Hello I have a query where I want to select the value of one of two fields depending if one is empty.
field1 and field2
I want to select them as complete_field
IF field1 is empty, then complete_field is field2
ELSE complete_field is field1
In php it would be done like:
$complete_field = $field1 == '' ? $field2 : $field1;
How would I do this in PostgreSQL?
I tried:
SELECT
(IF field1 = '' THEN field2 ELSE field1) AS complete_field
FROM
table
But it doesnt work.
Please help me :) Thanks
Use CASE WHEN .. THEN .. ELSE .. END, e.g.:
SELECT
(CASE WHEN (field1 IS NULL OR field1 = '') THEN field2 ELSE field1 END)
FROM table;
Check it out at the official docs.
Try COALESCE with NULLIF:
SELECT COALESCE(NULLIF(field1, ''), field2) AS the_field FROM my_table;
The ANSI SQL function Coalesce() is what you want.
select Coalesce(field1,field2)
from table;
Related
Say that you need to query some data and that there are three fields like the following (this is part of a larger query):
Field1, Field2, Field3.
So you select them like this:
SELECT Field1=MyTable.Field1, Field2=MyTable.Field2, Field3=MyTable.Field3
FROM MyTable
I need to compare these values and return the variable Result that is computed as follows:
0 if they are all the same
1/2 if two are the same and one is different
1 if they are all different.
How should I restructure my query? I think I need a subquery but I am not sure how to structure it.
You can use case:
select (case when field1 = field2 and field1 = field3 then 0
when field1 in (field2, field3) or field2 = field3 then 0.5
else 1
end) as result
Is it possible to add a select statement in case or if function in SQL
select case when :A='do' then (select col1 from table1) else 'n/a' end;
or
select if(:A='do',(select col1 from table1),'N/A');
If my parameter is 'do' it should display all the value in table or else it should just display 'N/A'.
Please help me. Thanks!
If this is Oracle, as per your tag, then neither of your queries are valid as they're not actually selecting from a table.
Perhaps what you're after is something like:
select case when :A='do' then col1 else 'n/a' end col
from table1;
Or maybe you're after something like:
select col1
from table1
where :A != 'do'
union all
select 'N/A' col1
from dual
where :A = 'do';
You didn't provide any example data, so I'm not sure if what you're trying to do is make all the values of col1 appear as 'N/A' if the bind variable is "do" or whether you only want a single row containing 'N/A'.
QUERY:
select ws_path from workpaths where
(
(ws_path like '%R_%') or
(ws_path like '%PB_%' ) or
(ws_path like '%ST_%')
)
OUTPUT:
/x/eng/users/ST_3609843_ijti4689_3609843_1601272247
/x/eng/users/ST_3610020_zozt5229_3610020_1601282033
/x/eng/users/ST_3611181_zozt5229_3611181_1601282032
/x/eng/users/ST_3611226_zozt5229_3611226_1601282033
/x/eng/users-random/john/N_3582168_3551186_1601040805
/x/eng/users-random/james/N_3582619_3551186_1601041405
/x/eng/users-random/jimmy/N_3582791_3551186_1601042005
/x/eng/users/R_3606462_3606462_1601251334
/x/eng/users/R_3611775_3612090_1601290909
/x/eng/users/R_3612813_3613016_1601292252
Is there way to group partially by ST_, N_ and R_?
i.e. group by ws_path wont work at the moment for the obvious reason
I need to only look at the last item in the path (split by '/') and then the front part of splitting with '_'
You can use regexp_substr to get the patterns being searched for and then group by the number of such occurrences.
select regexp_substr(ws_path,'\/R_|\/PB_|\/ST_'), count(*)
from workpaths
group by regexp_substr(ws_path,'\/R_|\/PB_|\/ST_')
Regexp is a good solution but can be expensive. A simpler substring might be cheaper and faster:
CREATE TABLE tbl (field1 VARCHAR(100));
INSERT INTO dbo.tbl
( field1 )
VALUES
('/x/eng/users/ST_3609843_ijti4689_3609843_1601272247'),
('/x/eng/users/ST_3610020_zozt5229_3610020_1601282033'),
('/x/eng/users/ST_3611181_zozt5229_3611181_1601282032'),
('/x/eng/users/ST_3611226_zozt5229_3611226_1601282033'),
('/x/eng/users-random/john/N_3582168_3551186_1601040805'),
('/x/eng/users-random/james/N_3582619_3551186_1601041405'),
('/x/eng/users-random/jimmy/N_3582791_3551186_1601042005'),
('/x/eng/users/R_3606462_3606462_1601251334'),
('/x/eng/users/R_3611775_3612090_1601290909'),
('/x/eng/users/R_3612813_3613016_1601292252');
SELECT
COUNT(CASE WHEN field1 LIKE '%/ST_%' THEN 1 ELSE NULL END) AS 'st_count',
COUNT(CASE WHEN field1 LIKE '%/N_%' THEN 1 ELSE NULL END) AS 'n_count',
COUNT(CASE WHEN field1 LIKE '%/R_%' THEN 1 ELSE NULL END) AS 'r_count'
FROM dbo.tbl
Is it possible in an SQL query to only show a field if another field has data? For example, if Field1 <> '', then show the value in Field2 else don't show the value?
It can be done using a case statement. (At least in SQL Server)
select case when Field1 <> ''
then Field2
end as Field2
from YourTable
Sure (this works in Oracle and SQLite):
select
field1,
(case
when field1 is null then null
else field2
end) field2_wrapped
from my_table
if 'has no data' means the empty string (''), you need to use this statement:
SELECT Filed2 FROM Table1 WHERE Filed1<>''
If 'no data' means NULL value, you need use
SELECT Filed2 FROM Table1 WHERE NOT (Filed2 IS NULL)
Take a look at Standard SQL functions COALESCE() and NULLIF():
COALESCE(NULLIF(Field1, ''), Field2)
I have a query in Access and would like to know if it were possible to use the where not exists clause to display a specific text for each field when there are no returned rows.
Example query:
Select Field1, Field2, Field3
From TableA
Where Field1 = "test";
If there are no returned results I would like the following to return:
Field1 = "test"
Field2 = "not provided"
Field2 = "not provided"
How about:
SELECT Field1, Field2
FROM Table
WHERE ID=3
UNION ALL SELECT DISTINCT "None","None" FROM AnyTableithAtLeastOneRow
WHERE 3 NOT IN (SELECT ID FROM Table)
The usual way to do what you're asking is:
Select Field1, isnull(Field2, 'Not Provided'), isnull(Field3, 'Not Provided')
edit
whoops, you're using Access, in that case the equivalent function is "nz" (what?! :p)
Select Field1, nz(Field2, 'Not Provided'), nz(Field3, 'Not Provided')