age | name | course | score
_________________________
10 |James | Math | 10
10 |James | Lab | 15
12 |Oliver | Math | 15
13 |William | Lab | 13
I want select record where math >= 10 and lab >11
I write this query
select * from mytable
where (course='Math' and score>10) and (course='Lab' and score>11)
but this query does not return any record.
I want this result
age | name
____________
10 |James
where condition (math >= 10 and lab >11) is dynamically generate and perhaps has 2 condition or 100 or more...
please help me
You query looks for records that satisfy both conditions at once - which cannot happen, since each record has a single course.
You want a condition that applies across rows having the same name, so this suggest aggregation instead:
select age, name
from mytable
where course in ('Math', 'Lab')
group by age, name
having
max(case when course = 'Math' then score end) > 10
and max(case when course = 'Lab' then score end) > 11
If you want the names, then use aggregation and a having clause:
select name, age
from mytable
where (course = 'Math' and score > 10) or
(course = 'Lab' and score > 11)
group by name, age
having count(distinct course) = 2;
If you want the detailed records, use window functions:
select t.*
from (select t.*,
(dense_rank() over (partition by name, age order by course asc) +
dense_rank() over (partition by name, age order by course desc)
) as cnt_unique_courses
from mytable t
where (course = 'Math' and score > 10) or
(course = 'Lab' and score > 11)
) t
where cnt_unique_courses = 2;
SQL Server doesn't support count(distinct) as a window function. But you can implement it by using dense_rank() twice.
If you formulate the problem as:
Select all unique (name, age) combinations
That have a row for course Math with a score >= 10
And that have a row for course Lab with a score > 11
Then you can translate this to something very similar in SQL:
select distinct t1.age, t1.name -- unique combinations
from mytable t1
where exists ( select top 1 'x' -- with a row math score >= 10
from mytable t2
where t2.name = t1.name
and t2.age = t1.age
and t2.course = 'math'
and t2.score >= 10 )
and exists ( select top 1 'x' -- with a row lab score > 11
from mytable t3
where t3.name = t1.name
and t3.age = t1.age
and t3.course = 'lab'
and t3.score > 11 );
i think either your data or your condition is not right to get your output. though based on your condition you can separately used your condition and then use Intersect from both selection and get your filtered data. like the code below.
select Age,Name
from Table_1
where Course ='Math' and Score>=10
INTERSECT
select Age,Name
from Table_1
where Course ='Lab' and Score>11
You can write query using co-related subquery
select * from table_1 t1
where score >11 and course ='lab'
and [name] in (select [name] from table_1 t2 where t1.[name] =t2.[name] and t1.age =t2.Age
and t2.Score >=10 and course = 'Math')
Related
I have a result set
name stage value
---- ----- -----
jim 1 4
jim 1 8
paul 1 8
paul 1 8
want to remove the rows where 8 is the only value against a person
keep the 2 jim rows and lose the 2 paul rows
You can use not exists. For a select query:
select t.*
from t
where not exists (select 1
from t t2
where t2.name = t.name and t2.value = 8
);
Similar logic (except using exists rather than not exists) can be used for a delete -- if you really want to delete the rows from the table.
If you have a complex query that you don't want to repeat, then window functions are helpful:
select t.*
from (select t.*,
sum(case when value = 8 then 1 else 0 end) over (partition by name) as cnt_8
from t
) t
where cnt_8 = 0;
If your database support analytical function then you can use count as follows:
Select * from
(Select t.*,
Count(case when value <> 8 then 1 end) over (partition by name) as cnt
From your_table t) t
Where cnt > 0
Assuming you also have an ID column (defined as an auto increment integer) defined in your table this query would select the row with the highest id for each unique combination:
select max(id) from t group by name,stage,value
In your example this would only return the latest id for rows having values paul,1,8 in columns name,stage,value respectively.
You can then use the prior query to filter out any duplciates using it in the where clause:
select * from t
where id in (select max(id) from t group by name,stage,value)
Finally you can also delete rows that are not unique if that's your goal:
delete from t
where not id in (select max(id) from t group by name,stage,value)
I spend day in hope to figure out how to solve this query.
I have following table
ID Name Pregnancy Gender
1 Raghad Yes Female
1 Raghad No Female
2 Ohoud no Male
What I need is to remove duplicate (in this case 1,1) and to keep one of these rows which has a pregnancy status of yes.
To clarify, I can't use delete since it's a restricted database. I can only retrieve data.
Using an exists clause:
DELETE
FROM yourTable t1
WHERE
pregnancy = 'no' AND
EXISTS (SELECT 1 FROM yourTable t2 WHERE t2.ID = t1.ID AND t2.pregnancy = 'yes');
There are other ways to go about doing this, e.g. using ROW_NUMBER, but as you did not tag your database, I offer the above solution which should work on basically any database.
If you want to just view your data with the "duplicates" removed, then use:
SELECT *
FROM yourTable t1
WHERE
pregnancy = 'yes' OR
NOT EXISTS (SELECT 1 FROM yourTable t2 WHERE t2.ID = t1.ID AND t2.pregnancy = 'yes');
If column Pregnancy have just two values "Yes" and "No", in that case you can use ROW_NUMBER() also to get the results.
