I am working with InterBase at the moment by Using Firebird. I need to get a record twice with a condition, but every time I run the query I keep getting an error message. Is there a way to get around the cross join in InterBase? Will there be a simpler way to get around to add this logic to the query. It seems like Firebird wont accept the cross join.
SELECT EMPLOYEE_NAME, LAST_NAME, CACODE
FROM EMPLOYEESTABLE
(
SELECT 889 AS CACODE, 8592-265-44444 AS STANDARDCACODE
UNION ALL
SELECT 695 AS CACODE, 8554-265-44578 AS STANDARDCACODE
) C
Select always needs from. The table RDB$DATABASE returns 1 row with the database name so we can use that. I think you are looking to do this:
SELECT EMPLOYEE_NAME, LAST_NAME, CACODE
FROM EMPLOYEESTABLE,
(
SELECT 889 AS CACODE, '8592-265-44444' AS STANDARDCACODE FROM RDB$DATABASE
UNION ALL
SELECT 695 AS CACODE, '8554-265-44578' AS STANDARDCACODE FROM RDB$DATABASE
) C
SELECT
889 AS CACODE,
/* '8592-265-44444' AS STANDARDCACODE, >>unused<< */
EMPLOYEE_NAME,
LAST_NAME
FROM EMPLOYEESTABLE
UNION ALL
SELECT
695 AS CACODE,
/* '8554-265-44578' AS STANDARDCACODE, >>unused<< */
EMPLOYEE_NAME,
LAST_NAME
FROM EMPLOYEESTABLE
Who need joins with such a simple query :-D
Related
I has two tables on MS SQL Server. First with the cities, second with student names and city ID from first table, where this student lives.
I need a sql query that will return results like this:
Stella Paris
Bob Moscow,New York
Mary Paris,New York
WITH CITIES(ID,CITY_NAME)AS
(
SELECT 1,'MOSCOW' UNION ALL
SELECT 2,'PARIS' UNION ALL
SELECT 3,'NEW YORK'
),
STUDENTS(ID,STUDENT_NAME,LIVE_IN) AS
(
SELECT 1,'STELLA','2' UNION ALL
SELECT 2,'BOB','1,3' UNION ALL
SELECT 3,'MARY','2,3'
)
SELECT X.ID,X.STUDENT_NAME,STRING_AGG(X.CITY_NAME,',')CITY_NAME FROM
(
SELECT S.ID,S.STUDENT_NAME,VALUE AS CITY_ID,C.CITY_NAME
FROM STUDENTS AS S
CROSS APPLY string_split(S.LIVE_IN,',')
JOIN CITIES AS C ON VALUE=C.ID
)X
GROUP BY X.ID,X.STUDENT_NAME;
For SQL Server 2017 and later, as pointed by #Squirrel, you can use STRING_SPLIT and STRING_AGG
Steps you can use:-
Use two different queries
first returning students records
then loop through to get cities records
update the array data as required
A better suggestion would be redesign database so as to be used table JOIN
I have a table of similar structure:
Name Movies_Watched
A Terminator
B Alien
A Batman
B Rambo
B Die Hard
....
I am trying to get this:
Name Movies_Watched
A Terminator;Batman
B Alien, Die Hard, Rambo
My initial guess was:
SELECT Name, Movies_Watched || Movies_Watched from TABLE
But obviously that's wrong. Can someone tell me how can I loop through the 2nd column and concatenate them? What's the logic like?
Got to know that group_concat is the right approach. But haven't been able to figure it out yet. When I've tried:
select name, group_concat(movies_watched) from table group by 1
But it throws an error saying User-defined transform function group_concat must have an over clause
You are looking for string_agg():
select name, string_agg(movie_watched, ';') as movies_watched
from t
group by name;
That said, you are using Postgres, so you should learn how to use arrays instead of strings for such things. For instance, there is no confusion with arrays when the movie name has a semicolon. That would be:
select name, array_agg(movie_watched) as movies_watched
from t
group by name;
use array_agg
SELECT Name, array_agg(Movies_Watched)
FROM data_table
GROUP BY Name
i think you need listagg or group_concat as you are using vertica upper is postgrey solution
SELECT Name, listagg(Movies_Watched)
FROM data_table
GROUP BY Name
or
select Name,
group_concat(Movies_Watched) over (partition by Name order by name) ag
from mytable
As already mentioned, in Vertica it's LISTAGG():
WITH
input(nm,movies_watched) AS (
SELECT 'A','Terminator'
UNION ALL SELECT 'B','Alien'
UNION ALL SELECT 'A','Batman'
UNION ALL SELECT 'B','Rambo'
UNION ALL SELECT 'B','Die Hard'
)
SELECT
nm AS "Name"
, LISTAGG(movies_watched) AS movies_watched
FROM input
GROUP BY nm;
-- out Name | movies_watched
-- out ------+----------------------
-- out A | Terminator,Batman
-- out B | Alien,Rambo,Die Hard
-- out (2 rows)
-- out
-- out Time: First fetch (2 rows): 12.735 ms. All rows formatted: 12.776 ms
When I try to select multiple ID's from a table, I am getting error ORA-02070.
