Let's say there is a table call ITEM and it contains 3 attributes(name, id, price):
name id price
Apple 1 3
Orange 1 3
Banana 2 4
Cherry 3 5
Mango 1 3
How should I write a query to use a constants selection operator to select those item that have same prices and same ids ? The first thing come into my mind is use a rename operator to rename id to id', and price to price', then union it with the ITEM table, but since I need to select 2 tuples (price=price' & id=id') from the table, how can I select them without using the conjunctions operator in relational algebra ?
Thank you.
I'm not quite sure but for me, it would be something like this in relational calculus:
and then in SQL:
SELECT name FROM ITEM i WHERE
EXISTS ITEM u
AND u.name != i.name
AND u.price=i.price
AND u.id = i.id
But still, I think your assumption is right, you can still do it by renaming. I do believe it is a bit longer than what I did above.
I have list of Ids 31165,31160,31321,31322,31199,31136 which is dynamic.
When I run query
select id,name from master_movievod where id in(31165,31160,31321,31322,31199,31136);
I get following result
31136|Independence Day
31160|Planet of the Apes
31165|Mrs. Doubtfire
31199|Moulin Rouge
31321|Adult Movie 2
31322|Adult Movie 3
This is sorted list in ascending order.
I want the list in the same order which I give as input like
31165|Mrs. Doubtfire
31160|Planet of the Apes
31321|Adult Movie 2
31322|Adult Movie 3
31199|Moulin Rouge
31136|Independece Day
Without an order by clause, there's no guarantee on the order a database returns the results to you. SQLite, unfortunately, doesn't have something like MySQL's field for custom sorting, but you can jimmy-rig something with a case expression:
SELECT id, name
FROM master_movievod
WHERE id IN (31165, 31160, 31321, 31322, 31199, 31136)
ORDER BY CASE ID WHEN 31165 THEN 0
WHEN 31160 THEN 1
WHEN 31321 THEN 2
WHEN 31322 THEN 3
WHEN 31199 THEN 4
WHEN 31136 THEN 5
END ASC
Unfortunately, SQLite does not have an option like MySQL's FIELD for doing a custom ordering. You are left with two options. The first is that you could create a custom table containing the ordering you want and use that to sort. This option isn't very attractive. The second (and easier) option is to use ORDER BY CASE to achieve the order you want:
SELECT id, name FROM master_movievod
WHERE id IN (31165,31160,31321,31322,31199,31136)
ORDER BY
CASE id
WHEN 31165 THEN 0
WHEN 31160 THEN 1
WHEN 31321 THEN 2
WHEN 31322 THEN 3
WHEN 31199 THEN 4
WHEN 31136 THEN 5
END ASC
I have data in a table like following
Name indicator
A 1
A 2
A 3
B 1
B 2
C 3
I want to get count of Names, for which both indicator 1,2 exists. In the preeceding example, this number is 2 (A & B both have indicator as 1, and 2).
The data I am dealing with is moderately large, and i need to get the similar information of some other permutations of (pre defined ) indicators (which i can change, once i get base query).
Try this:
SELECT Name
FROM Tablename
WHERE indicator IN(1, 2)
GROUP BY Name
HAVING COUNT(DISTINCT indicator) = 2;
See it in action here:
SQL Fiddle Demo
Suppose I have a table that looks like this:
id attribute
1 football
1 NFL
1 ball
2 football
2 autograph
2 nfl
2 blah
2 NFL
I would like to get a list of distinct ids where the attribute column contains the terms "football", "NFL", and "ball". So 1 would be included, but 2 would not. What's the most elegant/efficient way to do this in Terdata?
The number of attributes can vary for each id, and terms can repeat. For example, NFL appears twice for id 2.
You can use the following:
select id
from yourtable
where attribute in ('football', 'NFL', 'ball')
group by id
having count(distinct attribute) = 3
See SQL Fiddle with Demo (fiddle is showing MySQL, but this should work in TeraData)
Hi I know how to use the group by clause for sql. I am not sure how to explain this so Ill draw some charts. Here is my original data:
Name Location
----------------------
user1 1
user1 9
user1 3
user2 1
user2 10
user3 97
Here is the output I need
Name Location
----------------------
user1 1
9
3
user2 1
10
user3 97
Is this even possible?
The normal method for this is to handle it in the presentation layer, not the database layer.
Reasons:
The Name field is a property of that data row
If you leave the Name out, how do you know what Location goes with which name?
You are implicitly relying on the order of the data, which in SQL is a very bad practice (since there is no inherent ordering to the returned data)
Any solution will need to involve a cursor or a loop, which is not what SQL is optimized for - it likes working in SETS not on individual rows
Hope this helps
SELECT A.FINAL_NAME, A.LOCATION
FROM (SELECT DISTINCT DECODE((LAG(YT.NAME, 1) OVER(ORDER BY YT.NAME)),
YT.NAME,
NULL,
YT.NAME) AS FINAL_NAME,
YT.NAME,
YT.LOCATION
FROM YOUR_TABLE_7 YT) A
As Jirka correctly pointed out, I was using the Outer select, distinct and raw Name unnecessarily. My mistake was that as I used DISTINCT , I got the resulted sorted like
1 1
2 user2 1
3 user3 97
4 user1 1
5 3
6 9
7 10
I wanted to avoid output like this.
