I want to convert the decode function from oracle to postgres command.
example oracle command: select decode(p.statusgeometry,1,'pass','fail') as status
please help & guidance
The decode equivalent is CASE:
WITH p (statusgeometry) AS (VALUES (1),(2))
SELECT
CASE statusgeometry
WHEN 1 THEN 'pass'
WHEN 2 THEN 'fail'
END,
-- The following syntax is useful in case you need to do "something"
-- with the columns depending on the condition, e.g lower(), upper(), etc..
CASE
WHEN statusgeometry = 1 THEN 'pass'
WHEN statusgeometry = 2 THEN 'fail'
END
FROM p;
case | case
------+------
pass | pass
fail | fail
(2 rows)
Related
Segment
1
2
3
4
NUll
5
I want to impute 'Other' if the Segment value is null
expected output
Segment
1
2
3
4
Other
5
i have tried
select
case when segment is null then 'Other' else segment end as segment
from table;
It says invalid input syntax for type "numeric":Other
The case expression returns a single type. The problem is that segment is a number, but 'Other' is a string. The expression has to make a choice, and it chooses the numeric type (following standard SQL rules).
This is simple to fix. Just cast segment:
select (case when segment is null then 'Other' else segment::text end) as segment
from table;
It would be more natural to write this query using coalesce():
select coalesce(segment::text, 'Other') as segment
from table;
select
case when CAST(segment AS CHAR) IS NULL then 'Other' else CAST(segment AS CHAR) end as segment
from table
I have created a PL/SQL function where I have a case expression in a SQL query. This is working fine, but when I add another when condition it will not compile. Even if I use when ... and 2 > 1, this is also not compiling.
In the below code, the commented part is not working properly.
What I want is to add one more check in my when clause. Please advise.
create or replace function FUNCTION_NAME (date1 in varchar2,value1 in varchar2)
RETURN date
IS
date2 date;
BEGIN
SELECT D DATE2
INTO DATE2 FROM (SELECT CASE (SELECT TO_DATE(MAX(G.DATE3),'DD-MON-YYYY')
FROM TABLE1 G,
TABLE2 N
WHERE G.DATE3=N.DATE3)
WHEN LAST_DAY(TO_DATE(DATE1,'DD-MON-YYYY'))
/* AND MONTHS_BETWEEN (LAST_DAY(TO_DATE(SYSDATE)),
LAST_DAY(TO_DATE(TO_CHAR(DATE1),'DD-MON-YYYY'))) */
THEN LAST_DAY(TO_DATE(DATE1,'DD-MON-YYYY'))
ELSE
TO_DATE('31-DEC-99','DD-MON-YYYY')
END D
FROM DUAL);
RETURN DATE2;
END;
What you have is a case expression (not a case statement).
Case expressions are of two kinds: "simple" (case <expr> when val1 then ... when val2 then... etc.) and "searched" ( case when condition1 then ... when condition2 then ... etc.)
You wrote your case expression as a simple case expression. You can't, then, add conditions to the WHEN part. You must change the case expression to be "searched" all the way through.
case when (select ...) = last_day(...) AND <your commented condition> THEN .....
EDIT - copying part of a clarifying comment below my Answer.
Simple case expression:
case x when 1 then ....
Can also be written as searched case expression:
case when x = 1 then ....
These two are logically equivalent. However, if we want to add "AND 3 > 1" to the WHEN part, that works only in the searched form of the case expression.
There are two flavours of CASE.
Simple CASE:
select case dummy
when 'X' then 1
end as case_demo
from dual;
Searched CASE:
select case
when dummy = 'X' then 1
end as case_demo
from dual;
In your query you are mixing them like this, which won't work:
select case dummy
when 'X' and 1 = 1
then 1
end as case_demo
from dual;
If you switch to a "searched CASE", then you can add more when conditions:
select case
when dummy = 'X' and 1 = 1
then 1
end as case_demo
from dual;
I tried to:
select 1>2 from dual;
but got:
ORA-00923: FROM keyword not found where expected
Is there boolean type for column expression in Oracle SQL?
I able to do:
select case when 1>2 then 'T' else 'F' end from dual;
Originally I tried to compare date fields and the quickest way I found was getting difference and look to sign...
