I am getting trouble to write SQL sort Query,
I have table as follows
And I want to sort above data as, First should number and then alphabets and last special symbols like following table.
First numbers should sort like 1,2,3,11,22,33
then Alphabets should sort like a ,b,c,..z
and then symbols,
like following table
I have tried many ways but still not getting correct way, please help me to write query.
You can use a CASE WHEN on the ORDER BY to create some groups:
SELECT *
FROM table_name
ORDER BY
CASE WHEN LEFT(FilterParameterValue, 1) LIKE '[0-9]' OR LEFT(FilterParameterValue, 2) LIKE '-[0-9]' OR LEFT(FilterParameterValue, 2) LIKE '+[0-9]' THEN 1 ELSE 0 END DESC, -- group for numbers
CASE WHEN ISNUMERIC(FilterParameterValue) = 1 THEN CAST(FilterParameterValue AS MONEY) ELSE NULL END ASC, -- order the numeric values
CASE WHEN LEFT(FilterParameterValue, 1) LIKE '[A-Za-z]' THEN 1 ELSE 0 END DESC, -- group for chars from A to Z (capital / non capital)
colName
demo on dbfiddle.uk
Try using regex in order by clause.. For ex
ORDER BY IF(FilterParameterValue RLIKE '^[a-z]', 1, 2), FilterParameterValue
You can try to cast your Column to Numeric Type then ordered
SELECT * FROM table_name ORDER BY TRY_CAST(Column_Name as Type)
Related
SELECT cast ( SUBSTRING ( CAST ("ProcessingDate" AS text), 5, 2 ) as integer),
COUNT(*) INTO resultValue1,resultValue2
FROM "DemoLogs"."Project"
WHERE "Addr" = 'Y' AND "ProcessingDate" >= 20160110
GROUP BY 1
ORDER BY 1;
In my database, the ProcessingDate is stored as YYYYMMDD. So, I am extracting its month from it.
This query is working fine if we remove the INTO clause but, I want to store the result to use it further.
So what should be the datatype of the variable resultValue1 and resultValue2 how to store the data(because data will be multiple).
As I am new to PostgreSQL I don't know how to do this can anybody help me out.
Here resultValue1 & resultValue2 will be tables not variables. you can group by using the column names.
Give some column alias names for both columns and group by using them.
You probably want this.
SELECT cast ( SUBSTRING ( cast ("ProcessingDate" as text),5 , 2 ) as
integer) AS resultValue1, COUNT(*) AS resultValue2
INTO <NewTable> --NewTable will be created with those two columns
FROM "DemoLogs"."Project"
-- conditions
Group By 1
-- other contitions/clauses
;
Kindly refer this INTO Documentation.
Hope this helps.
Try this:
SELECT cast ( SUBSTRING ( cast ("ProcessingDate" as text),5 , 2 ) as integer)resultValue1,
COUNT(*)resultValue2
INTO <Table Name>
FROM "DemoLogs"."Project"
WHERE "Addr" = 'Y'
AND "ProcessingDate" >= '20160110'
Group By 1
Order By 1;
Store the above query in the variable and remove that INTO clause from it.
So, query will be now :
query = SELECT cast ( SUBSTRING ( CAST ("ProcessingDate" AS text), 5, 2 ) as
integer),
COUNT(*)
FROM "DemoLogs"."Project"
WHERE "Addr" = 'Y' AND "ProcessingDate" >= 20160110
GROUP BY 1
ORDER BY 1;
Now declare result_data of type record and use in the following manner:
We are looping over result_data because it gives number of rows as output
and I have declared resultset of type text to store the result (as I needed further)
FOR result_data IN EXECUTE query
LOOP
RAISE NOTICE 'result : %',to_json(result_data);
resultset = resultset || to_json(result_data);
END LOOP;
I want to ORDER BY data in the following sequence:
Number-Number
Alpha-Number
Alpha with no dash
I have the following sort:
120-1
120-2
120-10
Digital-1
Digital-10
Digital-2
Wedding
This is the order I'm looking for, except the Digital (Alpha with dash) entries are not sorted by numbers after the dash.
After many attempts, here's my current sql statement:
SELECT SessionID, Identifier FROM Session
ORDER BY
CASE
WHEN CAST(Identifier AS INTEGER) THEN SUBSTR(Identifier,0, INSTR(Identifier, '-')) + CAST(SUBSTR(Identifier, INSTR(Identifier, '-')+1, 999) AS INTEGER)
ELSE Identifier END
What am I doing wrong?
Thanks.
1) Order by strings having - in them so they appear first
2) Order by the first part before - in the strings containing - casting them as integers where applicable, Else order by the column itself
3) Order by the second part after - in the strings containing - casting them as integers where applicable, Else order by the column itself
select val
from t
order by
case when val like '%-%' then 1 else 2 end
,case when substr(val,1,instr(val,'-')-1) glob '*[0-9]*'
then cast(substr(val,1,instr(val,'-')-1) as integer)
when substr(val,1,instr(val,'-')-1) glob '*[a-zA-Z]*'
then substr(val,1,instr(val,'-')-1)
else val end
,case when val like '%-%' then cast(substr(val,instr(val,'-')+1) as integer)
else val end
SQL Fiddle
Your logic is somehow reconstructing the strings, but for the order by, you want something like this:
order by (case when CAST(Identifier AS INTEGER) <> 0 then 1
when Identifier like '%-%' then 2
else 3
end),
identifier
Note: This assumes that the numbers are never 0.
