I have strings in one of my table in which I need to replace some specials characters like ' _ ? ° and square brackets [ ].
When I try this it's working like expected :
SELECT REGEXP_REPLACE('BIG''EAST_?°[]', '[_?°'']', ' ') FROM DUAL;
I get:
BIG EAST []
Then I add the square brackets in my regex :
SELECT REGEXP_REPLACE('BIG''EAST_?°[]', '[_?°''\[\]]', ' ') FROM DUAL;
I expected this:
BIG EAST
But I get:
BIG'EAST_?°
How can I properly escape the square brackets in my regex?
You need to add a * to match multiple occurrences (and in any order) of characters from your pattern
SELECT REGEXP_REPLACE('BIG''EAST_?°[]', '[_?°''\[\]]*', ' ') FROM DUAL;
Related
How do I use an ORACLE REGEX function to remove all leading and trailing line break characters and spaces?
For example, assume I have the following string where refers to actual invisible carriage return line feed characters. Here's the input:
"
SELECT *
FROM
TABLE
"
And here's the desire output:
"SELECT *
FROM
TABLE"
This would do it if regex_replace() is a requirement:
select regexp_replace('
SELECT *
FROM
TABLE
', '^\s*|\s*$', '') as hello
from dual
See https://www.techonthenet.com/oracle/functions/regexp_replace.php for documentation.
A single regexp_replace is sufficient, eg.
select regexp_replace('
select frut
from prut
','^[[:space:]]*(.*[^[:space:]])[[:space:]]*$','\1',1,1,'mn') from dual;
results in
select frut
from prut
Can someone please help me to understand the following code:
REPLACE(LTRIM(RTRIM(dbo.UFN_SEPARATES_COLUMNS(CompletionDetails, 3, ','))), '.', '') AS BuildRequestID,
Does it say remove all trailing and leading spaces, then replace 3 with comma. Next, if there is ., replace it with ' '?
It does not at any point replace 3 with ,.
We can make all this easier to follow by formatting the full expression to cover multiple lines:
REPLACE(
LTRIM(RTRIM(
dbo.UFN_SEPARATES_COLUMNS(CompletionDetails, 3, ',')
))
,'.', ''
) AS BuildRequestID,
Expressions like this have to read from the inside out. So we start with this inner-most part:
dbo.UFN_SEPARATES_COLUMNS(CompletionDetails, 3, ',')
This UFN_SEPARATES_COLUMNS() function is not part of Sql Server, but was added by someone at your organization or as part of the vendor software package for the database you're looking at. But I'm confident based on inferences and the link (found via Google) it will treat CompletionDetails as delimited text, where the delimiter is a comma (based on the 3rd ',' argument) and returns the 3rd field (based on the 2nd 3 argument, where counting starts at 1 rather than 0). As CSV parsers go, this one is particularly naive, so be very careful what you expect from it.
Then we use LTRIM() and RTRIM() to remove both leading and trailing blanks from the field. Not all whitepsace is removed; only space characters. Tabs, line feeds, etc are not trimmed. Sql Server 2017 has a new TRIM() function that can handle wider character sets and do both sides of the string with one call.
The code then uses the REPLACE() function to remove all . characters from the result (replaces them with an empty string).
The code is trimming the leading and trailing spaces via the LTRIM() and RTRIM() functions of whatever is returned from the function dbo.x_COLUMNS... i.e. dbo.x_COLUMNS(CompletionDetails, 3, ','). LTRIM is left, RTRIM is right.
It then is replacing all periods (.) with nothing via the REPLACE() function.
So in summary, it's removing all periods from the string and the leading and trailing spaces.
The LTRIM removes leading spaces. RTRIM removes trailing spaces. REPLACE removes the period.
Declare #Val Char(20) = ' Frid.ay '
Select REPLACE(
LTRIM(
RTRIM(
#Val --dbo.x_COLUMNS(CompletionDetails, 3, ',')
)
), '.', ''
)
Result
BuildRequestID
--------------
Friday
remove all trailing and leading spaces, then replace 3 with comma.
Next, if there is ., replace it with ' '
No it does not say that.
But this does:
REPLACE(REPLACE(LTRIM(RTRIM(CompletionDetails)), '3', ','), '.', ' ')
it's not clear if you want . replaced by ' ' or ''.
I used the 1st case, you can change it as you like.
It's easier to understand like this:
remove all trailing and leading spaces: LTRIM(RTRIM(CompletionDetails))
replace 3 with comma: REPLACE( ?, '3', ',')
replace it with ' ': REPLACE(? , '.', ' ') or REPLACE(? , '.', '')
I can't see leading/trailing whitespace in the following following SQL statement executed with psql:
select name from my_table;
Is there a pragmatic way to see leading/trailing whitespace?
One of option is to use format() function.
