Null value in order by clause - sql

I have a weird scenario, in which I need to keep all the rows at top in which X column has NULL value else sort by Y column. Can you help me in writing query.

ORDER BY CASE WHEN X IS NULL THEN 0 ELSE 1 END, Y

You can use a CASE statement in ORDER BY:
ORDER BY
CASE WHEN X IS NULL THEN 0 ELSE 1 END ASC, Y

Here you go, this will work with any sql platform -- for a specific platform there might be a better way to do it.
SELECT * FROM
(
SELECT 1 AS orderC, *
FROM tableName
WHERE Xcolumn is null
UNION ALL
SELECT 2 AS orderC, *
FROM tableName
WHERE Xcolumn is not null
)
ORDER BY orderC ASC, columnY
Note, if you don't want orderC to be in the output, just specify all the other columns in the outer select.

Sharing what I learned before using:
ORDER BY FIELD(Xcolumn, NULL) DESC, Ycolumn DESC

SELECT * FROM TABLENAME ORDER BY X ,Y

You can use query like below:
SELECT * FROM Emp WHERE empId= 6 AND DELETED = 0
ORDER BY CASE WHEN DOB IS NULL THEN 0 ELSE 1 END
, CREATETIMESTAMP.
for more details you can see here

Related

SQL Query problem (2 different order by) in 1 table

This is the original table:
I have 2 different query and I want to make 1 query for these:
SELECT *
FROM SAMPLE
WHERE ORDER_PRIORITY<40
ORDER BY FS_GENERATE_DATE IS NOT NULL, FS_GENERATE_DATE,ORDER_PRIORITY,CREATE_ID,CR_DATE,ORDER_QTY;
second:
SELECT *
FROM SAMPLE
WHERE ORDER_PRIORITY>=40
ORDER BY FS_GENERATE_DATE IS NOT NULL, FS_GENERATE_DATE,ORDER_PRIORITY,CREATE_ID,CR_DATE,ORDER_QTY;
I need the next result in only 1 query:
if the order_priority<40 than the order will be the first according to the order by
but if order_priority>=40 than these data will be after the lower priority (first conditional /op<40/).
Result:
You can add this to your order by clause:
case when ORDER_PRIORITY<40 then 0 else 1 end
The final query will be:
SELECT
*
FROM SAMPLE
WHERE ORDER_PRIORITY>=40
ORDER BY
case when ORDER_PRIORITY<40 then 0 else 1 end,
FS_GENERATE_DATE IS NOT NULL, FS_GENERATE_DATE,ORDER_PRIORITY,CREATE_ID,CR_DATE,ORDER_QTY;
You are clearly using a database where booleans are allowed in the ORDER BY. So, you can just use:
SELECT S.*
FROM SAMPLE S
ORDER BY (ORDER_PRIORITY < 40) DESC,
FS_GENERATE_DATE IS NOT NULL,
FS_GENERATE_DATE,
ORDER_PRIORITY,
CREATE_ID, CR_DATE, ORDER_QTY;

SQL Group By and assign highest a value

I am struggling with the SQL (mssql) to manipulate my data as i need it. I have a table like this;
SOMEID, SOMEFIELD, DATE
5 True 01-01-2010
5 True 01-01-2011
5 False 05-05-2012
7 True 05-05-2011
7 False 06-07-2015
What I am trying to achieve is to add another column which assigns the value 1 if they are the most recent for that ID, and 0 if not. So in the above data example the new column values from top to bottom would be 0, 0, 1, 0, 1.
I know I need to group by date but am having trouble assigning the values.
Thanks for any pointers!
You can use row_number() in SQL Server like this:
select *
, case when (row_number() over (partition by SOMEID order by [Date] desc)) = 1 then 1 else 0 end seq
from
yourTable
order by
SOMEID, [Date];
SQL Fiddle Demo
You can use a self join to get the highest row per group then in update query use a case statement to assign value to new column
update a
set a.[somecol] = case when b.[SOMEID] is null then 1 else 0 end
from demo a
left join demo b on a.[SOMEID] = b.[SOMEID]
and a.[DATE] < b.[DATE]
DEMO
try this
SELECT SOMEID, SOMEFIELD, DATE
, CASE WHEN (SELECT MAX(SubTab.Date)
FROM myTable SubTab
WHERE SubTab.SOMEID = myTable.SOMEID
) = myTable.DATE
THEN 1 ELSE 0 END
FROM myTable

