I have a table having id and no field, what I really want is the result raw will be repeated no filed times, if the no field is 2 then that raw must be repeated twice in result.
this is my sample table structure:
id no
1 3
2 2
3 1
now I need to get a result like:
1 3
1 3
1 3
2 2
2 2
3 1
I tried to write mysql query to get the result like above, but failed.
You need a table of numbers to accomplish this. For just three values, this is easy:
select t.id, t.no
from t join
(select 1 as n union all select 2 union all select 3
) n
on t.no <= n.no;
This query must do what you want to achieve:
select t.id, t.no from test t cross join test y where t.id>=y.id
not completely solve your problem, but this one can help
set #i=0;
select
test_table.*
from
test_table
join
(select
#i:=#i+1 as i
from
any_table_with_number_of_rows_greater_than_max_no_of_test_table
where
#i < (select max(no) from test_table)) tmp on no >= i
order by
id desc
EDIT :
This is on SQL Server. I checked online and see that CTEs work on MySQL too. Just couldn't get them to work on SQLFiddle
Try this, remove unwanted columns
create table #temp (id int, no int)
insert into #temp values (1, 2),(2, 3),(3, 5)
select * from #temp
;with cte as
(
select id, no, no-1 nom from #temp
union all
select c.id, c.no, c.nom-1 from cte c inner join #temp t on t.id = c.id and c.nom < t.no and c.nom > 0
)
select * from cte order by 1
drop table #temp
Related
I have mysql table like this
I want to get row that has minimum 2 or more than 2 (multiple) row only from this table, so the result would be like this
What do i do?
thank you
Use GROUP BY and HAVING clauses
SELECT t.* FROM my_table t
JOIN (
SELECT cust_id, MIN(transaction_no) AS transaction_no
FROM my_table
GROUP BY cust_id
HAVING COUNT(cust_id) > 1
) agg ON t.transaction_no = agg.transaction_no
I have a table like this
I want to check the all rows in Column A with column B and get the count of duplicates.
For example, I want to get the
count of 12 as 3(2 times in A+1 time in B)
count of 11 as 2(2 times in A+0 time in B)
count of 13 as 2(1 time in A+0 time in B)
How can I acheive it?
You can calculate the total occurrences from a union all. A where clause can show only the values that occur in the A column:
select nr
, count(*)
from (
select A as nr
from YourTable
union all
select B
from YourTable
) sub
where nr in -- only values that occur at least once in the A column
(
select A
from YourTable
)
group by
nr
having count(*) > 1 -- show only duplicates
You can combine all values in A and B then do the group by.
Then only select those values found in column A.
Select A, count(A) as cnt
From (
Select A
from yourTable
Union All
Select B
from yourTable) t
Where t.A in
(select distinct A from yourTable)
Group by t.A
Order by t.A;
Result:
A cnt
11 2
12 3
13 1
See demo: http://sqlfiddle.com/#!9/9fcfe9/3
Considering I have the following two sets of rows (same type) in a WHERE clause:
A B
1 1
2 2
3 4
I need to find how many A is in B
For example, for the given table above, it would be 66% since 2 out of 3 numbers are in B
Another example:
A B
1 1
2 2
3 4
5
3
Would give 100% since all of the numbers in A are in B
Here is what I tried myself: (Doesn't work on all test cases..)
DROP PROCEDURE IF EXISTS getProductsByDate;
DELIMITER //
CREATE PROCEDURE getProductsByDate (IN d_given date)
BEGIN
SELECT
Product,
COUNT(*) AS 'total Number',
(SELECT
(SELECT COUNT(DISTINCT Part) FROM products WHERE Product=B.Product) - COUNT(*)
FROM
products AS b2
WHERE
b2.SOP < B.SOP AND b2.Part != B.Part) AS 'New Parts',
CONCAT(round((SELECT
(SELECT COUNT(DISTINCT Part) FROM products WHERE Product=B.Product) - COUNT(*)
FROM
products AS b2
WHERE
b2.SOP < B.SOP AND b2.Part != B.Part)/count(DISTINCT part)*100, 0), '%') as 'Share New'
FROM
products AS B
WHERE
b.SOP < d_given
GROUP BY Product;
END//
DELIMITER ;
CALL getProductsByDate (date("2018-01-01"));
Thanks.
