Assume there is a table called "allvalues" with a column named "column".
This column contains the values "A" to "J" while missing the "H".
I am given a set of values from "G" to "J".
How can I query the table to see which value of my set is missing in the column?
The following does not work:
select * from allvalues where column not in ('G', 'H', 'I', 'J')
This query would result in A, B, C, D, E, F, H which also contains values not included in the given set.
Obviously in such a small data pool the missing value is noticeable by eye, but imagine more entries in the table and a bigger set.
You need to start with a (derived) table with the values you are checking. One explicit method is:
with testvalues as (
select 'G' as val from dual union all
select 'H' as val from dual union all
select 'I' as val from dual union all
select 'J' as val from dual
)
select tv.val
from testvalues tv
where not exists (select 1 from allvalues av where av.column = tv.val);
Often, the values originate through a query or a table. So explicitly declaring them is unnecessary -- you can replace that part with a subquery.
Depends on which SQL syntax you can use, but basically you want to check your table allvalues + the extra values.
eg:
SELECT *
FROM ALLVALUES
WHERE COLUMN NOT IN (
( select s.column from allvalues s )
and column not in ('G', 'H', 'I', 'J')
this will work:
select * from table1;
G
H
I
J
select * from table1
minus
(select * from table1
intersect
select column from allvalues
)
sample input:
select * from ns_table10;
G
H
I
J
SELECT * FROM ns_table11;
A
B
C
D
E
F
G
J
I
select * from ns_table10
minus
(select * from ns_table10
intersect
select * from ns_table11
);
output:
H
Related
Say that I have the following table:
with data as (
select 'John' "name", 'A' "tag", 10 "count"
union all select 'John', 'B', 20
union all select 'Jane', 'A', 30
union all select 'Judith', 'A', 40
union all select 'Judith', 'B', 50
union all select 'Judith', 'C', 60
union all select 'Jason', 'D', 70
)
I know there are a number of distinct tag values, namely (A, B, C, D).
I would like to select the unique names that only have the tag A
I can get close by doing
-- wrong!
select
distinct("name")
from data
group by "name"
having count(distinct tag) = 1
however, this will include unique names that only have 1 distinct tag, regardless of what tag is it.
I am using PostgreSQL, although having more generic solutions would be great.
You're almost there - you already have groups with one tag, now just test if it is the tag you want:
select
distinct("name")
from data
group by "name"
having count(distinct tag) = 1 and max(tag)='A'
(Note max could be min as well - SQL just doesn't have single() aggregate function but that's different story.)
You can use not exists here:
select distinct "name"
from data d
where "tag" = 'A'
and not exists (
select * from data d2
where d2."name" = d."name" and d2."tag" != d."tag"
);
This is one possible way of solving it:
select
distinct("name")
from data
where "name" not in (
-- create list of names we want to exclude
select distinct name from data where "tag" != 'A'
)
But I don't know if it's the best or most efficient one.
I have the following problem: Show all rows in table where column first_name contains at least 2 vowels (a, e, i, o, u), and the number of occurences of each vowel is the same.
Valid example: Alexander, "e" appears 2 times, "a" appears 2 times. That is coreect.
Invalid example: Jonathan, it has 2 vowels (a, o), but "o" appears once, and "a" appears twice, the number of occurences is not equal.
I've solved this problem by calculating each vowel, and then verify every case (A E, A I, A O etc. Shortly, each combination of 2, 3, 4, 5). With that solution, I have a very long WHERE. Is there any shorter way and more elegant and simple?
This is how I solved it in TSQL in MS SQL Server 2019.
I know its not exactly what you wanted. Just an interesting thing to try. Thanks for that.
DROP TABLE IF EXISTS #Samples
SELECT n.Name
INTO #Samples
FROM
(
SELECT 'Ravi' AS Name
UNION
SELECT 'Tim'
UNION
SELECT 'Timothe'
UNION
SELECT 'Ian'
UNION
SELECT 'Lijoo'
UNION
SELECT 'John'
UNION
SELECT 'Jami'
) AS n
SELECT g.Name,
IIF(MAX (g.Repeat) = MIN (g.Repeat) AND SUM (g.Appearance) >= 2, 'Valid', 'Invalid') AS Validity
FROM
(
SELECT v.value,
s.Name,
SUM (LEN (s.Name) - LEN (REPLACE (s.Name, v.value, ''))) AS Repeat,
SUM (IIF(s.Name LIKE '%' + v.value + '%', 1, 0)) AS Appearance
FROM STRING_SPLIT('a,e,i,o,u', ',') AS v
CROSS APPLY #Samples AS s
GROUP BY v.value,
s.Name
) AS g
WHERE g.Repeat > 0
GROUP BY g.Name
Output
we can replace STRING_SPLIT with a temp table for supporting lower versions
DROP TABLE IF EXISTS #Vowels
SELECT C.Vowel
INTO #Vowels
FROM
(
SELECT 'a' AS Vowel
UNION
SELECT 'e'
UNION
SELECT 'i'
UNION
SELECT 'o'
UNION
SELECT 'u'
) AS C
SELECT g.Name,
IIF(MAX (g.Repeat) = MIN (g.Repeat) AND SUM (g.Appearance) >= 2, 'Valid', 'Invalid') AS Validity
FROM
(
SELECT v.Vowel,
s.Name,
SUM (LEN (s.Name) - LEN (REPLACE (s.Name, v.Vowel, ''))) AS Repeat,
SUM (IIF(s.Name LIKE '%' + v.Vowel + '%', 1, 0)) AS Appearance
FROM #Vowels AS v
CROSS APPLY #Samples AS s
GROUP BY v.Vowel,
s.Name
) AS g
WHERE g.Repeat > 0
GROUP BY g.Name
From Oracle 12, you can use:
SELECT name
FROM table_name
CROSS JOIN LATERAL(
SELECT 1
FROM (
-- Step 2: Count the frequency of each vowel
SELECT letter,
COUNT(*) As frequency
FROM (
-- Step 1: Find all the vowels
SELECT REGEXP_SUBSTR(LOWER(name), '[aeiou]', 1, LEVEL) AS letter
FROM DUAL
CONNECT BY LEVEL <= REGEXP_COUNT(LOWER(name), '[aeiou]')
)
GROUP BY letter
)
-- Step 3: Filter out names where the number of vowels are
-- not equal or the vowels do not occur at least twice
-- and there are not at least 2 different vowels.
