I have an interesting query here. I have a table that stores visitor's ip and page_id along with a timestamp (date). I would like to count the amount of visitors per day for each page_id so I can then take that output and calculate which page_id is trending (on a daily basis)
Table visitors_counter looks like this:
id|page_id | ip | date
1 | 37 |1.1.1.1| 2017-02-10 14:03:16
2 | 38 |1.2.1.1| 2017-02-10 11:04:16
3 | 39 |1.1.3.1| 2017-02-10 16:05:16
4 | 37 |1.5.1.1| 2017-02-10 17:08:16
5 | 37 |1.1.1.1| 2017-02-10 19:07:16
And what I would like to achieve would be something like:
id|page_id |visitors | date
1 | 37 |3 | 2017-02-10 14:03:16
2 | 38 |1 | 2017-02-10 11:04:16
3 | 39 |1 | 2017-02-10 16:05:16
So far I've been able to count the amount of unique visitors per day with
SELECT DATE(date) Date, COUNT(DISTINCT page_id) uniqueperday
FROM visitors_counter
GROUP BY DATE(date)
i know im close, but its not quite what I want as I don't know which page_id are the most visited ones
Thanks
for your result you should use the group by and count(*) (you have two times the same id for page 37 and want result 3)
SELECT DATE(date) Date, page_id, COUNT(*) visitperday
FROM visitors_counter
GROUP BY DATE(date), page_id
othewise the Giorgios is the right one
Simply add page_id in the GROUP BY clause:
SELECT DATE(date) Date, page_id, COUNT(DISTINCT ip) uniqueperday
FROM visitors_counter
GROUP BY DATE(date), page_id
The above query returns the number of unique ip visits per page per day.
You can use aggregation on page and date to find count of distinct visitors per page per day. Also, use user variables to generate sequence id.
set #id := 0;
select #id = #id + 1 id,
page_id,
count(distinct ip) visitors,
min(date)
from visitors_counter
group by page_id, DATE(date)
order by visitors desc, page_id;
Related
I am trying to make a "top purchaser" module on my store and I am a bit confused about the MySQL query.
I have a table with all transactions and I need to select the person (which could have one or many transactions) with the highest amount of money spent in the past month.
What I have:
name | money spent
------------------
john | 50
mike | 12
john | 10
jane | 504
carl | 99
jane | 12
jane | 1
What I want to see:
With a query, I need to see:
name | money spent last month
-----------------------------
jane | 517
carl | 99
john | 60
mike | 12
How do I do that?
I do not really seem to find many good solutions since my MySQL query skills are quite basic. I thought of making a table in which money is added to the user when he buys something.
That's a simple aggregated query :
SELECT t.name, SUM(t.moneyspent) money_spent_last_month
FROM mytable t
GROUP BY t.name
ORDER BY t.money_spent_last_month DESC
LIMIT 1
The query sums the total money sped by customer name. The results are ordered by descending total money spent, and only the first row is retained.
If you are looking to filter data over last month, you need a column in the table that keeps track of the transaction date, say transaction_date, and then you can just add a WHERE clause to the query, like :
SELECT t.name, SUM(t.moneyspent) money_spent_last_month
FROM mytable t
WHERE
t.transaction_date >=
DATE_ADD(LAST_DAY(DATE_SUB(NOW(), INTERVAL 2 MONTH)), INTERVAL 1 DAY)
AND t.transaction_date <=
DATE_SUB(NOW(), INTERVAL 1 MONTH)
GROUP BY t.name
ORDER BY t.money_spent_last_month DESC
LIMIT 1
This method is usually more efficient than using DATE_FORMAT to format dates as string and compare the results.
I have the following table
id | user_id | date | status
1 | 53 | 2018-09-18 06:59:54 | 1
2 | 62 | 2018-09-18 07:00:16 | 1
3 | 53 | 2018-09-18 09:34:12 | 2
4 | 53 | 2018-09-18 12:16:27 | 1
5 | 53 | 2018-09-18 18:03:19 | 2
6 | 62 | 2018-09-18 18:17:41 | 2
I would like to get the total working hours (from date range) and group them by user_id
UPDATE
The system does not "require" a check-out so if there is only one value can we set a default check out time lets say 19:00:00? IF not I can check every day at 21:00:00 if there is not a checkout time to manually insert it at 19:00:00
UPDATE 2
I have added a new field in the table "status" so the very first check-in of the date the status = 1 and every 2nd check-in the status = 2
So if a user check-ins for the 3rd time during the day the status will be 1 again etc.
I hope this will make things easier
Thanks
In case of multiple check-in and check-out happening within a day, for a user:
Utilizing Correlated Subquery, we can find corresponding "checkout_time" for every "checkin_time".
