I am trying to write this query:
SELECT DISTINCT createdcfgid FROM ab WHERE (createdcfgid ~ ‘^[0-9]+$’)
This results in
syntax error at or near "[" LINE 3: WHERE (createdcfgid ~ ‘^[0-9]+$’)
Anyone there who can give me a clue about what I am doing wrong?
Thanks in advance
It does just look like you are using the wrong quotes, try ' rather than ‘:
PostgreSQL 8.4 Schema Setup:
create table ab(createdcfgid text);
insert into ab(createdcfgid) values ('111');
Query:
SELECT DISTINCT createdcfgid FROM ab WHERE (createdcfgid ~ '^[0-9]+$')
Results:
| CREATEDCFGID |
----------------
| 111 |
this on SQL Fiddle
Related
I have a column of values where some values contain brackets with text which I would like to remove. This is an example of what I have and what I want:
CREATE TABLE test
(column_i_have varchar(50),
column_i_want varchar(50))
INSERT INTO test (column_i_have, column_i_want)
VALUES ('hospital (PWD)', 'hopistal'),
('nursing (LLC)','nursing'),
('longterm (AT)', 'longterm'),
('inpatient', 'inpatient')
I have only come across approaches that use the number of characters or the position to trim the string, but these values have varying lengths. One way I was thinking was something like:
TRIM('(*',col1)
Doesn't work. Is there a way to do this in postgres SQL without using the position? THANK YOU!
If all the values contain "valid" brackets, then you may use split_part function without any regular expressions:
select
test.*,
trim(split_part(column_i_have, '(', 1)) as res
from test
column_i_have | column_i_want | res
:------------- | :------------ | :--------
hospital (PWD) | hopistal | hospital
nursing (LLC) | nursing | nursing
longterm (AT) | longterm | longterm
inpatient | inpatient | inpatient
db<>fiddle here
You can replace partial patterns using regular expressions. For example:
select *, regexp_replace(v, '\([^\)]*\)', '', 'g') as r
from (
select '''hospital (PWD)'', ''nursing (LLC)'', ''longterm (AT)'', ''inpatient''' as v
) x
Result:
r
-------------------------------------------------
'hospital ', 'nursing ', 'longterm ', 'inpatient'
See example at db<>fiddle.
Could it be as easy as:
SELECT SUBSTRING(column_i_have, '\w+') AS column_i_want FROM test
See demo
If not, and you still want to use SUBSTRING() to get upto but exclude paranthesis, then maybe:
SELECT SUBSTRING(column_i_have, '^(.+?)(?:\s*\(.*)?$') AS column_i_want FROM test
See demo
But if you really are looking upto the opening paranthesis, then maybe just use SPLIT_PART():
SELECT SPLIT_PART(column_i_have, ' (', 1) AS column_i_want FROM test
See demo
The goal is to select all rows that contain some specific word, can be in the beginning or the end of the string and/or surrounded by white-space, should not be inside other word, so to speak.
Here are couple rows in my database:
+---+--------------------+
| 1 | string with test |
+---+--------------------+
| 2 | test string |
+---+--------------------+
| 3 | testing stringtest |
+---+--------------------+
| 4 | not-a-test |
+---+--------------------+
| 5 | test |
+---+--------------------+
So in this example, selecting word test, should return rows 1, 2 and 5.
Problem is that for some reason, SELECT * FROM ... WHERE ... RLIKE '(\s|^)test(\s|$)'; returns 0 rows.
Where am I wrong and maybe, how it could be done better?
Edit: Query should also select the row with just a word test.
The answer to my first question is:
I haven't escaped special characters, so \s should be \\s.
Working query: SELECT * FROM ... WHERE ... RLIKE '(\\s|^)test(\\s|$)';. (or just a space ( |^)/( |$), also works)
Hi you could grab with trailing space and with leading space
SELECT * from new_table
where text RLIKE(' test')
union
SELECT * from new_table
where text RLIKE('test ')
REGEXP_INSTR() function, which's is an extension of the INSTR() function, might be used for version 10.0.5+ case-insensitively as default :
SELECT *
FROM t
WHERE REGEXP_INSTR(str, 'TeSt ')>0
OR REGEXP_INSTR(str, ' tESt')>0
Demo
SELECT * FROM ...
WHERE ... LIKE 'test';
This should do the trick.
Is this what you want?
SELECT * FROM ... WHERE ... LIKE
'%test%';
Use word boundary tests:
Before MySQL 8.0, and in MariaDB:
WHERE ... REGEXP '[[:<:]]test[[:>:]]'
MySQL 8.0:
WHERE ... REGEXP '\btest\b'
(If that does not work, double up the backslashes; this depends on whether the client is collapsing backslashes before MySQL gets them.)
Note that this solution will also work with punctuation such as the comma in "foo, test, bar"
If I have something like this in SQL statement ('A','B','C'), how do I convert it into a column with multiple rows like this
col
---
A
B
C
I cannot change the way that string is created (as it is injected into SQL query from external program). For example, I cannot make it as ['A','B','C'] (replace with square brackets). I could wrap anything around it though like [('A','B','C')] or whatever.
Any help?