Hence I added the raw id and outer select
However , removing the DISTINCT solves the problem.
Hence only this much is enough
SELECT DECODE((LAG(YT.NAME, 1) OVER(ORDER BY YT.NAME)),
YT.NAME,
NULL,
YT.NAME) AS FINAL_NAME,
YT.LOCATION
FROM SO_BUFFER_TABLE_7 YT
Thanks Jirka
If you're using straight SQL*Plus to make your report (don't laugh, you can do some pretty cool stuff with it), you can do this with the BREAK command:
SQL> break on name
SQL> WITH q AS (
SELECT 'user1' NAME, 1 LOCATION FROM dual
UNION ALL
SELECT 'user1', 9 FROM dual
UNION ALL
SELECT 'user1', 3 FROM dual
UNION ALL
SELECT 'user2', 1 FROM dual
UNION ALL
SELECT 'user2', 10 FROM dual
UNION ALL
SELECT 'user3', 97 FROM dual
)
SELECT NAME,LOCATION
FROM q
ORDER BY name;
NAME LOCATION
----- ----------
user1 1
9
3
user2 1
10
user3 97
6 rows selected.
SQL>
I cannot but agree with the other commenters that this kind of problem does not look like it should ever be solved using SQL, but let us face it anyway.
SELECT
CASE main.name WHERE preceding_id IS NULL THEN main.name ELSE null END,
main.location
FROM mytable main LEFT JOIN mytable preceding
ON main.name = preceding.name AND MIN(preceding.id) < main.id
GROUP BY main.id, main.name, main.location, preceding.name
ORDER BY main.id
The GROUP BY clause is not responsible for the grouping job, at least not directly. In the first approximation, an outer join to the same table (LEFT JOIN below) can be used to determine on which row a particular value occurs for the first time. This is what we are after. This assumes that there are some unique id values that make it possible to arbitrarily order all the records. (The ORDER BY clause does NOT do this; it orders the output, not the input of the whole computation, but it is still necessary to make sure that the output is presented correctly, because the remaining SQL does not imply any particular order of processing.)
As you can see, there is still a GROUP BY clause in the SQL, but with a perhaps unexpected purpose. Its job is to "undo" a side effect of the LEFT JOIN, which is duplication of all main records that have many "preceding" ( = successfully joined) records.
This is quite normal with GROUP BY. The typical effect of a GROUP BY clause is a reduction of the number of records; and impossibility to query or test columns NOT listed in the GROUP BY clause, except through aggregate functions like COUNT, MIN, MAX, or SUM. This is because these columns really represent "groups of values" due to the GROUP BY, not just specific values.
If you are using SQL*Plus, use the BREAK function. In this case, break on NAME.
If you are using another reporting tool, you may be able to compare the "name" field to the previous record and suppress printing when they are equal.
If you use GROUP BY, output rows are sorted according to the GROUP BY columns as if you had an ORDER BY for the same columns. To avoid the overhead of sorting that GROUP BY produces, add ORDER BY NULL:
SELECT a, COUNT(b) FROM test_table GROUP BY a ORDER BY NULL;
Relying on implicit GROUP BY sorting in MySQL 5.6 is deprecated. To achieve a specific sort order of grouped results, it is preferable to use an explicit ORDER BY clause. GROUP BY sorting is a MySQL extension that may change in a future release; for example, to make it possible for the optimizer to order groupings in whatever manner it deems most efficient and to avoid the sorting overhead.
For full information - http://academy.comingweek.com/sql-groupby-clause/
SQL GROUP BY STATEMENT
SQL GROUP BY clause is used in collaboration with the SELECT statement to arrange identical data into groups.
Syntax:
1. SELECT column_nm, aggregate_function(column_nm) FROM table_nm WHERE column_nm operator value GROUP BY column_nm;
Example :
To understand the GROUP BY clauserefer the sample database.Below table showing fields from “order” table:
1. |EMPORD_ID|employee1ID|customerID|shippers_ID|
Below table showing fields from “shipper” table:
1. | shippers_ID| shippers_Name |
Below table showing fields from “table_emp1” table:
1. | employee1ID| first1_nm | last1_nm |
Example :
To find the number of orders sent by each shipper.
1. SELECT shipper.shippers_Name, COUNT (orders.EMPORD_ID) AS No_of_orders FROM orders LEFT JOIN shipper ON orders.shippers_ID = shipper.shippers_ID GROUP BY shippers_Name;
1. | shippers_Name | No_of_orders |
Example :
To use GROUP BY statement on more than one column.
1. SELECT shipper.shippers_Name, table_emp1.last1_nm, COUNT (orders.EMPORD_ID) AS No_of_orders FROM ((orders INNER JOIN shipper ON orders.shippers_ID=shipper.shippers_ID) INNER JOIN table_emp1 ON orders.employee1ID = table_emp1.employee1ID)
2. GROUP BY shippers_Name,last1_nm;
| shippers_Name | last1_nm |No_of_orders |
for more clarification refer my link
http://academy.comingweek.com/sql-groupby-clause/