distinct and sum if like - sql

Question

I have a table as the following
name
-----------
1#apple#1
2#apple#2
3#apple#4
4#box#4
5#box#5
and I want to get the result as:
name
--------------
apple 3
box 2
Thanks in advance for your help

Answer 1

This is what you need.
select
SUBSTRING(
name,
CHARINDEX('#', name) + 1,
LEN(name) - (
CHARINDEX('#', REVERSE(name)) + CHARINDEX('#', name)
)
),
count(1)
from
tbl
group by
SUBSTRING(
name,
CHARINDEX('#', name) + 1,
LEN(name) - (
CHARINDEX('#', REVERSE(name)) + CHARINDEX('#', name)
)
)

Answer 2

If your data does not contain any full stops (or periods depending on your vernacular), and the length of your string is less than 128 characters, then you can use PARSENAME to effectively split your string into parts, and extract the 2nd part:
DECLARE #T TABLE (Val VARCHAR(20));
INSERT #T (Val)
VALUES ('1#apple#1'), ('2#apple#2'), ('3#apple#4'),
('4#box#4'), ('5#box#5');
SELECT Val = PARSENAME(REPLACE(t.Val, '#', '.'), 2),
[Count] = COUNT(*)
FROM #T AS t
GROUP BY PARSENAME(REPLACE(t.Val, '#', '.'), 2);
Otherwise you will need to use CHARINDEX to find the first and last occurrence of # within your string (REVERSE is also needed to get the last position), then use SUBSTRING to extract the text between these positions:
DECLARE #T TABLE (Val VARCHAR(20));
INSERT #T (Val)
VALUES ('1#apple#1'), ('2#apple#2'), ('3#apple#4'),
('4#box#4'), ('5#box#5');
SELECT Val = SUBSTRING(t.Val, x.FirstPosition + 1, x.LastPosition - x.FirstPosition),
[Count] = COUNT(*)
FROM #T AS t
CROSS APPLY
( SELECT CHARINDEX('#', t.Val) ,
LEN(t.Val) - CHARINDEX('#', REVERSE(t.Val))
) AS x (FirstPosition, LastPosition)
GROUP BY SUBSTRING(t.Val, x.FirstPosition + 1, x.LastPosition - x.FirstPosition);

Answer 3

use case when
select case when name like '%apple%' then 'apple'
when name like '%box%' then 'box' end item_name,
count(*)
group by cas when name like '%apple%' then 'apple'
when name like '%box%' then 'box' end

Answer 4

No DBMS specified, so here is a postgres variant. The query does use regexps to simplify things a bit.
with t0 as (
select '1#apple#1' as value
union all select '2#apple#2'
union all select '3#apple#4'
union all select '4#box#4'
union all select '5#box#5'
),
trimmed as (
select regexp_replace(value,'[0-9]*#(.+?)#[0-9]*','\1') as name
from t0
)
select name, count(*)
from trimmed
group by name
order by name
DB Fiddle
Update
For Oracle DMBS, the query stays basically the same:
with t0 as (
select '1#apple#1' as value from dual
union all select '2#apple#2' from dual
union all select '3#apple#4' from dual
union all select '4#box#4' from dual
union all select '5#box#5' from dual
),
trimmed as (
select regexp_replace(value,'[0-9]*#(.+?)#[0-9]*','\1') as name
from t0
)
select name, count(*)
from trimmed
group by name
order by name
NAME | COUNT(*)
:---- | -------:
apple | 3
box | 2
db<>fiddle here
Update
MySQL 8.0
with t0 as (
select '1#apple#1' as value
union all select '2#apple#2'
union all select '3#apple#4'
union all select '4#box#4'
union all select '5#box#5'
),
trimmed as (
select regexp_replace(value,'[0-9]*#(.+?)#[0-9]*','$1') as name
from t0
)
select name, count(*)
from trimmed
group by name
order by name
name | count(*)
:---- | -------:
apple | 3
box | 2
db<>fiddle here

Answer 5

You can use case and group by to do the same.
select new_col , count(new_col)
from
(
select case when col_name like '%apple%' then 'apple'
when col_name like '%box%' then 'box'
else 'others' end new_col
from table_name
)
group by new_col
;