Naming your tables TA and TB respectively you could try something like this (test made on MSSQL and Mysql at moment)
SELECT ROUND(SUM(PERC) ,4)AS PERC_TOT
FROM (
SELECT DISTINCT TA.ID , 1.00/ (SELECT COUNT(DISTINCT ID) FROM TA) AS PERC
FROM TA
WHERE EXISTS ( SELECT DISTINCT ID FROM TB WHERE TB.ID=TA.ID)
) C;
Output with your first sample data set:
PERC_TOT
0,6667
Output with your second sample data set:
PERC_TOT
1,0000
Update (I made the original for two tables, as I was thinking at solution). This is for one single table (is almost the same than the former query): (I used ID1 for column A and ID2 for column B)
SELECT ROUND(SUM(PERC) ,4)AS PERC_TOT
FROM (
SELECT DISTINCT TA.ID1 , 1.00/ (SELECT COUNT(DISTINCT ID1) FROM TA) AS PERC
FROM TA
WHERE EXISTS ( SELECT DISTINCT ID2 FROM TA AS TB WHERE TB.ID2=TA.ID1)
) C;
I have a MySQL table like this
id Name count
1 ABC 1
2 CDF 3
3 FGH 4
using simply select query I get the values as
1 ABC 1
2 CDF 3
3 FGH 4
How I can get the result like this
1 ABC 1
2 CDF 3
3 FGH 4
4 NULL 0
You can see Last row. When Records are finished an extra row in this format
last_id+1, Null ,0 should be added. You can see above. Even I have no such row in my original table. There may be N rows not fixed 3,4
The answer is very simple
select (select max(id) from mytable)+1 as id, NULL as Name, 0 as count union all select id,Name,count from mytable;
This looks a little messy but it should work.
SELECT a.id, b.name, coalesce(b.`count`) as `count`
FROM
(
SELECT 1 as ID
UNION
SELECT 2 as ID
UNION
SELECT 3 as ID
UNION
SELECT 4 as ID
) a LEFT JOIN table1 b
ON a.id = b.id
WHERE a.ID IN (1,2,3,4)
UPDATE 1
You could simply generate a table that have 1 column preferably with name (ID) that has records maybe up 10,000 or more. Then you could simply join it with your table that has the original record. For Example, assuming that you have a table named DummyRecord with 1 column and has 10,000 rows on it
SELECT a.id, b.name, coalesce(b.`count`) as `count`
FROM DummyRecord a LEFT JOIN table1 b
ON a.id = b.id
WHERE a.ID >= 1 AND
a.ID <= 4
that's it. Or if you want to have from 10 to 100, then you could use this condition
...
WHERE a.ID >= 10 AND
a.ID <= 100
To clarify this is how one can append an extra row to the result set
select * from table union select 123 as id,'abc' as name
results
id | name
------------
*** | ***
*** | ***
123 | abc
Simply use mysql ROLLUP.
SELECT * FROM your_table
GROUP BY Name WITH ROLLUP;
select
x.id,
t.name,
ifnull(t.count, 0) as count
from
(SELECT 1 AS id
-- Part of the query below, you will need to generate dynamically,
-- just as you would otherwise need to generate 'in (1,2,3,4)'
UNION ALL SELECT 2
UNION ALL SELECT 3
UNION ALL SELECT 4
UNION ALL SELECT 5
) x
LEFT JOIN YourTable t
ON t.id = x.id
If the id does not exist in the table you're selecting from, you'll need to LEFT JOIN against a list of every id you want returned - this way, it will return the null values for ones that don't exist and the true values for those that do.
I would suggest creating a numbers table that is a single-columned table filled with numbers:
CREATE TABLE `numbers` (
id int(11) unsigned NOT NULL
);
And then inserting a large amount of numbers, starting at 1 and going up to what you think the highest id you'll ever see plus a thousand or so. Maybe go from 1 to 1000000 to be on the safe side. Regardless, you just need to make sure it's more-than-high enough to cover any possible id you'll run into.
After that, your query can look like:
SELECT n.id, a.*
FROM
`numbers` n
LEFT JOIN table t
ON t.id = n.id
WHERE n.id IN (1,2,3,4);
This solution will allow for a dynamically growing list of ids without the need for a sub-query with a list of unions; though, the other solutions provided will equally work for a small known list too (and could also be dynamically generated).
Columns a, b and c contain some values of the same nature. I need to select all the unique values. If I had just one column I'd use something like
SELECT DISTINCT a FROM mytable ORDER BY a;
but I need to treat a, b and c columns as one and gett all the unique values ever occurring among them.
As an example, let this be a CSV representation of mytable, the first row naming the columns:
a, b, c
1, 2, 3
1, 3, 4
5, 7, 1
The result of the query is to be:
1
2
3
4
5
7
UPDATE: I don't understand why do all of you suggest wrapping it in an extra SELECT? It seems to me that the answer is
(SELECT `a` AS `result` FROM `mytable`)
UNION (SELECT `b` FROM `mytable`)
UNION (SELECT `c` FROM `mytable`)
ORDER BY `result`;
isn't it?
So you want one column all with unique values from a, b and c? Try this:
(select a as yourField from d1)
union
(select b from d2)
union
(select c from d3)
order by yourField desc
limit 5
Working example
Edited after requirements changed... There you have the order by and limit you requested. Of course, you'll get only 5 records in this example
sorry i miss understood your question. here is updated query.
select a from my table
UNION
select b from my table
UNION
select c from my table
SELECT tmp.a
FROM
(SELECT column_1 AS a
FROM table
UNION
SELECT column_2 AS a
FROM table
UNION
SELECT column_3 AS a
FROM table) AS tmp
ORDER BY `tmp`.`a` ASC
try this:
SELECT b.iResult
FROM
(SELECT a as iResult FROM tableName
UNION
SELECT b as iResult FROM tableName
UNION
SELECT c as iResult FROM tableName) b
ORDER BY b.iResult
LIMIT BY 10 -- or put any number you want to limit.