I am using PostgreSQL with PostGis. I am executing a query like this:
select st_geomfromtext('point(22 232)',432)
It works fine. But now I want to take a value through a query. for example:
select st_geomfromtext('point((select x from data_name where id=1) 232)' , 432)
Here data_name is some table I am using and x stores some values. Now query inside is treated as a string and no value is returned.
Please help.
ERROR: syntax error at or near "select"
Try this:
select st_geomfromtext('point(' || x || ' 232)', 432) from data_name where id=1
Postgis has a function ST_MakePoint that is faster than ST_GeomFromText.
select ST_SetSRID(ST_MakePoint(x),432) from data_name where id=1;
While #muratgu answer is generally the way to go, one minor note:
A subquery gets you a different result when no row is found for id = 1. Then you get nothing back (no row), instead of:
select st_geomfromtext(NULL, 432)
If you need a drop-in replacement:
select st_geomfromtext('point('
|| (select x from data_name where id=1)
|| ' 232)' , 432)
Related
I need to find (exclude in fact) any results that contain '%' sign, wherever in a string field. That would mean ... WHERE string LIKE '%%%'. Googling about escaping gave me the following ideas. The first throws syntax error, the second returns rows but there are records actually contain '%'.
1st:
SELECT * FROM table
WHERE string NOT LIKE '%!%%' ESCAPE '!'
///tried with different escape characters
2nd:
SELECT * FROM table
WHERE string NOT LIKE '%[%]%'
Trying on GCP BigQuery.
Try:
SELECT *
FROM table
WHERE string NOT LIKE '%!%%' {ESCAPE '!'}
With curly braces as shown in microsoft sql server docs
Or also:
WITH indata(s) AS (
SELECT 'not excluded'
UNION ALL SELECT '%excluded'
UNION ALL SELECT 'Ex%cluded'
UNION ALL SELECT 'Excluded%'
)
SELECT * FROM indata WHERE INSTR(s,'%') = 0;
-- out s
-- out --------------
-- out not excluded
find (exclude in fact) any results that contain '%'
Consider below simple approach
select *
from your_table
where not regexp_contains(string , '%')
Why do I get an invalid identifier error in this query?
with
select
notif_set_code as notification_code
, (
select tab_to_string(
cast(collect(
case
when area_type is not null then
count(*) || ' ' || area_type
else 'default'
end
) as t_varchar2_tab), ' ,')
from my_table_2
where notif_set_code = nsd.notification_code
) areas
from my_table nsd
...
It sayd nsd.notification_code is not a valid identifier.
Using nsd.notif_set_code works if it's not in a with clause!
Presumably, there is no column called notification_code in nsd. You seem to intend:
where notif_set_code = nsd.notif_set_code
Renaming the column in the select doesn't change anything (although rather confusingly MySQL seems to allow column aliases in subqueries but that is an anomaly).
However, you should really qualify all column references:
where my_table_2.notif_set_code = nsd.notif_set_code
I would like to keep rows only with Numeric Character i.e. 0-9. My source data can have any type of character e.g. 2,%,( .
Input (postcode)
3453gds sdg3
454232
sdg(*d^
452
Expected Output (postcode)
454232
452
I have tried using WHERE REGEXP_LIKE(postcode, '^[[:digit:]]+$');
however in my version of Oracle I get an error saying
function regexp_like(character varying, "unknown") does not exist
You want regexp_like() and your version should work:
select t.*
from t
where regexp_like(t.postcode, '^[0-9]+$');
However, your error looks more like a Postgres error, so perhaps this will work:
t.postcode ~ '^[0-9]+$'
For Oracle 10 or higher you can use regexp functions. In earlier versions translate function will help you :
SELECT postcode
FROM table_name
WHERE length(translate(postcode,'0123456789','1')) is null
AND postcode IS NOT NULL;
OR
SELECT translate(postcode, '0123456789' || translate(postcode,'x123456789','x'),'0123456789') nums
FROM table_name ;
the above answer also works for me
SELECT translate('1234bsdfs3#23##PU', '0123456789' || translate('1234bsdfs3#23##PU','x123456789','x'),'0123456789') nums
FROM dual ;
Nums:
1234323
For an alternative to the Gordon Linoff answer, we can try using REGEXP_REPLACE:
SELECT *
FROM yourTable
WHERE REGEXP_REPLACE(postcode, '[0-9]+', '') IS NULL;
The idea here is to strip away all digit characters, and then assert that nothing were left behind. For a mixed digit-letter value, the regex replacement would result in a non-empty string.
I was wondering if is there a way to do a query like this one (where Z is a variable)
SELECT * FROM table t WHERE t.X = Z OR t.Y = Z
writing only one time the Z. To be more specific I would like to do the same query like so
SELECT * FROM table t WHERE (t.X OR t.Y) = Z
I am using an Oracle DB and it (obviously) gives me an error when I try to execute it BUT I really like a way to do it like in the second query.
To make you know my situation X and Y are both VARCHAR2 and I already try something like
SELECT * FROM table t WHERE (t.X || t.Y) like '%Z%'
But, unluckily, it is not as accurate as the first one.
You can use in:
WHERE Z IN (t.X, t.Y)
Columns can be in the IN list as well as constants.
Try Like This
SELECT * FROM table t WHERE (t.X like %Z% OR t.Y like %Z% )
SQLite 3.7.11
Given I have these two records in a SQLITE table, with the field called name:
"preview"
"view"
If I run the query:
SELECT * FROM table WHERE name LIKE "%" + $keyword + "%" ORDER BY name
with $keyword set to "vi"
I would get the results in this order:
1) "preview"
2) "view"
Is there a way to order so that the names whose first letter is the same as the first letter of the keyword (in this example "v") would come first?
You can sort by the position of your keyword in the search string
SELECT * FROM your_table
WHERE name LIKE concat('%', '$keyword', '%')
ORDER BY POSITION('$keyword' IN name)
name
Not sure if it can be done less verbose, but this should work. I used ? as a placeholder for your variable. The query as you posted it doesn't seem to be correct SQL.
SELECT * FROM table1
WHERE name LIKE '%' || ? || '%'
ORDER BY (CASE WHEN SUBSTR(Name,1,1) = SUBSTR(?,1,1) THEN 0 ELSE 1 END),
name