Here is the query that I am using:
select *
from hrs_employee_store
where employee_id in (13511677, 576000);
Here is the error which I am getting:
ORA-02070: database ODS_XSTORE does not support TO_NUMBER in this context
Also, when I use this query,
select * from hrs_employee_store
where employee_id in ('13511677', '576000');
I am just getting the row for 13511677.
Is there a way to fix this issue? Thanks
I suspect that EMPLOYEE_ID is not a number. Try:
select *
from hrs_employee_store
where EMPLOYEE_ID in ('13511677', '576000');
This returns matching employees, meaning there is no match for the second.
If you want NULL values for all the extra columns, you can use left join:
select *
from (select '13511677' as employee_id from dual union all
select '576000'
) eid left join
hrs_employee_store es
using (employee_id);
I have written a query which extracts the data from different columns group by city name.
My query is as follows:
select q.first_name
from (select employee_id as eid,first_name,city
from employees
group by city,first_name,employee_id
order by first_name)q
, employees e
where e.employee_id = q.eid;
The output of the query is employee names in a single column grouped by their cities.
Now I would like to enhance the above query to classify the employees by their city names in different columns.
I tried using pivot to make this work. Here is my pivot query:
select * from (
select q.first_name
from (select employee_id as eid,first_name,city
from employees
group by city,first_name,employee_id
order by first_name)q
, employees e
where e.employee_id = q.eid
) pivot
(for city in (select city from employees))
I get some syntax issue saying missing expression and I am not sure how to use pivot to achieve the below expected output.
Expected Output:
DFW CH NY
---- --- ---
TripeH John Hitman
Batista Cena Yokozuna
Rock James Mysterio
Appreciate if anyone can guide me in the right direction.
Unfortunately what you are trying to do is not possible, at least not in "straight" SQL - you would need dynamic SQL, or a two-step process (in the first step generating a string that is a new SQL statement). Complicated.
The problem is that you are not including a fixed list of city names (as string literals). You are trying to create columns based on whatever you get from (select city from employees). Thus the number of columns and the name of the columns is not known until the Oracle engine reads the data from the table, but before the engine starts it must already know what all the columns will be. Contradiction.
Note also that if this was possible, you almost surely would want (select distinct city from employees).
ADDED: The OP asks a follow-up question in a comment (see below).
The ideal arrangement is for the cities to be in their own, smaller table, and the "city" in the employees table to have a foreign key constraint so that the "city" thing is manageable. You don't want one HR clerk to enter New York, another to enter New York City and a third to enter NYC for the same city. One way or the other, first try your code by replacing the subquery that follows the operator IN in the pivot clause with simply the comma-separated list of string literals for the cities: ... IN ('DFW', 'CH', 'NY'). Note that the order in which you put them in this list will be the order of the columns in the output. I didn't check the entire query to see if there are any other issues; try this and let us know what happens.
Good luck!
select
(CASE WHEN CITY="DFW" THEN EMPLOYEE_NAME END) DFW,
(CASE WHEN CITY="CH" THEN EMPLOYEE_NAME END) CH,
(CASE WHEN CITY="NY" THEN EMPLOYEE_NAME END) NY
FROM employees
order by first_name
Maybe you need to transpose your result. See this link . I think DECODE or CASE works best for your case:
select
(CASE WHEN CITY="DFW" THEN EMPLOYEE_NAME END) DFW,
(CASE WHEN CITY="CH" THEN EMPLOYEE_NAME END) CH,
(CASE WHEN CITY="NY" THEN EMPLOYEE_NAME END) NY
FROM employees
order by first_name
Normally I would "edit" my first answer, but the question has changed so much, it's quite different from the original one so my older answer can't be "edited" - this now needs a completely new answer.
You can do what you want with pivoting, as I show below. Wondering why you want to do this in basic SQL and not by using reporting tools, which are written specifically for reporting needs. There's no way you need to keep your data in the pivoted format in the database.
You will see 'York' twice in the Chicago column; you will recognize that's on purpose (you will see I had a duplicate row in the "test" table at the top of my code); this is to demonstrate a possible defect of your arrangement.
Before you ask if you could get the list but without the row numbers - first, if you are simply generating a set of rows, those are not ordered. If you want things ordered for reporting purposes, you can do what I did, and then select "'DFW'", "'CHI'", "'NY'" from the query I wrote. Relational theory and the SQL standard do not guarantee the row order will be preserved, but Oracle apparently does preserve it, at least in current versions; you can use that solution at your own risk.
max(name) in the pivot clause may look odd to the uninitiated; one of the weird limitations of the PIVOT operator in Oracle is that it requires an aggregate function to be used, even if it's over a set of exactly one element.
