I used the following query and it returned what I wanted it to return, but I'm having a tough time wrapping my head around what the query is doing.
Query is nothing fancier than what's in the title: select distinct(count(*)) from table1
Distinct is not required in your SQL ,as you are going to get only result, count(*) without group by clause returns, count of all rows within that table.
Hence try this :
select count(*) from table1
Distinct is used for finding distinct values from a group of values:
say you have table1 , with column1 as :
Column1
----------
a
a
b
b
a
c
following sqls are run you will get output as :
1) select count(*) from table1
output :6
2) select distinct(count(*)) from table1
output :6
3) select count( distinct column1) from table1
output :3
Usually distinct is used inside count preferably with a particular column .
select count( distinct column_name_n ) from table1
The distinct is redundant... Select Count(*) with only one table can only generate one value, so distinct (which would eliminate duplicates) is irelelvant.
If you had multiple outputs, (if for example you were grouping on something) then it would cause the query to only display one output row for every distinct value of count(*) that would other wise be generated...
if, for example, you had
name
Bob
Bob
Bob
Bob
Mary
Mary
Mary
Mary
Dave
Dave
Al
George
then
select count(*)
From table
group By name
would result in
4
4
2
1
1
but
select distinct count(*)
From table
group By name
would result in
4
2
1
Related
I want a postgre SQL query that selects only first row from table against distinct LocalID and enter the result in another table.
Records:
ID| LocalID| Name
1 233 Tim
2 633 John
3 633 Alex
4 234 Mike
5 233 Dave
6 556 Kim
Wanted result:
ID| LocalID| Name
1 233 Tim
2 633 John
4 234 Mike
6 556 Kim
I tried using
CREATE TABLE Weeklylist AS (select distinct on (localid) * from Monthlylist)
But this query select the last distinct record and enters it into the table. All i want is the first occurrence of the row containing distinct localID should be entered in the table.
The use of distinct on in your existing statement indicates that you are using Postgres.
The problem with your query is that it is missing an ORDER BY clause. Without it, it is undefined which record will be selected (you are seeing the last record being picked, but this is not guaranteed to be consistent over subsequent executions of the same query). So, add the ORDER BY clause:
create table Weeklylist as
select distinct on (localid) * from Monthlylist order by localid, id
Side note: parentheses around the select statement are superfluous here.
You can use DISTINCT ON in PostgreSQL :
CREATE TABLE Weeklylist
AS
SELECT DISTINCT ON (LocalID) *
FROM Monthlylist ml
ORDER BY LocalID, ID -- Missing in your query
In MySQL older version correlated sub-query is one way :
SELECT ml.*
FROM Monthlylist ml
WHERE ml.id = (SELECT MIN(ml1.id) FROM Monthlylist ml1 WHERE ml1.LocalID = ml.LocalID);
This will give you what you need:
select *
from Monthlylist
where id in (
select min(id)
from Monthlylist
group by localid
)
create table WeeklyList as
select *
from Monthlylist
where id in (
select min(id)
from Monthlylist
group by localid
)
Demo on DB Fiddle
You need a subquery and join to get your desired output.
create table Weeklylist AS (
select t.* from Monthlylist t
inner join (select distinct on (localid) * from Monthlylist) t1 on t1.localid = t.localid and t.id = t1.id
order by id, localid)
see sqlfiddle
Is there any difference between COUNT(*) and COUNT(attribute_name)?
I used count(attribute_name) as I thought that it would be specific hence the searching process would be easier. Is that true?
It would be great to see any example with sql code with my issue to help me understand better
Imagine this table:
select Count(TelephoneNumber) from Calls -- returns 3
select Count(*) from Calls -- returns 4
count(column_name) also counts duplicate values. Consider:
select Count(TelephoneNumber) from Calls -- returns 4
COUNT(*) counts all the records in the group.
COUNT(column_name) only counts non-null values.
There is also another typical expression, COUNT(DISTINCT column_name), that counts non-null distinct values.
Since you asked for it, here is a demo on DB Fiddlde:
with t as (
select 1 x from dual
union all select 1 from dual
union all select null from dual
)
select count(*), count(x), count(distinct x) from t
COUNT(*) | COUNT(X) | COUNT(DISTINCTX)
-------: | -------: | ---------------:
3 | 2 | 1
COUNT(*) will count all the rows.
COUNT(column) will count non-NULLs only.
Your can use of COUNT(*) or COUNT(column) which should be based on the desired output only.
Consider below Example of employee table
ID Name Description
1 Raji Smart
2 Rahi Positive
3
4 Falle Smart
select count(*) from employee;
Count(*)
4
select count(name) from employee;
Count(Name)
3
count() only counts non-null values. * references the complete row and as such never excludes any rows. count(attribute_name) only counts rows where that column is no null.
So this:
select count(attribute_name)
from the_table
is equivalent to:
select count(*)
from the_table
where attribute_name is not null
The difference is simple: COUNT(*) counts the number of rows produced by the query, whereas COUNT(1) counts the number of 1 values. Note that when you include a literal such as a number or a string in a query, this literal is "appended" or attached to every row that is produced by the FROM clause.
For more detail this link would help you understand.
Does anyone happen to know a way of basically taking the 'Distinct' command but only using it on a single column. For lack of example, something similar to this:
Select (Distinct ID), Name, Term from Table
So it would get rid of row with duplicate ID's but still use the other column information. I would use distinct on the full query but the rows are all different due to certain columns data set. And I would need to output only the top most term between the two duplicates:
ID Name Term
1 Suzy A
1 Suzy B
2 John A
2 John B
3 Pete A
4 Carl A
5 Sally B
Any suggestions would be helpful.
select t.Id, t.Name, t.Term
from (select distinct ID from Table order by id, term) t
You can use row number for this
Select ID, Name, Term from(
Select ID, Name, Term, ROW_NUMBER ( )
OVER ( PARTITION BY ID order by Name) as rn from Table
Where rn = 1)
as tbl
Order by determines the order from which the first row will be picked.
Am trying to fetch the number of records in the table using Count(*) along with my query condition
Sample Table is
Table: STUD_NAME
Id Name
1 Steven
2 smith
2 Ben
1 Willy
My query is
select std.name
from STUD_Name where id='2'
for this it will display the output as "Smith" and "Ben", along with i need the total number of records in the STUD_NAME table.
By right it should display the total records as "4", please help me out to solve this issue and how to form the query in this case
SELECT name,
cnt as total_count
FROM (
SELECT id
name,
count(*) over () as cnt
FROM stud_name
) t
WHERE id = 2
Assuming that id is a numeric column the single quotes around the value 2 are not needed (and are actually harmful due to the implicit data type conversion that happens in the background)
What about:
select
std.name
,(select count(1) from STUD_Name) nrofstds
from STUD_Name std where std.id='2'
select STUD_NAME.name, CNT.count
from STUD_NAME
, (select count(*) COUNT from STUD_NAME) CNT
where id='2'
If I have a table such as
1 bob
1 ray
1 bob
1 ray
2 joe
2 joe
And I want to select distinct based on the two columns so that I would get
1 bob
1 ray
2 joe
How can I word my query? Is the only way to concatenate the columns and wrap them around a distinct function operator?
select distinct id, name from [table]
or
select id, name from [table] group by id, name
You can just do:
select distinct col1, col2 from your_table;
That's exactly what the distinct operator is for: removing duplicate result rows.
Keep in mind that distinct is usually a pretty expensive operation, since, after processing the query, the DB server might perform a sort operation in order to remove the duplicates.