HAVING MIN(frequency) >= 2
AND MIN(frequency) = MAX(frequency)
AND COUNT(*) >= 2
);
Which, for the sample data:
CREATE TABLE table_name (name) AS
SELECT 'Alexander' FROM DUAL UNION ALL
SELECT 'Johnaton' FROM DUAL UNION ALL
SELECT 'Anna' FROM DUAL;
Outputs:
NAME
Alexander
db<>fiddle here
I have a table with some input values and a table with lookup values like below:
select input.value, coalesce(mapping.value, input.value) result from (
select 'a' union all select 'c'
) input (value) left join (
select 'a', 'z' union all select 'b', 'y'
) mapping (lookupkey, value) on input.value = mapping.lookupkey
which gives:
value | result
--------------
a | z
c | c
i.e. I want to show the original values as well as the mapped value but if there is none then show the original value as the result.
The above works well so far with coalesce to determine if there is a mapped value or not. But now if I allow NULL as a valid mapped value, I want to see NULL as the result and not the original value, since it does find the mapped value, only that the mapped value is NULL. The same code above failed to achieve this:
select input.value, coalesce(mapping.value, input.value) result from (
select 'a' union all select 'c'
) input (value) left join (
select 'a', 'z' union all select 'b', 'y' union all select 'c', null
) mapping (lookupkey, value) on input.value = mapping.lookupkey
which gives the same output as above, but what I want is:
value | result
--------------
a | z
c | NULL
Is there an alternative to coalesce that can achieve what I want?
I think you just want a case expression e.g.
select input.[value]
, coalesce(mapping.[value], input.[value]) result
, case when mapping.lookupkey is not null then mapping.[value] else input.[value] end new_result
from (
select 'a'
union all
select 'c'
) input ([value])
left join (
select 'a', 'z'
union all
select 'b', 'y'
union all
select 'c', null
) mapping (lookupkey, [value]) on input.[value] = mapping.lookupkey
Returns:
value result new_result
a z z
c c NULL
I am using WinSQL to run a query on a table to count the number of occurrences of literal strings. When trying to do a count on a specific set of strings, I still want to see if some values return a count of 0. For example:
select letter, count(*)
from table
where letter in ('A', 'B', 'C')
group by letter
Let's say we know that 'A' occurs 3 times, 'B' occurs 0 times, and 'C' occurs 5 times. I expect to have a table returned as such:
letter count
A 3
B 0
C 5
However, the table never returns a row with a 0 count, which results like so:
letter count
A 3
C 5
I've looked around and saw some articles mentioning the use of joins, but I've had no luck in correctly returning a table that looks like the first example.
You can create an in-line table containing all letters that you look for, then LEFT JOIN your table to it:
select t1.col, count(t2.letter)
from (
select 'A' AS col union all select 'B' union all select 'C'
) as t1
left join table as t2 on t1.col = t2.letter
group by t1.col
on many platforms you can now use the values statement instead of union all to create your "in line" table - like this
select t.letter, count(mytable.letter)
from ( values ('A'),('B'),('C') ) as t(letter)
left join mytable on t.letter = mytable.letter
group by t.letter
I'm not that familiar with WinSQL, but it's not pretty if you don't have the values that you want in the left most column in a table somewhere. If you did, you could use a left join and a conditional. Without it, you can do something like this:
SELECT all_letters.letter, IFNULL(letter_count.letter_count, 0)
FROM
(
SELECT 'A' AS letter
UNION
SELECT 'B' AS letter
UNION
SELECT 'C' AS letter
) all_letters
LEFT JOIN
(SELECT letter, count(*) AS letter_count
FROM table
WHERE letter IN ('A', 'B', 'C')
GROUP BY letter) letter_count
ON all_letters.letter = letter_count.letter
I am trying to select all values that have a first name beginning with the letters a-d, however when I do this
select * from tblprofile where firstname between 'a' and 'd'
I get all values from a to c, not including d, how can I make sure it includes d?
It is inclusive.
You don't get the results you want because any string beginning with 'd' and longer than 1 character is greater than 'd'. For example 'da' > 'd'.
So, your query would return all values starting with 'a', 'b', 'c', and a value 'd'.
To get the results you want use
select * from tblprofile where firstname >= 'a' and firstname < 'e'
Try using Left() Function:
SELECT *
FROM tblprofile
WHERE LEFT(FirstName,1) between 'a' and 'd'
Another way is using a union select like this
SELECT * FROM tblprofile WHERE LEFT(FirstName,1) = 'a'
union
SELECT * FROM tblprofile WHERE LEFT(FirstName,1) = 'b'
union
SELECT * FROM tblprofile WHERE LEFT(FirstName,1) = 'c'
union
SELECT * FROM tblprofile WHERE LEFT(FirstName,1) = 'z'
The advantage of using union is if you need to get the results stating with A, K and X, strings out of sequence.