This s my first question on SO, so please bear if I am not super clear! I am trying to sum up values of 2 columns from 2 tables. Both table values have a common ID called 'imid'. These imids are further divided as 'pmid' (say 1 imid can have one or multiple pmids). Both tables have different structures. I would want to sum values on 1 column from table1 and another column from table(to use it for a php calculation). When I try JOIN it always gives me a timeout error. Query below.
SELECT F.`imid` AS imid, SUM( F.Impression ) AS si, D.accmgr AS accmgr, D.cmname AS cmname,sum(D.Item_Impression_Cap) AS sim
FROM trasactions F, rawdata D
WHERE F.`imid` = D.`imid`
AND F.EntryDate LIKE '2017-%-%'
GROUP BY D.`imid`, F.`imid` ORDER BY F.`imid` ASC
I get results, but not even close to the correct numbers.(eg) Impression as 6,557,824 instead of 1,233,287 for a particular imid.
Not sure where I am wrong! Any help would be great...
Edit: Thanks for the responses...I managed to write the query...
SELECT F.`imid`, F.si , F.imname,D.Item_Start_Date ,D.sim, D.Item_End_Date, D.accmgr, D.cmname
FROM (SELECT `imid` AS imid, SUM(Impression ) AS si, adname AS imname, EntryDate FROM trasactions GROUP BY imid) F LEFT JOIN
(SELECT imid,Item_Start_Date, Item_End_Date, accmgr AS accmgr, cmname AS cmname, sum(Item_Impression_Cap) AS sim FROM rawdata GROUP BY imid) D ON F.`imid`=D.`imid`
WHERE D.cmname IS NOT NULL
GROUP BY F.`imid`
ORDER BY F.`imid` ASC
Now there is a new question! I am doing some calculations based on the array values derived from the query above...(eg) $pacing = $row['si']/$avgday*100;
Say I am listing the 'Pacing' for all items active. Would it be possible to count values from variables. (eg) Would want to show the count of items where $pacing is less than 100%. Would it even be possible to do that!! Thanks again.
Related
I want make a leaderboard from my database, I already make a code but the sql says error
"SET #a1 = (SELECT SUM(d.poin_diskusi)
FROM diskusi d
GROUP BY d.id_akun)
MySQL said: Documentation
#1242 - Subquery returns more than 1 row"
heres my sql query :
SET #a1 = (SELECT SUM(d.poin_diskusi)
FROM diskusi d
GROUP BY d.id_akun);
SET #a2 = (SELECT SUM(j.nilai_jawaban)
FROM jawaban j
GROUP BY j.id_akun);
SET #hasil = #a1 + #a2;
SELECT #hasil
Thank you !
this is an example data in table jawaban:
this is an example data in table diskusi:
and my desired answer:
where tera123 have 135 TOTAL(d.poin_diskusi+j.nilai_jawaban) and david123 have 90 TOTAL(d.poin_diskusi+j.nilai_jawaban)
the point is to show TOTAL for every user
It looks to me (without looking at your images; if you want to show information, please provide it in text) like you want something like:
select id_akun, sum(point_value) as total_points from (
select id_akun, sum(poin_diskusi) as point_value
from diskusi group by id_akun
union all
select id_akun, sum(nilai_jawaban) as point_value
from jawaban group by id_akun
) as point_values
group by id_akun
The problem are the GROUP BY keywords. Inside the images you provided with example data, one can see that there are multiple entries for the columns jawaban.id_akun and diskusi.id_akun. Because of that, both SELECT will return multiple rows as result, for each different value of the id columns one summed up row. Try it without the GROUP BY clause.
I have two tables:
table 1.a
id--entry_date-amount
============================
2---2016-04-14--$400
3---2016-04-14--$400
4----2017-07-14--$200
5---2017-07-14--$500
6---2017-05-14--$600
7----2017-06-18--$100
table 2.b
id--entry_date
===========================
2---2016-04-14--$230
3---2016-04-14--$230
4----2017-07-14--$567
5---2017-07-14--$600
6---2017-05-14--$560
7----2017-06-18--$90
8---2016-04-14--$100
from the two tables how can i get count with montwise
my desired result:
month_name--total(count form table a)--total(count form table b)--amount(table a)--amount(table b)
========================================================
April,16-----------2-------------------3---$800-$500
May,17-----------1-------------------1 --$600--$700
June,17-----------2-------------------2--$100--$800
July,17-----------2-------------------2---$700-$400
this is the demo data.