Here's the code:
with t (city, name) as -- setting up input data for testing
(
select 'DFW', 'Smith' from dual union all
select 'CHI', 'York' from dual union all
select 'DFW', 'Matsumoto' from dual union all
select 'NY', 'Abu Osman' from dual union all
select 'DFW', 'Adams' from dual union all
select 'CHI', 'Wilson' from dual union all
select 'CHI', 'Arenas' from dual union all
select 'NY', 'Theodore' from dual union all
select 'CHI', 'McGhee' from dual union all
select 'NY', 'Zhou' from dual union all
select 'NY' , 'Simpson' from dual union all
select 'CHI', 'Narayanan' from dual union all
select 'CHI', 'York' from dual union all
select 'NY', 'Perez' from dual
)
select * from
(
select row_number() over (partition by city order by name) rn,
city, name
from t
)
pivot (max(name) for city in ('DFW', 'CHI', 'NY') )
order by rn
/
And the output:
RN 'DFW' 'CHI' 'NY'
---------- --------- --------- ---------
1 Adams Arenas Abu Osman
2 Matsumoto McGhee Perez
3 Smith Narayanan Simpson
4 Wilson Theodore
5 York Zhou
6 York
6 rows selected.
Assume we have loaded a flat file with patient diagnosis data into a table called “Data”. The table structure is:
Create table Data (
Firstname varchar(50),
Lastname varchar(50),
Date_of_birth datetime,
Medical_record_number varchar(20),
Diagnosis_date datetime,
Diagnosis_code varchar(20))
The data in the flat file looks like this:
'jane','jones','2/2/2001','MRN-11111','3/3/2009','diabetes'
'jane','jones','2/2/2001','MRN-11111','1/3/2009','asthma'
'jane','jones','5/5/1975','MRN-88888','2/17/2009','flu'
'tom','smith','4/12/2002','MRN-22222','3/3/2009','diabetes'
'tom','smith','4/12/2002','MRN-33333','1/3/2009','asthma'
'tom','smith','4/12/2002','MRN-33333','2/7/2009','asthma'
'jack','thomas','8/10/1991','MRN-44444','3/7/2009','asthma'
You can assume that no two patients have the same firstname, lastname, and date of birth combination. However one patient might have several visits on different days. These should all have the same medical record number.
The problem is this: Tom Smith has 2 different medical record numbers. Write a query that would always show all the patients
who are like Tom Smith – patients with more than one medical record number.
I came up with below query. It works perfectly fine, but wanted to know if there is a better way to write this query using Oracle Analytical function's. Thank you in advance
SELECT a.firstname,
a.lastname,
a.date_of_birth,
a.medical_record_number
FROM data a, data b
WHERE a.firstname = b.firstname
AND a.lastname = b.lastname
AND a.date_of_birth = b.date_of_birth
AND a.medical_record_number <> .medical_record_number
GROUP BY a.firstname,
a.lastname,
a.date_of_birth,
a.medical_record_number
It is possible to do via analytic functions, but whether it's faster than doing the join in your query* or not depends on what data you have. You'd need to test.
with data (firstname, lastname, date_of_birth, medical_record_number, diagnosis_date, diagnosis_code)
as (select 'jane','jones','2/2/2001','MRN-11111',to_date('3/3/2009', 'mm/dd/yyyy'),'diabetes' from dual union all
select 'jane','jones','2/2/2001','MRN-11111',to_date('1/3/2009', 'mm/dd/yyyy'),'asthma' from dual union all
select 'jane','jones','5/5/1975','MRN-88888',to_date('2/17/2009', 'mm/dd/yyyy'),'flu' from dual union all
select 'tom','smith','4/12/2002','MRN-22222',to_date('3/3/2009', 'mm/dd/yyyy'),'diabetes' from dual union all
select 'tom','smith','4/12/2002','MRN-33333',to_date('1/3/2009', 'mm/dd/yyyy'),'asthma' from dual union all
select 'tom','smith','4/12/2002','MRN-33333',to_date('2/7/2009', 'mm/dd/yyyy'),'asthma' from dual union all
select 'jack','thomas','8/10/1991','MRN-44444',to_date('3/7/2009', 'mm/dd/yyyy'),'asthma' from dual),
-- end of mimicking your table and its data
res as (select firstname,
lastname,
date_of_birth,
medical_record_number,
count(distinct medical_record_number) over (partition by firstname, lastname, date_of_birth) cnt_med_rec_nums
from data)
select distinct firstname,
lastname,
date_of_birth,
medical_record_number
from res
where cnt_med_rec_nums > 1;
*btw, the group by in your example query is not necessary; it would make much more sense to switch it out for a distinct - it makes your intent much clearer, since you're wanting to get a distinct set of records.
You can probably simplify the query a bit using a HAVING clause rather than doing a self-join
SELECT a.firstname,
a.lastname,
a.date_of_birth,
MIN(a.medical_record_number) lowest_medical_record_number,
MAX(a.medical_record_number) highest_medical_record_number
FROM data a
GROUP BY a.firstname,
a.lastname,
a.date_of_birth
HAVING COUNT( DISTINCT a.medical_record_number ) > 1
I'm returning the smallest and largest medical record number for each patient here (that's what I'd do if most of the patients with this problem have just two numbers rather than having dozens). You could return just one or you could return a comma-separated list of all the medical record numbers if you'd rather (which would probably make more sense if most of the bad folks have dozens of numbers).