What if I wanted sorting to have all the upper case first then the lower case?
A
B
C
D
a
b
c
d
I tried searching the net and all I could find was sorting that would make it
a
A
b
B
c
C
etc..
but I wanted all upper case values sorted first then lower case ones.
Any idea? thanks
Try to order by the BINARY value of your characters.
SELECT column
FROM my_table
ORDER BY NLSSORT(column, 'NLS_SORT = BINARY')
Use a case expression to conditionally order the column, based on upper or lower case. Then order by the original column.
select * from tablename
order by case when upper(col) = col then 1 else 2 end, col
Note: The ordering above works well when there is only one character in the string or the when the entire string is either upper case or lower case.
Related
I have data that looks like:
row
col1
col2
col3
...
coln
1
A
null
B
...
null
2
null
B
C
...
D
3
null
null
null
...
A
I want to condense the columns together to get:
row
final
1
A, B
2
B, C, D
3
A
The order of the letters doesn't matter, and if the solution includes the nulls eg. A,null,B,null ect. I can work out how to remove them later. I've used up to coln as I have about 200 columns to condense.
I've tried a few things and if I were trying to condense rows I could use STRING_AGG() example
Additionally I could do this:
SELECT
CONCAT(col1,", ",col2,", ",col3,", ",coln) #ect.
FROM mytable
However, this would involve writing out each column name by hand which isn't really feasible. Is there a better way to achieve this ideally for the whole table.
Additionally CONCAT returns NULL if any value is NULL.
#standardSQL
select row,
(select string_agg(col, ', ' order by offset)
from unnest(split(trim(format('%t', (select as struct t.* except(row))), '()'), ', ')) col with offset
where not upper(col) = 'NULL'
) as final
from `project.dataset.table` t
if to apply to sample data in your question - output is
Not in exact format that you asked for, but you can try if this simplifies things for you:
SELECT TO_JSON_STRING(mytable) FROM mytable
If you want the exact format, you can write a regex to extract values from the output JSON string.
I am trying to sort based on below condition where I have hidden one numeric value with '-' i want to sort in a manner that however I sort (desc,asc) '-' should come on top or at the bottom but it does not affect my sorting, below goes an example
and - can be a negative or positive value
before sorting
1
2
3
-
5
6
7
after sorting
-
1
2
3
5
6
7
You can set the sort order expression to be something like
=Val(Fields!myColumn.Value)
This will return zero for the - row and the numeric value for the other rows.
If this does not help, edit the question (rather than add more comments) and show what the data looks like that comes from your dataset query in the report.
It is quite unclear what the data type of the column is, because - is not a numeric value.
However, regardless of the type, this does what you want:
order by col
If the column is a number and - is NULL, this works because an ascending sort puts NULL values first.
If the column is a string and '-' is the value, this works because '-' sorts before numbers.
Here is a db<>fiddle.
Sort using a CASE expression:
SELECT *
FROM yourTable
ORDER BY
CASE WHEN col = '-' THEN 0 ELSE 1 END,
col;
Actually, it appears that you are mixing character and numeric data in the same column. In order to bear the value -, it must be text, so we should probably do a bit extra work to make sure the text numbers sort properly:
SELECT *
FROM yourTable
ORDER BY
CASE WHEN col = '-' THEN 0 ELSE 1 END,
TRY_CONVERT(NUMERIC(10), col);
Am trying to come up with a query where I can return back values where the the distance between the letters could be one or more than one for the chosen letter.
For example:
I have two columns which have letters in Column A and in Column B. I want to return back with rows when column B distance is more than Column A by one or more letters.
It's not clear to me, when you say "greater" if you mean that the distance between any two letters is 2 or 3 (Column B can be alphabetically before or after Column A, by a distance of 2 or 3).. Or if Column B has to be alphabetically after Column A, by a distance of 2 or 3
Because I'm not certain what you're talking about, I present two options. Read the "if" rule and choose the one that applies to your situation, then use the query under it:
If columnA is D and columnB can be any of: A B F G
SELECT * FROM table WHERE ABS(ASCII(columna) - ASCII(columnb)) IN (2,3)
If columnA is D and columnB can be any of: F G
SELECT * FROM table WHERE ASCII(columnb) - ASCII(columna) IN (2,3)
Edit1: Per your later comment, you are now saying that the distance is not just 2 or 3 letters (the first line of your question states "2 or 3") but any number of letters distance equal to or greater than 2:
SELECT * FROM table WHERE ASCII(columnb) - ASCII(columna) >= 2
Overall the technique isn't much different to the above queries and there are many ways to specify what you want:
SELECT * FROM table
WHERE
ASCII(columnb) - ASCII(columna)
BETWEEN <some_number_here> AND <other_number_here>
Ultimately the most important thing is to note the use of ASCII function, which gives us the ascii char code of the first letter in a string:
ASCII('ABCD') => 65
And we can use maths on this to work out if a letter distance from 'A' is more than 1 etc..