Also, note the usage of Ifnull(), Timestamp() functions etc, to consider default "checkout_time" as 19:00:00, in case of no corresponding entry.
Then, considering this enhanced data-set as Derived Table, we group the data-set based on the user_id and date. Date (yyyy-mm-dd) can be determined using Date() function.
Eventually, use Timestampdiff() function with Sum aggregation, to determine the total work seconds for a user_id at a particular date.
You can easily convert these total seconds to hours (either in your application code, or at the query itself (divide seconds by 3600).
The reason I have preferred to compute using seconds, as Timestampdiff() function returns integer only. So there may be truncation errors, in case of multiple checkin/checkout(s).
Use the following query (replace your_table with your actual table name):
SELECT inner_nest.user_id,
DATE(inner_nest.checkin_time) AS work_date,
SUM(TIMESTAMPDIFF(SECOND,
inner_nest.checkin_time,
inner_nest.checkout_time)) AS total_work_seconds
FROM
(
SELECT t1.user_id,
t1.date as checkin_time,
t1.status,
IFNULL( (
SELECT t2.date
FROM your_table AS t2
WHERE t2.user_id = t1.user_id
AND t2.status = 2
AND t2.date > t1.date
AND DATE(t2.date) = DATE(t1.date)
ORDER BY t2.date ASC LIMIT 1
),
TIMESTAMP(DATE(t1.date),'19:00:00')
) AS checkout_time
FROM `your_table` AS t1
WHERE t1.status = 1
) AS inner_nest
GROUP BY inner_nest.user_id, DATE(inner_nest.checkin_time)
Additional: Following solution will work for the case when there is a single check-in, and corresponding check-out on the same date.
You first need to group the dataset based on the user_id and date. Date (yyyy-mm-dd) can be determined using Date() function.
Now use aggregation functions like Min() and Max() to find the starting and closing time for a user_id at a particular date.
Eventually, use Timestampdiff() function to determine the working hours for a user_id at a particular date (difference between the closing and starting time)
Try the following query (replace your_table with your actual table name):
SELECT user_id,
DATE(`date`) AS working_date,
TIMESTAMPDIFF(HOUR, MIN(`date`), MAX(`date`)) AS working_hours
FROM your_table
GROUP BY
user_id,
DATE(`date`)
Use TIMESTAMPDIFF function
the query more like :
SELECT t1.user_id, TIMESTAMPDIFF(HOUR,t1.date,t2.date) as difference
FROM your_table t1
INNER JOIN your_table t2 on t1.user_id = t2.user_id
Group By t1.user_id
You can see this as preference TimeStampDiff
I have a table of urls. I need to select 1, in round robin. Hoping to display 1 per pageview OR per Visitor and show each 20 times if I have 100 pageviews OR Visitors.
Essentially: SELECT url FROM table LIMIT 1 (in rotation)
+-------+---------------------+
| id | url |
+-------+---------------------+
| 1 | http://google.com |
| 2 | http://yahoo.com |
| 3 | http://ebay.com |
| 4 | http://anything.com |
| 5 | http://other.com |
+-------+---------------------+
If I understand correctly what you mean by round-robin then you can do something along the lines of
SELECT id, url
FROM urls u CROSS JOIN
(
SELECT MIN(id) min_id, MAX(id) max_id
FROM urls
) m
WHERE id > IF(? >= max_id, 0, ?) -- last shown id goes here instead of placeholders
ORDER BY id
LIMIT 1;
Store (in session, file, another table, etc.) and pass to your query the last shown id or 0 for the initial query.
This will give you the next row or first again if you reached the last one.
This query will still work if you have gaps in ids.
Here is a SQLFiddle demo
You can do using order by rand()
SELECT * FROM table ORDER BY rand() LIMIT 1
You can use RAND() function in your ORDER BY clause:
SELECT * FROM tableName ORDER BY RAND() LIMIT 1
Is this what you want to achieve?
This question already has an answer here:
Closed 10 years ago.
Possible Duplicate:
SQL ORDER BY total within GROUP BY
UPDATE: I've found my solution, which I've posted here. Thanks to everyone for your help!
I'm developing a Facebook application which requires a leaderboard. Scores and time taken to complete the game are recorded and these are organised by score first, then in the case of two identical scores, the time is used. If a user has played multiple times, their best score is used.
The lower the score, the better the performance in the game.
My table structure is:
id
facebook_id - (Unique Identifier for the user)
name
email
score
time - (time to complete game in seconds)
timestamp - (unix timestamp of entry)
date - (readable format of timestamp)
ip
The query I thought would work is:
SELECT *
FROM entries
ORDER BY score ASC, time ASC
GROUP BY facebook_id
The problem I'm having is in some cases it's pulling in the user's first score in the database, not their highest score. I think this is down to the GROUP BY statement. I would have thought the ORDER BY statement would have fixed this, but apparently not.