I want to show data from two tables in a single query month wise.
How can i do this?
I tried:
SELECT MONTHNAME(r.entry_date),r.a_total FROM(
SELECT
IFNULL((SELECT COUNT(tr.id) AS amount FROM a AS tr WHERE MONTH(tr.entry_date)=MONTH(t.entry_date)),0) AS a_total
,t.entry_date
FROM(SELECT tr.id,tr.entry_date
FROM a AS tr
WHERE DATE(tr.entry_date) BETWEEN '2017-07-01' AND '2018-06-30') t
GROUP BY MONTH(t.entry_date)) r
But takes 58 seconds for simple query. How can i make this in a simple query?
You can get the counts and sum from each table individually, then use UNION to combine the two result sets into one result set. Something like this :
SELECT Month_name,
SUM(aCount) AS aCount,
SUM(bCount) AS bCount,
SUM(aAmount) AS aAmount,
SUM(bAmount) AS bAmount
FROM
(
SELECT
MONTHNAME(a.entry_date) AS Month_name,
COUNT(a.id) AS aCount,
0 AS bCount,
SUM(a.amount) AS aAmount,
0 AS bAmount
FROM a
GROUP BY MONTHNAME(a.entry_date)
UNION ALL
SELECT
MONTHNAME(b.entry_date) AS Month_name,
0 AS aCount,
COUNT(b.id) AS bCount,
0 AS aAmount,
SUM(b.amount) AS bAmount
FROM b
GROUP BY MONTHNAME(b.entry_date)
) AS t
GROUP BY Month_Name;
live demo
user9131497 has a good design for the big picture. However, I would suggest stuff like this for handling the dates:
SELECT DATE_FORMAT(entry_date, "%M,%y") AS 'Month',
COUNT(*) AS 'aCount'
FROM a
GROUP BY LEFT(entry_date, 7) -- eg, "2017-03"
Try that to see what I mean.
Note that this will work beyond a year. Or did you need January values from all years to be combined?? -- That's what your solution and user9131497's will do. Mine keeps them separate.
I’m designing a program for my school to keep student attendance records. So far I have the following query working fine and now I would like to add an IF statement to perform a percentage operation when a certain condition is given. As it is, the query is using INNER JOIN to search for data from two different tables (oxadmain and stuattend) and it’s displaying the results well on a results table:
SELECT o.name
, o.year
, o.photoID
, o.thumbs
, s.ID
, s.studid
, s.date
, s.teacher
, s.subject
, s.attendance
FROM stuattend s
JOIN oxadmain o
ON s.studid = o.stuid
ORDER
BY name ASC
Now I would like to add an “if” statement that
1) finds when stuattend.attendance is = Absent, calculates the percentage of absences the students may have in any given period of time, and then stores that (%) value in “percentage” and
2) ELSE assigns the value of 100% to “Percentage”.
So far I’ve been trying with the following:
<?php $_GET['studentID'] = $_row_RepeatedRS['WADAstuattend']; ?>
SELECT oxadmain.name , oxadmain.year , oxadmain.photoID , oxadmain.thumbs , stuattend.ID , stuattend.studid , stuattend.date , stuattend.teacher, stuattend.subject , stuattend.attendance
CASE
WHEN stuattend.attendance = Absent THEN SELECT Count (studentID) AS ClassDays, (SELECT Count(*) FROM stuattend WHERE studentID = stuattend.studid AND Absent = 1) AS ClassAbsent, ROUND ((ClassAbsent/ClassDays)*100, 2) AS Percentage
ELSE
Percentage = 100
END
FROM stuattend INNER JOIN oxadmain ON stuattend.studid=oxadmain.stuid
ORDER BY name ASC
Any suggestions on how to do this well?
Thank you for your attention
The base idea would be:
select stuattend.studid, sum(stuattend.attendance = `absent`) / count(*)
from stuattend
group by stuaddend.studid;
This very much depends on exactly one entry per student and per day, and of course gets 0 if no absence and 1 if always absent.
