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'm trying to create an extra column on my SQL where I could identify if the user_id generated a subsequent (or even third) row.
And if so, calculate the datediff between the connection time and previous row disconnection time.
If no duplicate user_id, the response should be NULL.
Here's a screenshot of my data and notes:
I tried the DATEDIFF formula but no success.
Could someone help me on this? I really would appreciate any input.
SELECT id,
user_id,
connected_at,
disconnected_at,
IFNULL(DateDiff('second', Lookup(disconnected_at), -1), connected_at)
FROM data
ORDER BY id, user_id, connected_at
When comparing values in different rows, you need to join the table to itself to get the second row. Try this:
SELECT
a.id,
a.user_id,
a.connected_at,
a.disconnected_at,
DateDiff('second',b.`connected_at`,a.`disconnected_at`) as `time_diff`
FROM `data` a
JOIN `data` b
ON a.`user_id` = b.`user_id` AND b.`connected_at` > a.`disconnected_at`
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
I have a table "A" with a "date" field. I want to make a select query and order the rows with previous dates in a descending order, and then, the rows with next dates in ascending order, all in the same query. Is it possible?
For example, table "A":
id date
---------------------
a march-20
b march-21
c march-22
d march-23
e march-24
I'd like to get, having as a starting date "march-22", this result:
id date
---------------------
c march-22
b march-21
a march-20
d march-23
e march-24
In one query, because I'm doing it with two of them and it's slow, because the only difference is the sorting, and the joins I have to do are a bit "heavy".
Thanks a lot.
You could use something like this -
SELECT *
FROM test
ORDER BY IF(
date <= '2012-03-22',
DATEDIFF('2000-01-01', date),
DATEDIFF(date, '2000-01-01')
);
Here is a link to a test on SQL Fiddle - http://sqlfiddle.com/#!2/31a3f/13
That's wrong, sorry :(
From documentation:
However, use of ORDER BY for individual SELECT statements implies nothing about the order in which the rows appear in the final result because UNION by default produces an unordered set of rows. Therefore, the use of ORDER BY in this context is typically in conjunction with LIMIT, so that it is used to determine the subset of the selected rows to retrieve for the SELECT, even though it does not necessarily affect the order of those rows in the final UNION result. If ORDER BY appears without LIMIT in a SELECT, it is optimized away because it will have no effect anyway.
This should do the trick. I'm not 100% sure about adding an order in a UNION...
SELECT * FROM A where date <= now() ORDER BY date DESC
UNION SELECT * FROM A where date > now() ORDER BY date ASC
I think the real question here is how to do the joining once. Create a temporary table with the result of joining, and make the 2 selects from that table. So it will be be time consuming only on creation (once) not on select query (twice).
CREATE TABLE tmp SELECT ... JOIN -- do the heavy duty here
With this you can make the two select statenets as you originally did.
Say I want to select rows with a date range or date_feild > 2009-06-01 && date_field < 2009-07-01. and I want to select the first_name_field and last_name_field but I only want the same name (F+L) to show up once per date. So the same name can show up, multiple times as long as their dates are different; but not if the names are on the same date. Does that makes sense? we are trying to track how many inquiries we got over a time period, but if the same user made multiple inquiries on the same day we want to count that as just 1.
I haven't even starting to program this yet so I am open to all suggestions.
Thanks!
use GROUP BY
SELECT date,name FROM table GROUP BY date,name
Don't know in mySQL but in SQL Server you can use COUNT DISTINCT.
SELECT
date_field,
COUNT(DISTINCT first_name + last_name)
FROM
YourTable
GROUP BY
date_field
The answer is inside your question:
select distinct date, concat(first_name, ' ', last_name) from table