This is my tokens_log table. I want to list ten users that have most tokens.
So result should be like this:
*# Username sumOfTokens*
#1 tolgay007 500
#2 hzrose 100
...
I tried following sql but didn't work.
SELECT * , SUM( token )
FROM `user_tokens_list`
GROUP BY username
ORDER BY SUM( token ) DESC
LIMIT 10
How can i achive this?
You need the column in group by instead of *
SELECT username , SUM( token )
FROM `user_tokens_list`
WHERE transaction ='sale'
GROUP BY username
ORDER BY SUM( token ) DESC
LIMIT 10
and use a proper where
or if you need all the column you need a join on subquery for username
SELECT user_tokens_list.*, t.sum_token
FROM `user_tokens_list`
INNER JOIN (
SELECT username , SUM( token ) sum_token
FROM `user_tokens_list`
transaction ='sale'
GROUP BY username
ORDER BY SUM( token ) DESC
LIMIT 10
) t on t.username = user_tokens_list.username
You would need to list all non-aggregated columns in the group by clause, otherwise you would get errors (unless option ONLY_FULL_GROUP_BY is disabled); this is one reason why select * with group by is not a good practice.
select username, sum(token) total_tokens
from mytable
group by username
order by total_tokens desc
Note: mysql allows table aliases in the order by clause; this can make the query easier to understand when the expressions become more complex than just a sum()
Related
Is it possible to order when the data comes from many select and union it together? Such as
In this statement, the vouchers data is not showing in the same sequence as I saved on the database, I also tried it with "ORDER BY v_payments.payment_id ASC" but won't be worked
( SELECT order_id as id, order_date as date, ... , time FROM orders WHERE client_code = '$searchId' AND order_status = 1 AND order_date BETWEEN '$start_date' AND '$end_date' ORDER BY time)
UNION
( SELECT vouchers.voucher_id as id, vouchers.payment_date as date, v_payments.account_name as name, ac_balance as oldBalance, v_payments.debit as debitAmount, v_payments.description as descriptions,
vouchers.v_no as v_no, vouchers.v_type as v_type, v_payments.credit as creditAmount, time, zero as tax, zero as freightAmount FROM vouchers INNER JOIN v_payments
ON vouchers.voucher_id = v_payments.voucher_id WHERE v_payments.client_code = '$searchId' AND voucher_status = 1 AND vouchers.payment_date BETWEEN '$start_date' AND '$end_date' ORDER BY v_payments.payment_id ASC , time )
UNION
( SELECT return_id as id, return_date as date, ... , time FROM w_return WHERE client_code = '$searchId' AND w_return_status = 1 AND return_date BETWEEN '$start_date' AND '$end_date' ORDER BY time)
Wrap the sub-select queries in the union within a SELECT
SELECT id, name
FROM
(
SELECT id, name FROM fruits
UNION
SELECT id, name FROM vegetables
)
foods
ORDER BY name
If you want the order to only apply to one of the sub-selects, use parentheses as you are doing.
Note that depending on your DB, the syntax may differ here. And if that's the case, you may get better help by specifying what DB server (MySQL, SQL Server, etc.) you are using and any error messages that result.
You need to put the ORDER BY at the end of the statement i.e. you are ordering the final resultset after union-ing the 3 intermediate resultsets
To use an ORDER BY or LIMIT clause to sort or limit the entire UNION result, parenthesize the individual SELECT statements and place the ORDER BY or LIMIT after the last one. See link below:
ORDER BY and LIMIT in Unions
(SELECT a FROM t1 WHERE a=10 AND B=1)
UNION
(SELECT a FROM t2 WHERE a=11 AND B=2)
ORDER BY a LIMIT 10;
I am trying to order the results of a MySQL query by order of 'Amount (descending)' and then UserName (ascending)' however I'm not getting the results I'm looking for.
SELECT UserName
, Sum(Amount)
FROM UserPowerUps
GROUP
BY UserName
HAVING COUNT(DISTINCT UserName) = 1
ORDER
BY Amount DESC
, UserName ASC;
Above is the MySQL code and below is the query output.
Query result
Assigning the sum to an alias should do the trick
SELECT UserName, Sum(Amount) as sumAmount
FROM UserPowerUps
GROUP BY UserName
HAVING COUNT(DISTINCT UserName) = 1
ORDER BY sumAmount DESC, UserName ASC;
Add the instruction sum in the order by clause:
SELECT UserName, Sum(Amount)
FROM UserPowerUps
GROUP BY UserName
HAVING COUNT(DISTINCT UserName) = 1
ORDER BY SUM(Amount) DESC, UserName ASC;
Does anyone know what's wrong with this query?
This works perfectly on its own:
SELECT * FROM
(SELECT * FROM data WHERE site = '".$id."'
