after is similar to MySQL's hidden columns in GROUP BY/HAVING statements, which is non-standard SQL, and can actually be quite confusing.
You could in fact use a GROUP BY link if it made sense to pick any row for a given link value.
Alternatively, you could use a sub-select to pick the row with the minimal id for a each link value (as described in this answer):
SELECT link, id, day, month FROM posted
WHERE (link, id) IN
(SELECT link, MIN(id) FROM posted ad='$key' GROUP BY link)
SELECT Id, Link, Day, Month FROM Posted
WHERE Id IN(
SELECT Min(Id) FROM Posted GROUP BY Link)
SELECT OTHER_COLUMNS FROM posted WHERE link in (
SELECT DISTINCT link FROM posted WHERE ad='$key' )
ORDER BY day, month
If what your asking is to only show rows that have 1 link for them then you can use the following:
SELECT * FROM posted WHERE link NOT IN
(SELECT link FROM posted GROUP BY link HAVING COUNT(LINK) > 1)
Again this is assuming that you want to cut out anything that has a duplicate link.
I think the best solution would be to do a subquery and then join that to the table. The sub query would return the primary key of the table. Here is an example:
select *
from (
SELECT row_number() over(partition by link order by day, month) row_id
, *
FROM posted
WHERE ad='$key'
) x
where x.row_id = 1
What this does is the row_number function puts a numerical sequence partitioned by each distinct link that results in the query.
By taking only those row_numbers that = 1, then you only return 1 row for each link.
The way you change what link gets marked "1" is through the order-by clause in the row_number function.
Hope this helps.
SELECT DISTINCT link,id,day,month FROM posted WHERE ad='$key' ORDER BY day, month
OR
SELECT link,id,day,month FROM posted WHERE ad='$key' ORDER BY day, month
If you want all columns where link is unique:
SELECT * FROM posted WHERE link in
(SELECT link FROM posted WHERE ad='$key' GROUP BY link);
What you want is the following:
SELECT DISTINCT * FROM posted WHERE ad='$key' GROUP BY link ORDER BY day, month
if there are 4 rows for example where link is the same, it will pick only one (I asume the first one).
I had a similar problem, maybe that help someone, for example - table with 3 columns
SELECT * FROM DataTable WHERE Data_text = 'test' GROUP BY Data_Name ORDER BY Data_Name ASC
or
SELECT Data_Id, Data_Text, Data_Name FROM DataTable WHERE Data_text = 'test' GROUP BY Data_Name ORDER BY Data_Name ASC
Two ways work for me.
SELECT a.* FROM orders a INNER JOIN (SELECT course,MAX(id) as id FROM orders WHERE admission_id=".$id." GROUP BY course ) AS b ON a.course = b.course AND a.id = b.id
With the Above Query you will get unique records with where condition
In MySQL you can simply use "group by". Below will select ALL, with a DISTINCT "col"
SELECT *
FROM tbl
GROUP BY col
Select the datecolumn of month so that u can get only one row per link, e.g.:
select link, min(datecolumn) from posted WHERE ad='$key' ORDER BY day, month
Good luck............
Or
u if you have date column as timestamp convert the format to date and perform distinct on link so that you can get distinct link values based on date instead datetime
This sounds quite simple but I just can't figure it out.
I have a table orders (id, username, telephone_number).
I want to get number of orders from one user by comparing the last 8 numbers in telephone_number.
I tried using SUBSTR(telephone_number, -8), I've searched and experimented a lot, but still I can't get it to work.
Any suggestions?
Untested:
SELECT
COUNT(*) AS cnt,
*
FROM
Orders
GROUP BY
SUBSTR(telephone_number, -8)
ORDER BY
cnt DESC
The idea:
Select COUNT(*) (i.e., number of rows in each GROUPing) and all fields from Orders (*)
GROUP by the last eight digits of telephone_number1
Optionally, ORDER by number of rows in GROUPing descending.
1) If you plan to do this type of query often, some kind of index on the last part of the phone number could be desirable. How this could be best implemented depends on the concrete values stored in the field.
//Memory intensive.
SELECT COUNT(*) FROM `orders` WHERE REGEXP `telephone_number` = '(.*?)12345678'
OR
//The same, but better and quicker.
SELECT COUNT(*) FROM `orders` WHERE `telephone_number` LIKE '%12345678'
You can use the below query to get last 8 characters from a column values.
select right(rtrim(First_Name),8) FROM [ated].[dbo].[Employee]