UPDATE 1
I have PostgreSQL 8.4.20
You could create an ARRAY from VALUES and then unnest it:
SELECT
unnest(ARRAY[col_a, col_b, col_c])
FROM
(VALUES('A','B','C')) AS x(col_a, col_b, col_c)
Result:
| unnest |
|--------|
| A |
| B |
| C |
Edit: you could also tweak jspcal's answer by using dollar quotes ($$) like this so you can concatenate your string into the SQL statement:
SELECT * FROM regexp_split_to_table(
regexp_replace(
$$('A','B','C','D','foo')$$,
'^\(''|''\)+', '', 'g'),
''','''
);
The built-in regexp_split_to_table function will do this for you. Since you plan to inject it directly without escaping, use $$ (dollar quoting) from thibautg's answer.
select * from regexp_split_to_table(
regexp_replace($$('A','B','C')$$, '^\(''|''\)+', '', 'g'),
''','''
);
I have a table and I'd like to pull one row per id with field values concatenated.
In my table, for example, I have this:
TM67 | 4 | 32556
TM67 | 9 | 98200
TM67 | 72 | 22300
TM99 | 2 | 23009
TM99 | 3 | 11200
And I'd like to output:
TM67 | 4,9,72 | 32556,98200,22300
TM99 | 2,3 | 23009,11200
In MySQL I was able to use the aggregate function GROUP_CONCAT, but that doesn't seem to work here... Is there an equivalent for PostgreSQL, or another way to accomplish this?
Since 9.0 this is even easier:
SELECT id,
string_agg(some_column, ',')
FROM the_table
GROUP BY id
This is probably a good starting point (version 8.4+ only):
SELECT id_field, array_agg(value_field1), array_agg(value_field2)
FROM data_table
GROUP BY id_field
array_agg returns an array, but you can CAST that to text and edit as needed (see clarifications, below).
Prior to version 8.4, you have to define it yourself prior to use:
CREATE AGGREGATE array_agg (anyelement)
(
sfunc = array_append,
stype = anyarray,
initcond = '{}'
);
(paraphrased from the PostgreSQL documentation)
Clarifications:
The result of casting an array to text is that the resulting string starts and ends with curly braces. Those braces need to be removed by some method, if they are not desired.
Casting ANYARRAY to TEXT best simulates CSV output as elements that contain embedded commas are double-quoted in the output in standard CSV style. Neither array_to_string() or string_agg() (the "group_concat" function added in 9.1) quote strings with embedded commas, resulting in an incorrect number of elements in the resulting list.
The new 9.1 string_agg() function does NOT cast the inner results to TEXT first. So "string_agg(value_field)" would generate an error if value_field is an integer. "string_agg(value_field::text)" would be required. The array_agg() method requires only one cast after the aggregation (rather than a cast per value).
SELECT array_to_string(array(SELECT a FROM b),', ');
Will do as well.
Try like this:
select field1, array_to_string(array_agg(field2), ',')
from table1
group by field1;
Assuming that the table your_table has three columns (name, id, value), the query is this one:
select name,
array_to_string(array_agg(id), ','),
array_to_string(array_agg(value), ',')
from your_table
group by name
order by name
;
"TM67" "4,9,72" "32556,98200,22300"
"TM99" "2,3" "23009,11200"
KI
and the version to work on the array type:
select
array_to_string(
array(select distinct unnest(zip_codes) from table),
', '
);
My sugestion in postgresql
SELECT cpf || ';' || nome || ';' || telefone
FROM (
SELECT cpf
,nome
,STRING_AGG(CONCAT_WS( ';' , DDD_1, TELEFONE_1),';') AS telefone
FROM (
SELECT DISTINCT *
FROM temp_bd
ORDER BY cpf DESC ) AS y
GROUP BY 1,2 ) AS x
In my experience, I had bigint as column type. So The below code worked for me. I am using PostgreSQL 12.
Type cast is happening here. (::text).
string_agg(some_column::text, ',')
Hope below Oracle query will work.
Select First_column,LISTAGG(second_column,',')
WITHIN GROUP (ORDER BY second_column) as Sec_column,
LISTAGG(third_column,',')
WITHIN GROUP (ORDER BY second_column) as thrd_column
FROM tablename
GROUP BY first_column
I am using Postgres/Redshift to query a table of URLs and am trying to use
SELECT regex_substr to select a string that is between the second and third forward slash in the column.
For example I need the second slash delimited string in the following data:
/abc/required_string/5856365/
/abc/required_string/2/
/abc/required_string/l-en/
/abc/required_string/l-en/
Following some of the regexs in this this thread:
SELECT regexp_substr(column, '/[^/]*/([^/]*)/')
FROM table
None seem to work. I keep getting:
/abc/required_string/
/abc/required_string/
What about split_part?
SELECT split_part(column, '/', 3) FROM table
Example:
select split_part ('/abc/required_string/2/', '/', 3)
Returns: required string
This may work :
SQL Fiddle
PostgreSQL 9.3 Schema Setup:
CREATE TABLE t
("c" varchar(29))
;
INSERT INTO t
("c")
VALUES
('/abc/required_string/5856365/'),
('/abc/required_string/2/'),
('/abc/required_string/l-en/'),
('/abc/required_string/l-en/')
;
Query 1:
SELECT substring("c" from '/[^/]*/([^/]*)/')
FROM t
Results:
| substring |
|-----------------|
| required_string |
| required_string |
| required_string |
| required_string |