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on which day most people login?
Use DAYNAME() like that.
SELECT DAYNAME([column_with_date])
FROM [table]
GROUP BY DAYNAME([column_with_date])
order by count(DAYNAME([column_with_date])) desc limit 1;
You can use below code to find max frequency
select date,max(freq)
from (SELECT date,count(date) as freq from TableName
group by date);
this will give the Date and max frequency of that date in column.
or you can use below code
SELECT date,count(date) as freq from TableName
group by date
limit 1;
here i'm assuming date as column name.
Related
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how to rank each country based on number of people in SQL.
Try
SELECT * FROM (
SELECT country, count(people) as count FROM your_table GROUP BY country
) ORDER BY count DESC
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How can I write this query without TOP and Limit only use the SQL
You just sort the data one way, grab the records you want, then sort it the other way:
SELECT SOME_FIELD
FROM (SELECT st.*
FROM SOME_TABLE st
ORDER BY SOME_FIELD DESC)
WHERE ROWNUM <= 7
ORDER BY SOME_FIELD ASC
dbfiddle here
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I have data with column createdate like '24/04/2019 14:52:38',24/04/2019 14:52:37,24/04/2019 14:52:35,24/03/2019 14:52:38 etc.
how to get data based on max date and time in SQL query.
select
top(1) *
from
xx
order by
createdate desc
something like this? (works if createdate column is of date/datetime/timestamp type)
select
*
from
<table_name>
where
createdate = (select
max(createdate)
from
<table_name>)
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How can I put 1,2,3,4.... in sequence order in column salary where salary is null?? in oracle
I tried many things but could not find any accurate result
select nvl(column,1) as columnName from yourTable
Use ROWNUM pseudocolumn with nvl function:
select nvl(salary, ROWNUM ) from yourTable
ROWNUM pseudocolumn returns a number indicating the order in which Oracle selects the row from a table or set of joined rows. The first row selected has a ROWNUM of 1, the second has 2, and so on.
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What's the simplest SQL statement that will return the duplicate values for a given column for example ORDERS_NUMBERS and group it by oldest CREATION_DATE in an Oracle database table?
SELECT ORDER_NUMBERS, count(*)
FROM YourTableName
GROUP_BY ORDER_NUMBERS
HAVING count(*) > 1
ORDER BY CREATION_DATE
select ORDERS_NUMBERS,CREATION_DATE,count(*)
from
table
group by ORDERS_NUMBERS,CREATION_DATE
order by CREATION_DATE desc
having count(*) > 1
select
t.orders_numbers,
t.creation_date
from table t
group by t.orders_numbers,
t.creation_date
having (count (*) >1)
;