Why do I have an error in this query?
My request:
SELECT * FROM CURVES c WHERE c.TYPE_CURVES in ({0}, {10}, {20}, {30})
Error:
ORA-00911 invalid character
Because it should read:
SELECT * FROM CURVES c WHERE c.TYPE_CURVES in (0)
This is a good site for understanding it.
EDIT
Adding multiple pieces of data...
SELECT * FROM CURVES c WHERE c.TYPE_CURVES in (0,20,30,40)
Or as strings...
SELECT * FROM CURVES c WHERE c.TYPE_CURVES in ('0','20','30','40')
Related
Does anyone know why I am receiving this error? I am using SQL in IMPALA and it wont run. Theres a yellow underline under mem_register_hsty_view and transparency_services_summary_2018.
Here is my code:
use sndbx_dx;
SELECT
r.member_identifier,
n.fst_nme
FROM mem_register_hsty_view n
JOIN transparency_services_summary_2018 r
ON RIGHT(TRIM(r.member_identifier),4) = LEFT(n.fst_nme,4)
ORDER BY
r.id_key,
r.group_number,
n.fst_nme;
Here is the error:
AnalysisException: Syntax error in line 1:undefined: ...ervices_summary_2018 r ON RIGHT(TRIM(r.member_identifi... ^ Encountered: RIGHT Expected: CASE, CAST, DEFAULT, EXISTS, FALSE, IF, INTERVAL, NOT, NULL, REPLACE, TRUNCATE, TRUE, IDENTIFIER CAUSED BY: Exception: Syntax error
From the current Impala documentation the functions for taking some number of characters from the left or right of the string appear to actually be STRLEFT and STRRIGHT, respectively. Apply this to your current query gives:
SELECT
r.member_identifier,
n.fst_nme
FROM mem_register_hsty_view n
INNER JOIN transparency_services_summary_2018 r
ON STRRIGHT(TRIM(r.member_identifier), 4) = STRLEFT(n.fst_nme, 4)
ORDER BY
r.id_key,
r.group_number,
n.fst_nme;
Why do I get this error? I'm trying to do queries and for this i need of several selects from different tables, without any combination, but I'm getting this error and I don't known why.
Syntax error(s) [missing EOF at 'select']
Code:
select
d.tdok typ,
d.nr numer,
d.symbmg symbol,
d.data dataDok,
d.dokumwe dokumWe,
d.datawe datawe,
d.idkntrh idKth,
d.kwotadok kwotadok,
d.kwotavat,
k.nazwa1 nazwa1,
k.nazwa2 nazwa2
from dokum0 d
join kontrahent0 k on d.idkntrh=k.idkntrh
select t.kwota_n kwota_n
from dokumterm t
where t.dokum_id = id_dokum;
select p.id_pozdok0 AS settledCount
from pozdok p
where p.dokum_id = id_dokum;
select p.id_pozdok0 AS posCount
from pozdok p
where p.dokum_id=id_dokum;
select t.kupspr kupspr
from sltdok t
where t.tdok=tdok;
Can anyone help me?
You dont have semicolon (;) after end of first select... So this:
join kontrahent0 k on d.idkntrh=k.idkntrh
should be:
join kontrahent0 k on d.idkntrh=k.idkntrh;
I'm trying to execute the following query in Oracle:
SELECT c.id_cliente, c.nombre_cliente, c.apellidos_cliente
FROM cliente c
WHERE not exists (SELECT f.id_finca
FROM finca f
WHERE f.habitaciones = 3
EXCEPT
SELECT v.id_fincas
FROM visitas v
WHERE v.id_cliente = c.id_cliente)
But I am getting the error "missing right parenthesis".
The parenthesis are well-balanced, how can I solve this error?
Use MINUS instead of EXCEPT.
SELECT c.id_cliente,
c.nombre_cliente,
c.apellidos_cliente
FROM cliente c
WHERE NOT EXISTS (SELECT f.id_finca
FROM finca f
WHERE f.habitaciones = 3
MINUS
SELECT v.id_fincas
FROM visitas v
WHERE v.id_cliente = c.id_cliente)
there is no operator EXCEPT in Oracle. Use MINUS instead of it.
Additionally, your query seems weird: better try to make it with two not exists, I think it can give some performance.
I want to find out whether a point is within the circle or not using postgresql.
For point within polygon, I have used following query. i need some equivalent query for circle too.
SELECT a
FROM a_table
WHERE
ST_within(a::geometry,ST_GeomFromText('Polygon((50 -80.98 , 20.99 -90.99 , 90.98 -99.99 , 50 -80.98))'));
for circle, i tried this below query :
SELECT a
FROM a_table
WHERE
ST_within(a::geometry,ST_GeomFromText('POINT(10 20)',10));
and
SELECT a
FROM a_table
WHERE
ST_within(a::geometry,ST_GeomFromText('circle((10 20),10)'));
but both of these gives errors like this :
ERROR: parse error - invalid geometry
SQL state: XX000
Hint: "714" <-- parse error at position 4 within geometry
and
ERROR: parse error - invalid geometry
SQL state: XX000
Hint: "ci" <-- parse error at position 2 within geometry
select a
from a_table
where a <# circle '((10, 20),10))';
Geometric Functions
select point '(1,1)' <# circle '((0,0), 1)';
?column?
----------
f
select point '(1,1)' <# circle '((0,0), 1.5)';
?column?
----------
t
I am getting an error indicating the wrong number of arguments when I run the following query:
SELECT
population_postcodes.*,
target_postcodes.*,
SQR( EXP(population_postcodes.longitude- target_postcodes.longitude, 2) + EXP(population_postcodes.latitude-target_postcodes.latitude, 2) ) as distance
FROM population_postcodes INNER JOIN target_postcodes on Population_postcodes.Population_postcode = Target_postcodes.Target_postcode;
Could anyone please suggest how I can fix this?
I have also tried the following code:
SELECT Population_postcodes.*, Target_postcodes.*
FROM population_postcodes
INNER JOIN target_postcodes
ON Population_postcodes.Population_postcode = Target_postcodes.Target_postcode
SQR( (population_postcodes.longitude- target_postcodes.longitude)^2 + (population_postcodes.latitude-target_postcodes.latitude)^2 ) as distance;
And this code:
SELECT Population_postcodes.*, Target_postcodes.*, SQR( (population_postcodes.longitude- target_postcodes.longitude)^2 + (population_postcodes.latitude-target_postcodes.latitude)^2 ) as distance
FROM population_postcodes
INNER JOIN target_postcodes
ON Population_postcodes.Population_postcode = Target_postcodes.Target_postcode;
Exp needs one parameter, you give two.
Old: EXP(population_postcodes.longitude- target_postcodes.longitude, 2)
New: (population_postcodes.longitude- target_postcodes.longitude)*(population_postcodes.longitude- target_postcodes.longitude)
Try replacing...
EXP(<expression>, 2)
...to...
<expression>^2
In Access, the EXP function returns e (the base of natural logarithms) raised to a power. To raise an expression to a power, use the ^ operator.
In your case, be careful to put brackets around the expression, for example...
(population_postcodes.longitude- target_postcodes.longitude)^2
...to force the power to be applied last. By default, the ^ operator is evaluated before the - operator.