I have this situation in a certain table:
id | name
1 'Test'
2 'Test'
3 'Test'
How can I make a query to SELECT by distinct the name? I also need the ID column, even if I get the first occurrence of the element, e.g. "if the name column repeats, give me the first record with this repetition."
select name, MIN(ID)
from aCertainTable
group by name
Related
Suppose, there is a table and I need to sort one of its column (name) alphabetically and at the same time I want to sort it by using ID column in asc order based on the condition ( rows that have same name). So, I failed to understand how this will work. Once the records will be sorted by column (name) then will it sort all rows by using id column?
Can someone explain how actually order by clause works in this case
select name,
id
from hack h
order by name,
id
use order by name, id
select name,
id
from hack
order by name,
id
I just tried to understand what you want to know, you want to realize how it happens when the order by clause have two or more columns ,am I right? Let's go to an example,
the first column is id and the second is name,
2 A
5 B
6 A
3 A
1 B
the result of SQL "select name,id from hack order by name,id" will get the result as below
A 2
A 3
A 6
B 1
B 5
see, it will sort first by name column, and then sort id in the same name value group.
That's it ,did I make it clear?
This answers the original question.
In the code you posted:
substring(name, len(name) - 2, len(name))
returns the last 3 characters of the name.
So you are sorting by these last 3 characters and not by name.
When there are 2 names with the same last 3 characters these will be sorted by id.
If there are more than one column names after "order by" keyword, the system orders the records according to the first column just after order by.
I have a SQL table with "name" as one column, date as another, and location as a third. The location column supports null values.
I am trying to write a query to determine the number of times a null value occurs in the location column for each distinct value in the name column.
Can someone please assist?
One method uses conditional aggregation:
select name, sum(case when location is null then 1 else 0 end)
from t
group by name;
Another method that involves slightly less typing is:
select name, count(*) - count(location)
from t
group by name;
use count along with filters, as you only requires Null occurrence
select name, count(*) occurances
from mytable
where location is null
group by name
From your question, you'll want to get a distinct list of all different 'name' rows, and then you would like a count of how many NULLs there are per each name.
The following will achieve this:
SELECT name, count(*) as null_counts
FROM table
WHERE location IS NULL
GROUP BY name
The WHERE clause will only retrieve records where the records have NULL as their location.
The GROUP BY will pivot the data based on NAME.
The SELECT will give you the name, and the COUNT(*) of the number of records, per name.
I want to return all the rows from a table which are unique. I.e. if a certain field in two rows contain the same name, that name shouldn't be shown.
Since you want only the uniques names (and not an unique row for every names like you could have with DISTINCT), you have to use a GROUP BY and a HAVING (instead of a WHERE, because your parameter is the result of a function, not a variable) :
SELECT name FROM myTable GROUP BY name HAVING COUNT(name) = 1
SELECT DISTINCT column_name FROM table
If you want the complete rows, then use row_number() or distinct on:
select distinct on (name) t.*
from table t
order by name;
i need help concerning a sql query. first of all, i have a database with the following structure (example):
ID NAME VALUE
123 ABC_A Text Text Text
123 ABC_A Some more Text
123 ABC_A Even more Text
123 ABC_B some other text
now, i want to get all the different values of rows with the name "ABC_A". i tried to get those via group by and having, without success.
IS this what you want?
SELECT DISTINCT Value
FROM tableName
WHERE ID = 123 AND Name = 'ABC_A'
but if the value of the ID and Name are unique then you can omit distinct (to avoid overkilling the server)
SELECT Value
FROM tableName
WHERE ID = 123 AND Name = 'ABC_A'
Additional to John, if you use the keyword Distinct you onle get DIFFERENT Values, therefore
SELECT DISTINCT Value
FROM tableName
WHERE ID = 123 AND
Name = 'ABC_A'
i want to get all the different values of rows with the name "ABC_A"
This would be for example:
SELECT value, count(value) FROM tbl WHERE name = 'ABC_A' GROUP BY value;
If you do not need the count of times one value appears, remove it, or use DISTINCT:
SELECT DISTINCT value FROM tbl WHERE name = 'ABC_A';
If you want the different values of rows by ID also,
SELECT value, count(value)
FROM tbl
WHERE id = 123 AND name = 'ABC_A'
GROUP BY value;
Or if you want "ALL" the different values (with duplicates too) remove the GROUP BY (and you no longer can use the count(), which would be always 1):
SELECT value FROM tbl WHERE id = 123 AND name = 'ABC_A';
The GROUP BY statement is used in conjunction with the aggregate functions to group the result-set by one or more columns.
In above case why dont you use simple where clause
select * from <tableName> where name ='ABC_A'
Imagine you've got a table with 2 columns: ID and NAME. ID is simply a number, incrementing for each row (as you'd expect). NAME is some random varchar string. NAME can be same for different rows. Now, imagine you want to get the 3 latest occurences in this table, where NAME only may occur once.
For example, if you've got this data:
ID NAME
1 HELLO
2 TEST
3 HELLO
4 HELLO
5 QWERTY
6 HELLO
Then the result of the question should be:
6 HELLO
5 QWERTY
2 TEST
Is it possible achieve this on SQL level?
SELECT
MAX(ID),
Name
FROM
table
GROUP BY
Name
ORDER BY
MAX(ID) desc
LIMIT 3
SELECT MAX(ID), NAME
FROM THAT_TABLE
GROUP BY NAME
See: GROUP BY (Aggregate) Functions
I suppose, you need to use "DISTINCT" for the "name" column:
SELECT DISTINCT name, id FROM table_name ORDER BY id DESC LIMIT 3;
Another way to achieve this is to use "GROUP BY" for "name" (see another answer)