How can I find the most frequent value within a table, then add up the value from those columns for outputting data?
player | money
player001 | 120
player001 | 140
player002 | 100
player003 | 200
The end result should output something like Player001 has 260 money - sort of like a leaderboard.
Based on your result, you don't want the most frequent value. You want the highest sum per player. That would be:
select player, sum(money)
from t
group by player
order by sum(money) desc
limit 1;
If you want a single string from this, I would recommend doing that at the application layer. However, you could use concat() to format a string in the SQL itself.
Related
I have Database table payment such as below
level_count | amount
__________________________
650 | 12
1000 | 35
1700 | 50
__________________________
Now Wanted to check if I supplied input as 650 which is level_count column value. Then I should get amount as 12. Then If I supplied input as 999 I should still get 12. Means It should compare its successive rows and compare. Suppose If I enter 1200 then I should get 35 and If I enter 1700 or above I should get 50.
I have tried flowing but didn't got any success.
Where I am going wrong.
SELECT * FROM payment T1
INNER JOIN payment T2 on T1.level_count>=T2.level_count AND T1.level_count<T2.level_count
WHERE T1.level_count = '650'
When I execute above query I get no results.
You may try using a LIMIT query here:
SELECT *
FROM payment
WHERE level_count <= 999 -- or 650, or another input value
ORDER BY level_count DESC
LIMIT 1;
The logic of the above query works in two parts. First, the WHERE clause removes all records for which the level_count is greater than the input value. But this still leaves us potentially with more than one record (which include the record we actually want). The LIMIT trick then keeps the single remaining record with the highest level_count.
I am trying to query a dataset from a single table, which contains quiz answers/entries from multiple users. I want to pull out the highest scoring entry from each individual user.
My data looks like the following:
ID TP_ID quiz_id name num_questions correct incorrect percent created_at
1 10154312970149546 1 Joe 3 2 1 67 2015-09-20 22:47:10
2 10154312970149546 1 Joe 3 3 0 100 2015-09-21 20:15:20
3 125564674465289 1 Test User 3 1 2 33 2015-09-23 08:07:18
4 10153627558393996 1 Bob 3 3 0 100 2015-09-23 11:27:02
My query looks like the following:
SELECT * FROM `entries`
WHERE `TP_ID` IN('10153627558393996', '10154312970149546')
GROUP BY `TP_ID`
ORDER BY `correct` DESC
In my mind, what that should do is get the two users from the IN clause, order them by the number of correct answers and then group them together, so I should be left with the 2 highest scores from those two users.
In reality it's giving me two results, but the one from Joe gives me the lower of the two values (2), with Bob first with a score of 3. Swapping to ASC ordering keeps the scores the same but places Joe first.
So, how could I achieve what I need?
You're after the groupwise maximum, which can be obtained by joining the grouped results back to the table:
SELECT * FROM entries NATURAL JOIN (
SELECT TP_ID, MAX(correct) correct
FROM entries
WHERE TP_ID IN ('10153627558393996', '10154312970149546')
GROUP BY TP_ID
) t
Of course, if a user has multiple records with the maximal score, it will return all of them; should you only want some subset, you'll need to express the logic for determining which.
MySql is quite lax when it comes to group-by-clauses - but as a rule of thumb you should try to follow the rule that other DBMSs enforce:
In a group-by-query each column should either be part of the group-by-clause or contain a column-function.
For your query I would suggest:
SELECT `TP_ID`,`name`,max(`correct`) FROM `entries`
WHERE `TP_ID` IN('10153627558393996', '10154312970149546')
GROUP BY `TP_ID`,`name`
Since your table seems quite denormalized the group by name-par could be omitted, but it might be necessary in other cases.
ORDER BY is only used to specify in which order the results are returned but does nothing about what results are returned - so you need to apply the max()-function to get the highest number of right answers.