For example:
----------------------------------------------------------------------------
| ID | NAME | SCORE | TIME | TIMESTAMP | DATE | IP |
----------------------------------------------------------------------------
| 1 | Joe Bloggs | 65 | 300 | 1234567890 | XXX | XXX |
----------------------------------------------------------------------------
| 2 | Jane Doe | 72 | 280 | 1234567890 | XXX | XXX |
----------------------------------------------------------------------------
| 3 | Joe Bloggs | 55 | 285 | 1234567890 | XXX | XXX |
----------------------------------------------------------------------------
| 4 | Jane Doe | 78 | 320 | 1234567890 | XXX | XXX |
----------------------------------------------------------------------------
When I use the query above, I get the following result:
1. Joe Bloggs - 65 - 300 - (Joes First Entry, not his best entry)
2. Jane Doe - 72 - 280
I would have expected...
1. Joe Bloggs - 55 - 285 - (Joe's best entry)
2. Jane Doe - 72 - 280
It's like the Group By is ignoring the Order - and just overwriting the values.
Using MIN(score) with the group by selects the lowest score, which is correct - however it merges the time from the users first record in the database, so often returns incorrectly.
So, how can I select a user's highest score and the associated time, name, etc and order the results by score, then time?
Thanks in advance!
Your query does not actually make sense, because the order by should be after the group by. What SQL engine are you using? Most would give an error.
I think what you want is more like:
select e.facebookid, minscore, min(e.time) as mintime -- or do you want maxtime?
from entries e join
(select e.facebookid, min(score) as minscore
from entries e
group by facebookid
) esum
on e.facebookid = esum.facebookid and
e.score = e.minscore
group by e.facebookid, minscore
You can also do this with window functions, but that depends on your database.
One approach would be this:
SELECT entries.facebook_id, MIN(entries.score) AS score, MIN(entries.time) AS time
FROM entries
INNER JOIN (
SELECT facebook_id, MIN(score) AS score
FROM entries
GROUP BY facebook_id) highscores
ON highscores.facebook_id = entries.facebook_id
AND entries.score = highscores.score
GROUP BY entries.facebook_id
ORDER BY MIN(entries.score) ASC, MIN(entries.time) ASC
If you need more information from the entries table, you can then use this as a subquery, and join again on the information presented (facebook_id, score, time) to get one row per user.
You need to aggregate twice, is the crux of this; once to find the minimum score for the user, and again to find the minimum time for that user and score. You could reverse the order of the aggregation, but I would expect that this will filter most quickly and thus be most efficient.
You might also want to check which is faster, aggregating the second time: using the minimum score or grouping using the score as well.
You need to min the score
SELECT
facebook_id,
name,
email,
min(score) as high_score
FROM
entries
GROUP BY
facebook_id,
name,
email
ORDER BY
min(score) ASC
Thanks for your help. #Penguat had the closest answer.. Here was my final Query for anyone who might have the same issue...
SELECT f.facebook_id, f.name, f.score, f.time FROM
(SELECT facebook_id, name, min(score)
AS highscore FROM golf_entries
WHERE time > 0
GROUP BY facebook_id)
AS x
INNER JOIN golf_entries as f
ON f.facebook_id = x.facebook_id
AND f.score = x.highscore
ORDER BY score ASC, time ASC
Thanks again!
If you want their best time, you want to use the MIN() function - you said that the lower the score, the better they did.
SELECT facebook_id, MIN(score), time, name, ...
FROM entries
GROUP BY facebook_id, time, name, ...
ORDER BY score, time
With PHP and MySQL I am trying to display items with most votes over a certain period of time (24 hour, 1 week, 1 year, etc). When someone votes on an item, a table records the user, item id, vote, and time, like so:
Table1
username | itemid | vote | time
asdf | 127 | 1 | 1306726126
asdf | 124 | -1 | 1306726123
bob | 127 | 1 | 1306726129
bob | 124 | 1 | 1306726123
Now, I have another table with item details.
Table2
itemid | name | category | date | etc
What I WANT to do is call a table to display all the data from table 2 for only items with votes in the last 24hours and sort it by the votes. This means I need to SUM votes with TIME < 24 hours, then RIGHT JOIN (?) to my other database? I don't know, I am having difficulty figuring out how I should go about doing this. Any suggestiongs?
Something like this should work.
SELECT SUM(Table1.vote) as votes, Table2.* FROM Table2
LEFT JOIN Table1 ON Table1.itemid=Table2.itemid
WHERE Table1.`time`>=DATE_SUB(Table1.`time`, INTERVAL 24 HOUR)
GROUP BY Table1.itemid
ORDER BY Table1.votes DESC