To make this a bit more stable I would suggest to write a calendar day table, which simply keeps a list of all days and a column if this is a school day, so workday=1 means they should have been there and workday=0 means sunday or holiday. Then you could left join from this table to the presence and absence days, and even would give good results when presence is not contained in your table.
Just ask if you decide which way to go.
Hi I am new for developing.Kindly bear my codings. I have created a table arlog with id(auto increment), status, ticket number and code. Ticket and code number is set as unique. That is the duplicate of this combination cannot inserted again. But individually ticket number or cpt can be inserted as many times.It works fine. Now I want to use select query with another table with respect to the arlog table ticket number and code.Here is the select statement
$result = mysql_query("SELECT * FROM `ar` C WHERE provider='".$_SESSION['PROVIDER']
."' AND C.`TicketNo` IN ( SELECT TicketNo FROM `arlog` L where L.status NOT IN('New','Completed','Completed(Ar_aging)
','Completed(Rework)','Rework','Completed_Followup','Completed_Supervising' )
and L.assign='".$_SESSION['NAME']."' ) order by id desc") or die(mysql_error());
The query check the ticket number in arlog and displays correcly. But I want to combine TicketNo and Code in the arlog. I have made research but could not find solution. First of all is it possible?
Please try following sql:
SELECT L.TicketNo ,L.Code,C.* FROM `ar` C left join `arlog` L ON C.TicketNo = L.TicketNo where C.provider='your condition' and L.status NOT IN('New','Completed','Completed(Ar_aging)','Completed(Rework)','Rework','Completed_Followup','Completed_Supervising' ) and L.assign='your condition' order by by C.id desc
Hope this can help you!
I think you need to use CONCAT_WS()
There is a nice example of its usage in below link
MySQL SELECT AS combine two columns into one
I am trying to calculate the difference of values list coming from a database.
I would like to achieve it using php or mysql, but I do not know how to proceed.
I have a table named player_scores. One of its rows contains the goals scored.
Ex.
pl_date pl_scores
03/11/2014 18
02/11/2014 15
01/11/2014 10
I would like to echo the difference between the goals scored during the matches played in different dates.
Ex:
pl_date pl_scores diff
03/11/2014 18 +3
02/11/2014 15 +5
01/11/2014 10 no diff
How can I obtain the desired result?
You seem to want to compare a score against the score on a previous row.
Possibly simplest if done using a a sub query that gets the max pl_date that is less than the pl_date for the current row, then joining the results of that sub query back against the player_scores table to get the details for each date:-
SELECT ps1.pl_date, ps1.pl_scores, IF(ps2.pl_date IS NULL OR ps1.pl_scores = ps1.pl_scores, 'no diff', ps1.pl_scores - ps1.pl_scores) AS diff
FROM
(
SELECT ps1.pl_date, MAX(ps2.pl_date) prev_date
FROM player_scores ps1
LEFT OUTER JOIN player_scores ps2
ON ps1.pl_date > ps2.pl_date
GROUP BY ps1.pl_date
) sub0
INNER JOIN player_scores ps1
ON sub0.pl_date = ps1.pl_date
LEFT OUTER JOIN player_scores ps2
ON sub0.prev_date = ps2.pl_date
There are potentially other ways to do this (for example, using variables to work through the results of an ordered sub query, comparing each row with the value stored in the variable for the previous row)
SELECT score FROM TABLE WHERE DATE = TheDateYouWant
$score = $data['score'];
SELECT score FROM TABLE WHERE date = dateYouWant
$difference = $score - $data['score'];
Something like this?
You could use two queries, one to get the value to use in the comparison (in the example below is the smaller number of scores) and the second one to get the records with a dedicated column with the difference:
SELECT MIN(pl_scores);
SELECT pl_date, pl_scores, (pl_scores - minScore) as diff FROM player_scores;
Or, using a transaction (one query execution php side):
START TRANSACTION;
SELECT MIN(Importo) FROM Transazione INTO #min;
SELECT Importo, (Importo - #min) as diff FROM Transazione;
select *,
coalesce(
(SELECT concat(IF(t1.pl_scores>t2.pl_scores,'+',''),(t1.pl_scores-t2.pl_scores))
FROM tableX t2 WHERE t2.pl_date<t1.pl_date ORDER BY t2.pl_date DESC LIMIT 1)
, 'no data' ) as diff
FROM tableX t1
WHERE 1
order by t1.pl_date DESC