Okay I'm still fairly new to MS Access, but have got some of the bases down. My next issue is pulling data from two different queries but still needing them to show.
Here's what I have
I have one query with the following information
| ID Number | Points |
The other query has the following
| ID Number | Points over 1000 |
In this new query I need to do display the following
| ID Number | Points | Points over 1000 | Total Points |
There's going to be some rows where Points over 1000 doesn't exist and needs to be empty or a 0, but I need the ID Number In Points over 1000 to match and check the ID Number in just the points column.
and in the end add them up in the Points total
I hope that makes sense?
Thanks again
In theory this Query should work the way you want it to.
SELECT
tmpQ.ID,
Sum(tmpQ.Points) As ActualPoints,
Sum(tmpQ.PointsOver1000) As Over1000,
[ActualPoints] + [Over1000] As TotalPoints
FROM
(
SELECT
qryA.[ID Number] As ID,
Sum(qryA.Points) As Points,
Sum(0) As PointsOver1000
FROM
qryA
GROUP BY
qryA.[ID Number]
UNION ALL
SELECT
qryB.[ID Number] As ID,
Sum(0) As Points,
Sum(qryB.PointsOver1000) As PointsOver1000
FROM
qryB
GROUP BY
qryB.[ID Number]
) As tmpQ
GROUP BY
tmpQ.ID;
Where qryA and qryB are the two queries you have that will give you the result of two different Points.
I think it will be easiest to start with the table I have and the result I am aiming for.
Name | Date
A | 03/01/2012
A | 03/01/2012
B | 02/01/2012
A | 02/01/2012
B | 02/01/2012
A | 02/01/2012
B | 01/01/2012
B | 01/01/2012
A | 01/01/2012
I want the result of my query to be:
Name | 01/01/2012 | 02/01/2012 | 03/01/2012
A | 1 | 2 | 2
B | 2 | 2 | 0
So basically I want to count the number of rows that have the same date, but for each individual name. So a simple group by of dates won't do because it would merge the names together. And then I want to output a table that shows the counts for each individual date using php.
I've seen answers suggest something like this:
SELECT
NAME,
SUM(CASE WHEN GRADE = 1 THEN 1 ELSE 0 END) AS GRADE1,
SUM(CASE WHEN GRADE = 2 THEN 1 ELSE 0 END) AS GRADE2,
SUM(CASE WHEN GRADE = 3 THEN 1 ELSE 0 END) AS GRADE3
FROM Rodzaj
GROUP BY NAME
so I imagine there would be a way for me to tweak that but I was wondering if there is another way, or is that the most efficient?
I was perhaps thinking if the while loop were to output just one specific name and date each time along with the count, so the first result would be A,01/01/2012,1 then the next A,02/01/2012,2 - A,03/01/2012,3 - B,01/01/2012,2 etc. then perhaps that would be doable through a different technique but not sure if something like that is possible and if it would be efficient.
So I'm basically looking to see if anyone has any ideas that are a bit outside the box for this and how they would compare.
I hope I explained everything well enough and thanks in advance for any help.
You have to include two columns in your GROUP BY:
SELECT name, COUNT(*) AS count
FROM your_table
GROUP BY name, date
This will get the counts of each name -> date combination in row-format. Since you also wanted to include a 0 count if the name didn't have any rows on a certain date, you can use:
SELECT a.name,
b.date,
COUNT(c.name) AS date_count
FROM (SELECT DISTINCT name FROM your_table) a
CROSS JOIN (SELECT DISTINCT date FROM your_table) b
LEFT JOIN your_table c ON a.name = c.name AND
b.date = c.date
GROUP BY a.name,
b.date
SQLFiddle Demo
You're asking for a "pivot". Basically, it is what it is. The real problem with a pivot is that the column names must adapt to the data, which is impossible to do with SQL alone.