AND disabled = '0'
AND carvotes NOT LIKE '0'
AND (time > ( now( ) - INTERVAL 14 DAY ))
GROUP BY car ORDER BY carvotes DESC LIMIT 0 , 10)
X order by time DESC
So does this:
SELECT * FROM data WHERE site = '".$id."' AND disabled = '0' GROUP BY car DESC ORDER BY time desc LIMIT 0 , 30
But combining them like this:
SELECT * FROM data WHERE site = '".$id."' AND disabled = '0' AND car NOT IN (SELECT * FROM
(SELECT * FROM data WHERE site = '".$id."'
AND disabled = '0'
AND carvotes NOT LIKE '0'
AND (time > ( now( ) - INTERVAL 14 DAY ))
GROUP BY car ORDER BY carvotes DESC LIMIT 0 , 10)
X order by time DESC) GROUP BY car DESC ORDER BY time desc LIMIT 0 , 30
Gives errors. Any ideas?
Please try the following...
$result = mysqli_query( $con,
"SELECT *
FROM data
WHERE site = '" . $id .
"' AND disabled = '0'
AND car NOT IN ( SELECT car
FROM ( SELECT car,
carvotes
FROM data
WHERE site = '" . $id .
"' AND disabled = '0'
AND carvotes NOT LIKE '0'
AND ( time > ( NOW( ) - INTERVAL 14 DAY ) )
GROUP BY car
ORDER BY carvotes DESC
LIMIT 10 ) X
)
GROUP BY car
ORDER BY time DESC
LIMIT 30" );
The main cause of your problem is that with car NOT IN ( SELECT * FROM ( SELECT *... you are trying to compare each record's value of car with each row returned by your subquery. IN requires you to have the same number of fields on both sides of the comparison. By using SELECT * at both levels of the subquery you were ensuring that the right side of the comparison had however many fields are in data versus your single field on the left, which confused MySQL.
Since you are aiming to compare to a single field, namely car, our subquery has to select just the car field from its dataset. Since the sort order of the subquery's results has no effect upon the IN comparison, and since our innermost query will be returning just car, I have removed the outer level of the subquery.
Beyond changing the first part of the subquery to SELECT car, the only other change that I have made to the subquery is to change LIMIT 0, 10 to LIMIT 10. The former means limit to the the 10 records that are offset by 0 from the first record. This is useful if you want records 6 to 15, but redundant for 1 to 10 as LIMIT 10 has the same affect and is slightly simpler. Ditto for LIMIT 0, 30 at the end of your overall statement.
As for the main body of the statement, I have not made any attempt to specify what fields (or aggregate functions of those fields) should be returned since you have made no statement indicating what your requirements / preferences are. If you are satisfied that GROUP BY has left you with a still valid set of values, then all the good, but if not then I recommend that you rewrite your Question to be specific about that detail.
By default, MySQL sorts the data subjected to a GROUP BY into ascending order, but if an ORDER BY clause is also present then it overrides the GROUP BY's sort pattern. As such, there is no benefit to specifying DESC after either of your GROUP BY car clauses, so I have removed it where it occurs.
Interesting Sidenote : You can override a GROUP BY's sort by specifying ORDER BY NULL.
If you have any questions or comments, then please feel free to post a Comment accordingly.
Further Reading
https://dev.mysql.com/doc/refman/5.7/en/order-by-optimization.html - on optimising your ORDER BY sorting
https://dev.mysql.com/doc/refman/5.7/en/select.html - on the SELECT statement's syntax - specifically the parts to do with LIMIT.
https://www.w3schools.com/php/php_mysql_select_limit.asp - a simpler explanation of LIMIT
This is your query:
SELECT *
FROM data
WHERE site = '".$id."' AND disabled = '0' AND
car NOT IN (SELECT *
FROM (SELECT *
FROM data
WHERE site = '".$id."' AND
disabled = '0' AND
carvotes NOT LIKE '0' AND
(time > ( now( ) - INTERVAL 14 DAY ))
GROUP BY car
ORDER BY carvotes DESC
LIMIT 0 , 10
) x
ORDER BY time DESC
)
GROUP BY car DESC
ORDER BY time desc
LIMIT 0 , 30 ;
Several comments:
Do not wrap integer constants in single quotes. This can mislead people. This can mislead optimizers.
Do not use string functions on integers (such as like). Same reason.
NOT IN with subqueries is dangerous. The construct does not handle NULL values the way you expect. Use NOT EXISTS or LEFT JOIN instead.
When using subqueries, ORDER BY is almost never appropriate.