Probably also worth noting that ASCII() works on single byte ascii characters. If your data is multibyte (Unicode), you might need to use ORD() instead:
Edit2: Your latest edit to the question revises the limit to "B greater than A by one or more" which is equivalent to >= 1 ..
The question seems not to have a clear spec, please treat the answer as a guide for the general technique:
--for an open ended distance, ascii chars
SELECT * FROM table WHERE ASCII(columnb) - ASCII(columna) >= <some_distance>
--for an open ended distance, unicode
SELECT * FROM table
WHERE ORD(columnb) - ORD(columna) >= <some_distance>
--for a definite range of distances (replace … appropriately)
SELECT * FROM table
WHERE ... BETWEEN <some_distance> AND <some_other_distance>
this will work indeed:
select * from table_name where ascii(col_1)+2=ascii(col_2);
You can use something like this if you need it to be exactly 2 or 3 letters greater
select Column A, ColumnB from table name where ASCII(ColumnB) - ASCII(ColumnA) in (2,3)
If you want all those rows where the the difference is equal more than 2, then use this
select Column A, ColumnB from table name where ASCII(ColumnB) - ASCII(ColumnA) >=2
this is where you can make ascii in action..
select * from SampleTable where (ASCII(sampleTable.ColumnB) - ASCII(ColumnA)) >= 2;
I have this table
id value
1 OK
2 xminimum
3 NO
4 YES
I want to sort this table by value where minimum is always first then the rest according to alphabetic order of value column
Meaning:
xminimum
NO
OK
YES
I wrote this query:
Select *
from table_a
order by case when value='xminimum' then 1 else ????? end
I don't know what to put in the else... conceptually it should be else value end so it means alphabetic order.. but I can not combine integer with text.
How do I fix it?
As requested, copied from my comment:
Select *
from table_a
order by case when value='xminimum' then 1 else 2 end, value
Another solution:
SELECT *
FROM table_a
ORDER BY value <> 'xminimum', value;
Do it like you have and add the value column as second column to sort by:
SELECT *
FROM table_a
ORDER BY CASE WHEN value='xminimum' THEN 1 ELSE 2 END, value
I am working in Teradata with some descriptive data that needs to be transformed from a gerneric varchar(60) into the different field lengths based on the type of data element and the attribute value. So I need to take whatever is in the Varchar(60) and based on field 'ABCD' act on field 'XYZ'. In this case XYZ is a varchar(3). To do this I am using CASE logic within my select. What I want to do is
eliminate all occurances of non alphabet/numeric data. All I want left are upper case Alpha chars and numbers.
In this case "Where abcd = 'GROUP' then xyz should come out as a '000', '002', 'A', 'C'
eliminate extra padding
Shift everything Right
abcd xyz
1 GROUP NULL
2 GROUP $
3 GROUP 000000000000000000000000000000000000000000000000000000000000
4 GROUP 000000000000000000000000000000000000000000000000000000000002
5 GROUP A
6 GROUP C
7 GROUP r
To do this I have tried TRIM and SUBSTR amongst several other things that did not work. I have pasted what I have working now, but I am not reliably working through the data within the select. I am really looking for some options on how to better work with strings in Teradata. I have been working out of the "SQL Functions, Operators, Expressions and Predicates" online PDF. Is there a better reference. We are on TD 13
SELECT abcd
, CASE
-- xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
WHEN abcd= 'GROUP'
THEN(
CASE
WHEN SUBSTR(tx.abcd,60, 4) = 0
THEN (
SUBSTR(tx.abcd,60, 3)
)
ELSE
TRIM (TRAILING FROM tx.abcd)
END
)
END AS abcd
FROM db.descr tx
WHERE tx.abcd IS IN ( 'GROUP')
The end result should look like this
abcd xyz
1 GROUP 000
2 GROUP 002
3 GROUP A
4 GROUP C
I will have to deal with approx 60 different "abcd" types, but they should all conform to the type of data I am currently seeing.. ie.. mixed case, non numeric, non alphabet, padded, etc..
I know there is a better way, but I have come in several circles trying to figure this out over the weekend and need a little push in the right direction.
Thanks in advance,
Pat
The SQL below uses the CHARACTER_LENGTH function to first determine if there is a need to perform what amounts to a RIGHT(tx.xyz, 3) using the native functions in Teradata 13.x. I think this may accomplish what you are looking to do. I hope I have not misinterpreted your explanation:
SELECT CASE WHEN tx.abcd = 'GROUP'
AND CHARACTER_LENGTH(TRIM(BOTH FROM tx.xyz) > 3
THEN SUBSTRING(tx.xyz FROM (CHARACTER_LENGTH(TRIM(BOTH FROM tx.xyz)) - 3))
ELSE tx.abcd
END
FROM db.descr tx;
EDIT: Fixed parenthesis in SUBSTRING