UPDATE I tried SIGN function, I don't know if it is vendor specific extension:
select SIGN(1-2) from dual;
select SIGN(DATE '2017-01-02' - DATE '2017-02-12') from dual;
but this trick doesn't work for strings...
No there is not, you can use 0 and 1 just as yes/no.
If you need to get the result 1 if something is true and 0 if it is false, you can use a case expression:
select case when (any_logical_condition_here) then 1 else 0 end as my_col
from ....
where ....
For example:
select case when 1 > 2 then 1 else 0 end as bool_result
from dual;
BOOL_RESULT
---------------------------------------
0
NOTE though - "Boolean" refers strictly to the TRUE/FALSE logic, it has no place for UNKNOWN. When you deal with null, as you must in SQL, you need three-valued logic. The case expression as written above returns 1 when the logical condition is true and 0 otherwise. Try it with 1 > null - the truth value is UNKNOWN, the case expression will return 0.
I'm trying to read a column from a database using a SQL query. The column consists of empty string or numbers as strings, such as
"7500" "4460" "" "2900" "2640" "1850" "" "2570" "9050" "8000" "9600"
I'm trying to find the right sql query to extract all the numbers (as integers) and removing the empty ones, but I'm stuck. So far I've got
SELECT *
FROM base
WHERE CONVERT(INT, code) IS NOT NULL
Done in program R (package sqldf)
If all columns are valid integers, you could use:
select * , cast(code as int) IntCode
from base
where code <> ''
To prevent cases when field code is not a valid number, use:
select *, cast(codeN as int) IntCode
from base
cross apply (select case when code <> '' and not code like '%[^0-9]%' then code else NULL end) N(codeN)
where codeN is not null
SQL Fiddle
UPDATE
To find rows where code is not a valid number, use
select * from base where code like '%[^0-9]%'
select *
from base
where col like '[1-9]%'
Example: http://sqlfiddle.com/#!6/f7626/2/0
If you don't need to test for the number being valid, ie. a string such as '909XY2' then this may run marginally faster, more or less depending on the size of the table
Is this what you want?
SELECT (case when code not like '%[^0-9]%' then cast(code as int) end)
FROM base
WHERE code <> '' and code not like '%[^0-9]%';
The conditions are repeated in the where and case on purpose. SQL Server does not guarantee that where filters are applied before logic in the select, so you can get an error with conversions. More recent versions of SQL Server have try_convert() to fix this problem.
Using sqldf with the default sqlite database and this test data:
DF <- data.frame(a = c("7500", "4460", "", "2900", "2640", "1850", "", "2570",
"9050", "8000", "9600"), stringsAsFactors = FALSE)
try this:
library(sqldf)
sqldf("select cast(a as aint) as aint from DF where length(a) > 0")
giving:
aint
1 7500
2 4460
3 2900
4 2640
5 1850
6 2570
7 9050
8 8000
9 9600
Note In plain R one could write:
transform(subset(DF, nchar(a) > 0), a = as.integer(a))
I have the following Searched Case field selection in a Oracle 10g SELECT query
(case
when LOADER_CELLS.CELL_MODE='RW' then 1
when LOADER_CELLS.CELL_MODE='R' then 2
end) as CELL_EDIT_MODE_ID
but if I write it as a Simple Case expression, as follows:
(case LOADER_CELLS.CELL_MODE
when 'RW' then 1
when 'R' then 2
end) as CELL_EDIT_MODE_ID
I get a ORA-12704: character set mismatch error on the when 'RW' line.
I gave a look to the Oracle documentation, and it seems my syntax is correct. http://docs.oracle.com/cd/B19306_01/server.102/b14200/expressions004.htm
Can someone help me on this?
" I supposed that it could be a encoding problem but I don't know how to "cast" the constant strings to a NVARCHAR"
you do it with "N" syntax.
case LOADER_CELLS.CELL_MODE
when n'RW' then 1
when n'R' then 2
end
eg
SQL> select case a when 'a' then 1 end from foo;
select case a when 'a' then 1 end from foo
*
ERROR at line 1:
ORA-12704: character set mismatch
SQL> select case a when n'a' then 1 end from foo;
CASEAWHENN'A'THEN1END
---------------------