I am running this query:
Select column_name
from table
where column_name ~ '%[A-Za-z]%'
group by column_name
but I am not getting any results. What am I doing wrong?
Goal: This is a column that includes phone numbers. I am trying to find any values that contain string characters.
I don't understand why ilike does not support regex
This is what I found here
The operator ~~ is equivalent to LIKE, and ~~* corresponds to ILIKE. There are also !~~ and !~~* operators that represent NOT LIKE and NOT ILIKE, respectively. All of these operators are PostgreSQL-specific.
Doesn't this mean I can use ilike by using ~~*?
Edit: What I have learned so far
Don't use like, use tilda.
Where column_name ~ '%[A]%'
Where column_name ~ $$[A]$$ does work.
Theory: It has something to with the dollar signs or the apostrophe.
Result: It was the % signs.
According to RegexBuddy, the correct syntax for the WHERE clause is:
WHERE mycolumn ~ $$[A-Z]$$
Related
I have a scenario wherein I want to replace a period when its surrounded by Alphabets and not when surrounded by Numbers. I figured out a Regular Expression pattern that can identify only the periods in Key names but the pattern is not working in SQL
SELECT REGEXP_REPLACE("Amount.fee:0.75,Amount.tot:645.55","(?<!\d)(\.)(?!\d)","_","ig");
Expected output: Amount_fee:0.75,Amount_tot:645.55
Note, I am trying this because, In MemSQL I couldn't access JSON key when it has period in it.
Also verified the pattern "(?<!\d)(.)(?!\d)" using https://coding.tools/regex-replace and it working fine. But, SQL is not working. Am using MemSQL 7.1.9 and POSIX Enhanced Regular expression are supposed to be work. Any help is much appreciated.
Since it looks like you are trying to workaround accessing a JSON key with a period, I will show you how to do that.
This can be done by either surrounding the json key name with backtics while using the shorthand json extract syntax:
select col::%`Amount.fee` from (select '{"Amount.fee":0.75,"Amount.tot":645.55}' col);
+--------------------+
| col::%`Amount.fee` |
+--------------------+
| 0.75 |
+--------------------+
or by using the json_extract_ builtins directly:
select json_extract_double('{"Amount.fee":0.75,"Amount.tot":645.55}', 'Amount.fee');
+------------------------------------------------------------------------------+
| json_extract_double('{"Amount.fee":0.75,"Amount.tot":645.55}', 'Amount.fee') |
+------------------------------------------------------------------------------+
| 0.75 |
+------------------------------------------------------------------------------+
Assuming you only want to target dots that are in between two non digit characters, where the dot is not the first or last character in the string, you may match on ([^\d])\.([^\d]) and replace with \1_\2:
SELECT REGEXP_REPLACE("Amount.fee:0.75,Amount.tot:645.55", "([^\d])\.([^\d])", "\1_\2", "ig");
Here is a regex demo showing that the replacement is working. Note that you might have to use $1_$2 instead of \1_\2 as the replacement, depending on the regex flavor of your SQL tool.
In PostgreSQL, I have the following text type in a column value.
{186=>15.55255158, 21=>5123.43494408, 164=>0.0}
I would like to select the numbers before the => character and use the ouput in a subquery. So the output should be:
186
21
164
I tried several regex statement but it does not work. Any help would be appreciated.
You need to both use a regular expression, and a function to extract the values matched from the regular expression into a data set. The ~ operator is only used to match in a where clause. You need the REGEXP_MATCHES function.
SELECT REGEXP_MATCHES(your_column_name, '(\d+)=>', 'g')
FROM your_table_name
The 'g' option will return multiple matches, rather than just the first.
SQL Fiddle
you can use the simple substring(column_name from '(\d+)=>') to extract the appropriate data.
I have 2 tables in Redshift, one of them has a column containing Regex strings. And I want to join them like so:
select *
from one o
join two t
on o.value ~ t.regex
But this query throws an error:
[Amazon](500310) Invalid operation: The pattern must be a valid UTF-8 literal character expression
Details:
-----------------------------------------------
error: The pattern must be a valid UTF-8 literal character expression
code: 8001
context:
query: 412993
location: cgx_impl.cpp:1911
process: padbmaster [pid=5211]
-----------------------------------------------;
As far as I understood from searching in the docs, the right side of a regex operator ~ must be a string literal.