I have been asked to sort a locations table in ascending order. The data in the table currently looks something like this:
100F01
105B02
10B01
GK1-A01
201E12
20A01
However, when this data is displayed on the screen I want it to sort the numeric results in ascending order and then the string results in alphabetic order. The output should look something like this:
10B01
20A01
100F01
105B02
201E12
GK1-A01
I have tried using the following code
SELECT location FROM freelocations
ORDER BY CAST(SUBSTRING(location, 1, 2) AS INT)
however, as expected, this returns an error message because some locations don't start with a numeric:
Conversion failed when converting the varchar value 'GK' to data type int.
Any ideas or tips will be greatly appreciated
Maybe something like this?
select location
from freelocations
order by
case
when patindex('%[^0-9]%', location) = 1 then 9999999999
else cast(substring(location, 1, patindex('%[^0-9]%', location) - 1) as int)
end,
location
(It's a bit clumsy due to SQL Server lacking a regex replace function)
order by FIELD(string+0,1),string+0
but note that '10e20' will be misinterpreted!
Before casting you should check if the first 2 or 3 characters really are digits:
SELECT
location
FROM
freelocations
ORDER BY
CASE
WHEN location like '[0-9][0-9][0-9]%' THEN cast(substring(location,1,3) AS INT)
WHEN location like '[0-9][0-9]%' THEN cast(substring(location,1,2) AS INT)
ELSE 9999
END
Try this to avoid casting:
SELECT
x,
RIGHT('000' +LEFT (x,patindex('%[^0-9]%',x)-1),3) sort1,
RIGHT (x,LEN(x)-patindex('%[^0-9]%',x)+1) sort2
FROM
(
SELECT '123asdf' x
UNION
SELECT '12asdf'
UNION
SELECT '13asdf'
UNION
SELECT '12zsdf'
) X
ORDER BY sort1 ,sort2
I have a column named Code that is varchar(3).
It contains numbers and strings as well. For example: ' 1', '234', 'Xxx', '9 ','Aa ' etc.
Is there way -just like in MS EXCEL- ORDER anything that looks like a number, as a number?
So that output for the given example above will be:
1. 1
2. 234
3. 9
4. Aa
5. Xxx
ORDER BY CASE WHEN ISNUMERIC(YourField) = 1 THEN CONVERT(INT, YourField) - 500 ELSE ASCII(LOWER(YourField)) END
If the field can be converted to a number it is sorted by number otherwise it uses ASCII coding to sort. I have used "- 500" just so there is no cross over in the sort, and to ensure numbers are sorted ahead of text.
ADDENDUM:
Brian Arsuaga has posted a more robust solution to this which I actually prefer, but since this has already been marked as the answer I am adding his solution to this for the benefit of anyone reading this in the future.
ORDER BY
ISNUMERIC(YourField) DESC,
CASE WHEN ISNUMERIC(YourField) = 1 THEN CONVERT(INT, YourField) ELSE 0 END,
YourField
If you don't like using an arbitrary sentinel (500), which might cause sorting issues depending on the range of numbers you expect, you can use multiple expressions for the ordering.
-- put the numbers at the top
ORDER BY ISNUMERIC(YourField) DESC,
-- sort the numbers as numbers, sort the strings as nothing
CONVERT(INT, CASE WHEN ISNUMERIC(YourField) = 1 THEN YourField ELSE '0' END),
-- sort the strings
YourField
The last term is only a tiebreaker when either two terms are both numbers with the same value ('01', '1') or two terms are both non-numbers. For non-numbers, their first and second terms will always be 0.
More complicated, but maybe a little more safe.
Edited to add a nice comparison with the help of the guy below
create table #t
(
YourField varchar(4)
)
insert into #t(YourField) Values('1'), ('3'), ('234'), ('0'), ('00'),
('09'), ('9'), ('1a'), ('aaa'), ('aba'), ('-500')
Select YourField from #t
ORDER BY ISNUMERIC(YourField) DESC,
CONVERT(INT, CASE WHEN ISNUMERIC(YourField) = 1 THEN YourField ELSE '0' END),
YourField
drop table #t
I'm trying to compose an SQL SELECT query with multiple search words. But I want the result be ordered by number of words matches.
For example, let the search string is "red green blue". I want the results which contains all these three words on top, after that the results, which contains two of them, and at the end - only one word matches.
SELECT
*
FROM
table
WHERE
(col LIKE '%red%') OR
(col LIKE '%green%') OR
(col LIKE '%blue%')
ORDER BY
?????
Thanks in advance!
ORDER BY
(
CASE
WHEN col LIKE '%red%' THEN 1
ELSE 0
END CASE
+
CASE
WHEN col LIKE '%green%' THEN 1
ELSE 0
END CASE
+
CASE
WHEN col LIKE '%blue%' THEN 1
ELSE 0
END CASE
) DESC
If your DB vendor has IF, you can use it instead of CASE (e.g., for Mysql you can write
IF (col LIKE '%red% , 1,0) + IF(....'
What platform are you using? if SQL Server, then it sounds like a Full Text Search archtecture would be your best fit.
http://msdn.microsoft.com/en-us/library/ms142583.aspx