Here's how you do it:
SELECT
Name,
SUM(`Date` = '01/01/2012') AS `01/01/2012`,
SUM(`Date` = '02/01/2012') AS `02/01/2012`,
SUM(`Date` = '03/01/2012') AS `03/01/2012`
FROM mytable
GROUP BY Name
Note the cool way you can SUM() a condition in mysql, becasue in mysql true is 1 and false is 0, so summing a condition is equivalent to counting the number of times it's true.
It is not more efficient to use an inner group by first.
Just in case anyone is interested in what was the best method:
Zane's second suggestion was the slowest, I loaded in a third of the data I did for the other two and it took quite a while. Perhaps on smaller tables it would be more efficient, and although I am not working with a huge table roughly 28,000 rows was enough to create significant lag, with the between clause dropping the result to about 4000 rows.
Bohemian's answer gave me the least amount to code, I threw in a loop to create all the case statements and it worked with relative ease. The benefit of this method was the simplicity, besides creating the loop for the cases, the results come in without the need for any php tricks, just simple foreach to get all the columns. Recommended for those not confident with php.
However, I found Zane's first suggestion the quickest performing and despite the need for extra php coding it seems I will be sticking with this method. The disadvantage of this method is that it only gives the dates that actually have data, so creating a table with all the dates becomes a bit more complicated. What I did was create a variable that keeps track of what date it is supposed to be compared to the table column which is reset on each table row, when the result of the query is equal to that date it echoes the value otherwise it does a while loop echoing table cells with 0 until the dates do match. It also had to do a check to see if the 'Name' value is still the same and if not it would switch to the next row after filling in any missing cells with 0 to the end of that row. If anyone is interested in seeing the code you can message me.
Results of the two methods over 3 months of data (a column for each day so roughly 90 case statements) ~ 12,000 rows out of 28,000:Bohemian's Pivot - ~0.158s (highest seen ~0.36s)Zane's Double Group by - ~0.086s (highest seen ~0.15s)
I want to log certain activities in MySql with a timecode using time(). Now I'm accumulating thousands of records, I want to output the data by sets of hours/days/months etc.
What would be the suggested method for grouping time codes in MySQL?
Example data:
1248651289
1248651299
1248651386
1248651588
1248651647
1248651700
1248651707
1248651737
1248651808
1248652269
Example code:
$sql = "SELECT COUNT(timecode) FROM timecodeTable";
//GROUP BY round(timecode/3600, 1) //group by hour??
Edit:
There's two groupings that can be made so I should make that clearer: The 24 hours in the day can be grouped but I'm more interested in grouping over time so returning 365 results for each year the tracking is in place, so total's for each day passed, then being able to select a range of dates and see more details on hours/minutes accessed over those times selected.
This is why I've titled it as using PHP, as I'd expect this might be easier with a PHP loop to generate the hours/days etc?
Peter
SELECT COUNT(*), HOUR(timecode)
FROM timecodeTable
GROUP BY HOUR(timecode);
Your result set, given the above data, would look as such:
+----------+----------------+
| COUNT(*) | HOUR(timecode) |
+----------+----------------+
| 10 | 18 |
+----------+----------------+
Many more related functions can be found here.
Edit
After doing some tests of my own based on the output of your comment I determined that your database is in a state of epic fail. :) You're using INT's as TIMESTAMPs. This is never a good idea. There's no justifiable reason to use an INT in place of TIMESTAMP/DATETIME.
That said, you'd have to modify my above example as follows:
SELECT COUNT(*), HOUR(FROM_UNIXTIME(timecode))
FROM timecodeTable
GROUP BY HOUR(FROM_UNIXTIME(timecode));
Edit 2
You can use additional GROUP BY clauses to achieve this:
SELECT
COUNT(*),
YEAR(timecode),
DAYOFYEAR(timecode),
HOUR(timecode)
FROM timecodeTable
GROUP BY YEAR(timecode), DAYOFYEAR(timecode), HOUR(timecode);
Note, I omitted the FROM_UNIXTIME() for brevity.