Never use SELECT * with GROUP BY. It is just wrong. Happily, MySQL 5.7 has changed its defaults to reject this anti-pattern
So, a better way to write this query is something like this:
SELECT d.car, MAX(time) as time
FROM data d LEFT JOIN
(SELECT d2.*
FROM data d2
WHERE d2.site = '".$id."' AND
d2.disabled = 0 AND
d2.carvotes NOT LIKE 0 AND
(d2.time > ( now( ) - INTERVAL 14 DAY ))
GROUP BY d2.car
ORDER BY carvotes DESC
LIMIT 0 , 10
) car10
ON d.car = car10.car
WHERE d.site = '".$id."' AND d.disabled = 0' AND
car10.car IS NOT NULL
GROUP BY car DESC
ORDER BY MAX(time) desc
LIMIT 0 , 30 ;
Alternatively, use SELECT * and remove the GROUP BY in the outer query.
I have a list of login logs from our website, however I am needing to see which user ID has had the most IP's logged into it. Our table is as follows:
userid, ip, date (unix)
I need it to output which userid's have had the most IP's logged into them.
I've tried something such as:
SELECT
userID
FROM loginLogs
GROUP BY userID
HAVING COUNT( DISTINCT ip ) > 1
But that just shows a list of user ID's.
Select userID, count(distinct ip)
from loginLogs
Group by 1
Order by 2 desc
Maybe like this?
SELECT `userID`, count(`ip`) cnt FROM `loginLogs` GROUP BY `userID` HAVING cnt > 1
You can just order by distinct values, descending;
SELECT userID, COUNT(DISTINCT ip) `distinct IP#s`
FROM loginLogs
GROUP BY userID
ORDER BY `distinct IP#s` DESC;
An SQLfiddle to test with.
SELECT userID, COUNT(*) AS count FROM loginLogs
GROUP BY userId ORDER BY count DESC
This will give you all of your users from most logged in to the least. Use LIMIT 1 if you want to limit the results.
You have to order those results order by COUNT( DISTINCT ip ) desc and take the first Limit 0, 1
SELECT `userID`
FROM `loginLogs`
GROUP BY `userID`
ORDER BY COUNT( DISTINCT `ip` ) desc
LIMIT 0, 1
You could wrap what you have in a subquery to get the list of userIDs and distinct IPs, as well.
SELECT DISTINCT ll.`userID`, ll.`ip`
FROM ( SELECT `userID`, COUNT( 1 ) AS Cnt
FROM `loginLogs`
GROUP BY `userID`
HAVING COUNT( DISTINCT `ip` ) > 1 ) id
LEFT JOIN `loginLogs` ll
ON id.`userID` = ll.`userID`
ORDER BY id.`Cnt`;
If you just want to see the user with the most ips and you also want to see the list of ips, you can use GROUP_CONCAT():
SELECT `userID`, group_concat(DISTINCT `ip`)
FROM `loginLogs`
GROUP BY `userID`
ORDER BY COUNT( DISTINCT `ip` ) DESC
LIMIT 1
This error occurs while I run the select query, please advise.
Error No : 1111
Error : Invalid use of group function
Query:
SELECT cm.contactid, cm.firstname, cm.surname, cm.mobile, COUNT( * ) total
FROM sales_master as sm, contact_master as cm
WHERE sm.contactid = cm.contactid
AND cm.mobile != '' AND orderdate>='2012-12-18' AND orderdate<='2013-03-18'
GROUP BY sm.contactid
HAVING COUNT(*) >= 1
ORDER BY COUNT(*) DESC, cm.firstname ASC
Change your query like this. Use join instead of cartisian. If type of cm.orderdate is DAte than remove DATE() from the below query. And use alias total in GROUP BY and ORDER BY
instead of Count(*) .
SELECT
cm.contactid,
cm.firstname,
cm.surname,
cm.mobile,
COUNT(cm.contactid) total
FROM sales_master as sm
LEFT JOIN contact_master as cm
ON sm.contactid = cm.contactid
WHERE
AND cm.mobile != ''
AND DATE(cm.orderdate) >= '2012-12-18'
AND DATE(cm.orderdate) <= '2013-03-18'
GROUP BY cm.contactid
HAVING total >= 1
ORDER BY total DESC, cm.firstname ASC
There will be two reason:
1) Version compatibility Problem
2) Your query syntax is not ok,I am sending you the solution of second option.
For Version compatibility Problem you must go to the link mentioned by some one on
comment.
You have to use subquery concept for that.
Modify your query something like that:
SELECT sm.contactid ,cm.contactid as contactid , cm.firstname as firstname, cm.surname as surname , cm.mobile as mobile , COUNT( * ) total
FROM (
SELECT sm.contactid,contactid , firstname,surname ,mobile ,total
FROM sales_master as sm, contact_master as cm
WHERE sm.contactid = cm.contactid
AND cm.mobile != '' AND orderdate>='2012-12-18' AND orderdate<='2013-03-18'
GROUP BY sm.contactid
) q
GROUP BY sm.contactid HAVING COUNT(*) >= 1
ORDER BY COUNT(*) DESC, cm.firstname ASC
I am not sure that i made exact query or not but concept is some thing like that means
"Subquery"...