rows with max value - sql

I'm wanting to retrieve all rows and columns which contain the MAX value in one column.
For example: 3 columns - ID (not unique), name, age
I'm wanting all rows and columns which contain the max value in age.
Thanks

SELECT * FROM table WHERE age = (SELECT MAX(age) FROM table)

Select * from table where value=max(value)
Maybe you need to exchange where with having

Related

In Snowflake, I want to count duplicates in a table based on all the columns in the table without typing out every column name

I have a table with 60 columns in it. I would like to identify how many duplicates there are in the table based on all the columns being identical.
I don't want to have to type out every field name in the SELECT or GROUP BY clauses. Is there a way to do that?
You can use an approach like this for each table:
SELECT
MD5(OBJECT_CONSTRUCT(SRC.*)::VARCHAR) DUP_MD5, SUM(1) AS TOTAL_COUNT
FROM <table> SRC
GROUP BY 1
HAVING SUM(1) > 1;

Determine the number of times a null value occurs in column B for a distinct value in column A, SQL table

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.

Split a column in postgres in select query

I have a column with multiple employee id's with comma separation. Need to retrieve data with select query with where condition of that employee id's column but need to pass one id at a time. In the below employee id column have multiple id's.
SELECT lt.*
FROM leave as lt
WHERE true
and employee_id IN ($employee_id)
So, try something like this:
SELECT * FROM ... WHERE id = ANY (string_to_array(ids, ',')::integer[])

find unique rows using SQL?

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;

How to select 10 rows below the result returned by the SQL query?

Here is the SQL table:
KEY | NAME | VALUE
---------------------
13b | Jeffrey | 23.5
F48 | Jonas | 18.2
2G8 | Debby | 21.1
Now, if I type:
SELECT *
FROM table
WHERE VALUE = 23.5
I will get the first row.
What I need to accomplish is to get the first and the next two rows below. Is there a way to do it?
Columns are not sorted and WHERE condition doesn't participate in the selection of the rows, except for the first one. I just need the two additional rows below the returned one - the ones that were entered after the one which has been returned by the SELECT query.
Without a date column or an auto-increment column, you can't reliably determine the order the records were entered.
The physical order with which rows are stored in the table is non-deterministic.
You need to define an order to the results to do this. There is no guaranteed order to the data otherwise.
If by "the next 2 rows after" you mean "the next 2 records that were inserted into the table AFTER that particular row", you will need to use an auto incrementing field or a "date create" timestamp field to do this.
If each row has an ID column that is unique and auto incrementing, you could do something like:
SELECT * FROM table WHERE id > (SELECT id FROM table WHERE value = 23.5)
If I understand correctly, you're looking for something like:
SELECT * FROM table WHERE value <> 23.5
You can obviously write a program to do that but i am assuming you want a query. What about using a Union. You would also have to create a new column called value_id or something in those lines which is incremented sequentially (probably use a sequence). The idea is that value_id will be incremented for every insert and using that you can write a where clause to return the remaining two values you want.
For example:
Select * from table where value = 23.5
Union
Select * from table where value_id > 2 limit 2;
Limit 2 because you already got the first value in the first query
You need an order if you want to be able to think in terms of "before" and "after".
Assuming you have one you can use ROW_NUMBER() (see more here http://msdn.microsoft.com/en-us/library/ms186734.aspx) and do something like:
With MyTable
(select row_number() over (order by key) as n, key, name, value
from table)
select key, name, value
from MyTable
where n >= (select n from MyTable where value = 23.5)