So this would work:
select *
from one o
where o.value ~ 'regex'
And this would fail:
select *
from one o
where 'regex' ~ o.value
Is there any way around this? Anything I missed?
Thanks!
Here's a workaround I am using. Maybe it's not super fast, but it works:
First create a function:
CREATE FUNCTION is_regex_match(pattern text, s text) RETURNS BOOLEAN IMMUTABLE AS $$
import re
return True if re.search(pattern, s) else False
$$ LANGUAGE plpythonu;
Then use it like this (o.value contains a regex pattern):
select *
from one o
where is_regex_match(o.value, 'some string');
You could try using the built-in function regexp_substr()
https://docs.aws.amazon.com/redshift/latest/dg/REGEXP_SUBSTR.html
select *
from one o
join two t
on regexp_substr(o.value, t.regex) <> ''
Edit example added of raw query
It appears that the fields must be explicitly cast as varchars when built.
with fake_table as (
SELECT 'sample value'::varchar as value, '[a-z]'::varchar as pattern
)
SELECT *
, regexp_substr(value, pattern)
FROM
fake_table
WHERE
regexp_substr(value, pattern) <>''
I'm trying to create a regex to find (and then eventually replace) parts of strings in a PG DB. I'm using PSQL 9.0.4
I've tested my regex outside of PG and it works perfectly. However, it isn't playing well with PG. If anyone can help me understand what I'm doing wrong it would me much appreciated.
Regex:
{php}.*\n.*\n.*'mister_xx']\)\);.*\n} \n{\/php}
Postgres Query:
SELECT COUNT(*) FROM (SELECT * FROM "Table" WHERE "Column" ~ '{php}.*\n.*\n.*'mister_xx']\)\);.*\n} \n{\/php}') as x;
Postgres Response:
WARNING: nonstandard use of escape in a string literal
LINE 1: ...M (SELECT * FROM "Table" WHERE "Column" ~ '{php}.*\n...
^
HINT: Use the escape string syntax for escapes, e.g., E'\r\n'.
ERROR: syntax error at or near "mister_xx"
LINE 1: ..."Table" WHERE "Column" ~ '{php}.*\n.*\n.*'mister_x...
In SQL, quotes are delimited as two quotes, for example:
'Child''s play'
Applying this to your regex makes it work:
SELECT COUNT(*)
FROM "Table"
WHERE "Column" ~ '{php}.*\n.*\n.*''mister_xx'']\)\);.*\n} \n{\/php}' as x;
Note also how the redundant subquery .
You need to double escape the backslashes and add an E before the statement:
SELECT * FROM "Table" WHERE "Column" ~ E'{php}\n.\n.*''mister_xx'']\)\);.*\n} \n{\/php}'
im trying to reject all inputs not in the format "03 xxxx xxxx" so i created a table like
create table records
(
....
num varchar(255) NOT NULL,
...
CONSTRAINT num_check CHECK (num like '03 [0-9]{4} [0-9]{4}')
)
which should (i think?) accept for example "03 1234 1234". but if i try to add this via sql manager i get an error with the message:
"the INSERT statement conflicted with the CHECK constraint "num_check" "
at first i thought my Regex was off but ive tried it in a few other places and it accepts the example above.
any ideas?
like does not work with regular expressions, it has its own, much simpler wildcard patterns, which only support %, _ , [a-z], and [^a-z]. That's it. {4} would not works, just like most regex features.
You should be able to use:
like '03 [0-9][0-9][0-9][0-9] [0-9][0-9][0-9][0-9]'
Another option, a little less repetitive:
declare #digitChar nvarchar(12)
set #digitChar = '[0-9]'
Where clause:
like '03 ' + replicate(#digitChar,4) + ' ' + replicate(#digitChar,4)
Example: http://sqlfiddle.com/#!3/d41d8/3251