;WITH CTE
AS (
SELECT *,ROW_NUMBER() OVER (PARTITION BY id ORDER BY Pregnancy DESC) RN
FROM TABLE_NAME
)
SELECT *
FROM CTE
WHERE RN = 1
In case of multiple values when you want to give highest priorty to "Yes", you can write your query like following
;WITH CTE
AS (
SELECT *,ROW_NUMBER() OVER
(PARTITION BY id ORDER BY CASE WHEN Pregnancy = 'Yes' then 0 else 1 end) RN
FROM TABLE_NAME
)
SELECT *
FROM CTE
WHERE RN= 1
For this sample data you can group by ID, Name, Gender and return the maximum value of the column Pregnancy for each group since Yes is greater compared to No:
SELECT ID, Name, MAX(Pregnancy) Pregnancy, Gender
FROM tablename
GROUP BY ID, Name, Gender
See the demo.
Results:
> ID | Name | Pregnancy | Gender
> -: | :----- | :-------- | :-----
> 1 | Raghad | Yes | Female
> 2 | Ohoud | No | Male
Here is how you could do it in MySQL 8.
Similar Common Table Expressions exist in SQL Server and Oracle.
There you may need to add a comma after then closing parentheses that
ends the CTE (with) definition.
with dups as (
Select id from test
group by id
Having count(1) > 1
)
select * from test
where id in (select id from dups)
and Pregnancy = 'Yes'
union all
select * from test where id not in (select id from dups);
You can see it in action, by running it here
Note this does it without deleting the original.
But it gives you a result set to work with that has what you want.
If you wanted to delete, then you could use this instead, after the dups CTE definition:
delete from test
where id in (select id from dups) and Pregnancy = 'No'
Or distill this into:
delete from test
where id in (Select id from test
group by id
Having count(1) > 1) and Pregnancy = 'No'
1) First of all, update design of your table. ID must be primary key. This would automatically restrict the duplicate rows having same ID.
2) You can use Group by and having clause to remove duplicates
delete from table where pregnancy='no' and exists (SELECT
id
FROM table
GROUP BY id
HAVING count(id)>1)
I have 2 tables containing different data, linked by a column "id", except the id is repeated multiple times
For example,
Table 1:
id grade
1 A
1 C
Table 2:
Id company
1 Alpha
1 Beta
1 Charlie
The number of rows would be inconsistent, table 1 may sometimes have more/less/equal rows compared to table 2. How am I able to combine/merge them into this outcome:
id grade company
1 A Alpha
1 C Beta
1 Charlie
I am using Microsoft access' query.
This is a real pain in MS Access. But you can do it by using a subquery to generate sequence numbers. Here is one method assuming that the rows are unique:
select id, max(grade) as grade, max(company) as company
from ((select id, grade, null as company,
(select count(*)
from table1 as tt1
where tt1.id = t1.id and tt1.grade <= t1.grade
) as seqnum
from table1 as tt1
) union all
(select id, null as grade, company,
(select count(*)
from table2 as tt2
where tt2.id = t2.id and tt2.company <= t1.company
) as seqnum
from table2 as tt2
)
) t12
group by id, seqnum;
This would be much simpler in almost any other database.
some sample data:
Id name value ref
1 ab xy
2 aba z
3 ab xy
4 abc def
5 gxr mdy
what I am trying to do is to get the two column that appeared more than once
so row 1 and row 3 would be selected.
select name, value from table_x
where value is not null group by name having count(name) >= 2
and having count(value) >= 2;
got stucked.....
#vkp's answer is correct if you only care about finding the distinct name/value pairs that appear more than once. But if you actually want the individual rows that satisfy the criteria, try this:
SELECT t1.Name, t1.[Value]
FROM Table_X t1
JOIN
(
SELECT Name, [Value]
FROM Table_X
where [Value] IS NOT NULL
GROUP BY Name, [Value]
HAVING COUNT(1) >= 2
) t2 ON t1.Name = t2.Name AND t1.[Value] = t2.[Value]
Your syntax is incorrect. group by name and value and check for count >=2 thereafter.
select name, value
from table_x
where value is not null
group by name, value
having count(*) >= 2;
Given the SQL table
id date employee_type employee_level
1 10/01/2015 other 2
1 09/13/2011 full-time 1
1 09/25/2010 intern 1
2 09/25/2013 full-time 3
2 09/25/2011 full-time 2
2 09/25/2008 full-time 1
3 09/23/2015 full-time 5
3 09/23/2013 full-time 4
Is it possible to search for ids that have one row with employee_type "intern", and the row above it in the table (same id with later date) with employee_type "full-time".
In this case, id 1 meets my requirement.
Thanks a lot!
Assuming that you mean the same id with the previous date, then you can use lag(), an ANSI standard function supported by most databases:
select t.*
from table t
where t.id in (select id
from (select t.*,
lag(employee_type) over (partition by id order by date) as prev_et
from table t
) tt
where tt.employee_type = 'intern' and tt.prev_et = 'full-time'
);
If your database doesn't support lag(), you can do something similar with correlated subqueries.
I believe the request isn't as described in the question; instead what you appear to be wanting is list all rows for folks who have been interns.
SELECT
t1.*
FROM yourtable AS t1
INNER JOIN (
SELECT DISTINCT
id
FROM yourtable
WHERE employee_type = 'intern'
) AS t2 ON t1.id = t2.id
;
Alternatively you might be wanting only those folks who have been both 'intern' and 'full-time' in which case you could use the query below that uses a HAVING clause:
SELECT
t1.*
FROM yourtable AS t1
INNER JOIN (
SELECT id
FROM yourtable
WHERE employee_type = 'intern'
OR employee_type = 'full-time'
GROUP BY id
HAVING COUNT(DISTINCT employee_type) > 1
) AS t2 ON t1.id = t2.id
;