SQL: Order by statement with start value

I have a field like this:
1月~3月
12月~1月
3月~12月
4月~12月
9月~8月
6月~7月
How can i sort that column following:
4月~12月
6月~7月
9月~8月
12月~1月
1月~3月
3月~12月
It start by 4 and end by 3 (4-5-6-7-8-9-10-11-12-1-2-3)(month)
This will do it, you need to separate out the numeric portion of your field, and also use a CASE statement:
SELECT *
FROM Table1
ORDER BY CASE WHEN CAST(LEFT(Col1, CHARINDEX('月',Col1)-1)AS INT) >= 4 THEN 1 END DESC
,CAST(LEFT(Col1, CHARINDEX('月',Col1)-1)AS INT)
Demo: SQL Fiddle
SQL Server syntax above, might vary depending on database.
order by case when col < 4 then 1 else 0 end, col
or if it's really a varchar
order by case when convert(int,substring(col,1,1)) < 4 then 1 else 0 end, col

T-Sql: turn multiple rows into one row

How does one turn these multiple rows into one row? N and Y are bool values.
Id IsPnt IsPms, IsPdt
1 N Y N
1 N Y N
1 Y N N
into this
Id IsPnt IsPms, IsPdt
1 Y Y N
Edit:
The query that produces the resultset looks like this
select b.id,
CASE mpft.PlanIndCd WHEN 'PBMN' THEN 1 ELSE 0 END AS IsPnt,
CASE mpft.PlanIndCd WHEN 'PBMT' THEN 1 ELSE 0 END AS IsPbt,
CASE mpft.PlanIndCd WHEN 'PBMS' THEN 1 ELSE 0 END AS IsPms
from vw_D_SomveViewName pb
-- bunch of joins
where mpft.PlanIndCd in ('HANR', 'PBMN','PBMT','PBMS','HAWR')
You can simply use MAX() on this if the values are really Y and N only.
SELECT ID, MAX(IsPnt) IsPnt, MAX(IsPms) IsPms, MAX(IsPdt) IsPdt
FROM tableName
GROUP BY ID
UPDATE 1
SELECT b.id,
MAX(CASE mpft.PlanIndCd WHEN 'PBMN' THEN 1 ELSE 0 END) AS IsPnt,
MAX(CASE mpft.PlanIndCd WHEN 'PBMT' THEN 1 ELSE 0 END) AS IsPbt,
MAX(CASE mpft.PlanIndCd WHEN 'PBMS' THEN 1 ELSE 0 END) AS IsPms
FROM vw_D_SomveViewName pb
-- bunch of joins
WHERE mpft.PlanIndCd in ('HANR', 'PBMN','PBMT','PBMS','HAWR')
GROUP BY b.ID
Will this work?
select
id,
max(IsPnt),
max(IsPms),
max(IsPdt)
from
table
GROUP BY
id
After the edit of your question, you can simply use the PIVOT table operator directly instead of using the MAX expression, something like:
SELECT
Id,
PBMN AS IsPnt,
PBMT AS IsPbt,
PBMS AS IsPms
FROM
(
SELECT
id,
mpft.PlanIndCd,
ROW_NUMBER() OVER(PARTITION BY id
ORDER BY ( SELECT 1)) AS RN
from vw_D_SomveViewName pb
-- bunch of joins
where mpft.PlanIndCd in ('HANR', 'PBMN','PBMT','PBMS','HAWR')
) AS t
PIVOt
(
MAX(RN)
FOR PlanIndCd IN ([PBMN], [PBMT], [PBMS])
) AS p;
You can see it in action in the following demo example:
Demo on SQL Fiddle
select Id, MAX(IPnt), MAX(IsPms), MAX(IsPdt)
from table etc

Order By a field being equal to a specific value?

Let's say I have this MySQL query:
SELECT * FROM A WHERE x='abc' OR y=0;
How can I prioritize the rows so that cases where x='abc' are ordered FIRST? If y=0 but x!='abc', I want those rows to come after cases where x='abc'.
Can this be accomplished with a simple ORDER BY clause?
Thanks!
SELECT *
FROM A
WHERE x='abc'
OR y=0
order by case when x='abc' then 0 else 1 end;
I would do it with a case statement:
SELECT *,
CASE WHEN column_one = 0 THEN 0
WHEN column_two = 'ADMIN' THEN 1
END AS multi_column
FROM sometable
ORDER BY multi_column;
ORDER BY FIELD(x,'abc',x)