With given query case: select format( '"%s"', name ) from my_table;
PoC:
SELECT format( '"%s"', name )
FROM ( VALUES ( ' a ' ), ( ' b ' ) ) v(name);
format
--------
" a "
" b "
(2 rows)
Turn off "aligned mode" in psql: \a
\a
select * from my_table;
id|col1|col2
12|foo|bar
I'd append surrounding quotes:
select '"' || name || '"' from my_table;
If you don't mind substituting all whitespace characters whether or not they are leading/trailing, something like the following will do it:
SELECT REPLACE(REPLACE(REPLACE(REPLACE(txt, ' ', '_'),
E'\t', '\t'),
E'\r', '\r'),
E'\n', '\n') AS txt
FROM test;
This is using an underscore to mark the spaces but of course you are free to choose your own. See SQL fiddle demo.
If you strictly only want to show up the leading/trailing ones it will get more complex - but if this is really desired, something may be possible using regex_replace.
You can try this:
select name from my_table
where name like '% '
How can I remove any white space from substring of string?
For example I have this number '+370 650 12345'. I need all numbers to have this format country_code rest_of_the_number or in that example: +370 65012345. How could you achieve that with PostgreSQL?
I could use trim() function, but then it would remove all whitespace.
Assuming the column is named phone_number:
left(phone_number, strpos(phone_number, ' '))
||regexp_replace(substr(phone_number, strpos(phone_number, ' ') + 1), ' ', '', 'g')
It first takes everything up to the first space and then concatenates it with the result of replacing all spaces from the rest of the string.
If you also need to deal with other whitespace than just a space, you could use '\s' for the search value in regexp_replace()
If you are able to assume that a country code will always be present, you could try using a regular expression to capture the parts of interest. Assuming that your phone numbers are stored in a column named content in a table named numbers, you could try something like the following:
SELECT parts[1] || ' ' || parts[2] || parts[3]
FROM (
SELECT
regexp_matches(content, E'^\\s*(\\+\\d+)\\s+(\\d+)\\s+(\\d+)\\s*$') AS parts
FROM numbers
) t;
The following will work even if the country code is absent (see SQL Fiddle Demo here):
SELECT TRIM(REPLACE(REPLACE(REGEXP_REPLACE('+370 650 12345', '^((\+\d+)\s+)?(.*)$', '\1|\3'), ' ', ''), '|', ' '));
Returns: +370 65012345
SELECT TRIM(REPLACE(REPLACE(REGEXP_REPLACE('370 650 12345', '^((\+\d+)\s+)?(.*)$', '\1|\3'), ' ', ''), '|', ' '));
Returns: 37065012345
It looks for a country code (a set of numbers starting with a + sign) at the beginning, and replaces any whitespace following that code with a pipe |. It then replaces all the spaces in the resulting string with the empty string, then replaces occurrences of | with spaces! The choice of | is arbitrary, I suppose it could be any non-digit character.
I have used REPLACE function in order to delete email addresses from hundreds of records. However, as it is known, the semicolon is the separator, usually between each email address and anther. The problem is, there are a lot of semicolons left randomly.
For example: the field:
123#hotmail.com;456#yahoo.com;789#gmail.com;xyz#msn.com
Let's say that after I deleted two email addresses, the field content became like:
;456#yahoo.com;789#gmail.com;
I need to clean these fields from these extra undesired semicolons to be like
456#yahoo.com;789#gmail.com
For double semicolons I have used REPLACE as well by replacing each ;; with ;
Is there anyway to delete any semicolon that is not preceded or following by any character?
If you only need to replace semicolons at the start or end of the string, using a regular expression with the anchor '^' (beginning of string) / '$' (end of string) should achieve what you want:
with v_data as (
select '123#hotmail.com;456#yahoo.com;789#gmail.com;xyz#msn.com' value
from dual union all
select ';456#yahoo.com;789#gmail.com;' value from dual
)
select
value,
regexp_replace(regexp_replace(value, '^;', ''), ';$', '') as normalized_value
from v_data
If you also need to replace stray semicolons from the middle of the string, you'll probably need regexes with lookahead/lookbehind.
You remove leading and trailing characters with TRIM:
select trim(both ';' from ';456#yahoo.com;;;789#gmail.com;') from dual;
To replace multiple characters with only one occurrence use REGEXP_REPLACE:
select regexp_replace(';456#yahoo.com;;;789#gmail.com;', ';+', ';') from dual;
Both methods combined:
select regexp_replace( trim(both ';' from ';456#yahoo.com;;;789#gmail.com;'), ';+', ';' ) from dual;
regular expression replace can help
select regexp_replace('123#hotmail.com;456#yahoo.com;;456#yahoo.com;;789#gmail.com',
'456#yahoo.com(;)+') as result from dual;
Output:
| RESULT |
|-------------------------------|
| 123#hotmail.com;789#gmail.com |