Issue with comparison in Postgres (Version 11) while comparing _ sign.
I have a string (shown below) I want it to compare it with a word and wants to check whether this word _WIN_ exists in the string. If yes it should give true.
But when I search like this '%_WIN_%' it is giving as TRUE even though the searched string doesn't exactly contains this word.
Can anyone please suggest what I'm doing wrong?
select 'New_vit_Vitamin_D_IND-tonline_WINTERSEAS_2020.02.09' ILIKE '%_WIN_%'
Note: Expected RESULT should be FALSE but giving as TRUE
The underscore is the wildcard for a single character in SQL. If you want to search for the character itself you need to escape it:
ILIKE '%\_WIN\_%' ESCAPE '\'
Related
I need a regrex query to match any string having given character. So i tried for example
SELECT wt.CHGUSER FROM "CDB"."WTBALL" wt where REGEXP_LIKE (wt.CHGUSER, '^\d*115*$');
So i am expecting to fetch all the strings having 115 somewhere in between each string. I tried many combinations but i am getting empty column or weird combination.
Are you sure You need a regex? You write "all the strings having 115 somewhere in between each string", but test for a all-digit string with "115" somewhere...
Btw. this could be done also without regex:
WHERE LOCATE('115', wt.CHGUSER) > 0
AND TRANSLATE(wt.CHGUSER, '', '0123456789') --if You really want to test all-digit string
why not use the native "LIKE" expression?
where wt.CHGUSER like '%115%'
This will give different results than your regexp because your expression is looking for '115' so long as there is a digit immediate before and after it. A more generic regexp, which matches your question, would be '.*115.*'
What about -
REGEXP_LIKE (wt.CHGUSER, '^*\d115\d*$');
This may sound like a duplicate, but existing solutions does not work.
I need to remove all non-alphanumerics from a varchar field. I'm using the following but it doesn't work in all cases (it works with diamond questionmark characters):
select TRANSLATE(FIELDNAME, '?',
TRANSLATE(FIELDNAME , '', 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789'))
from TABLENAME
What it's doing is the inner translate parse all non-alphanumeric characters, then the outer translate replace them all with a '?'. This seems to work for replacement character�. However, it throws The second, third or fourth argument of the TRANSLATE scalar function is incorrect. which is expected according to IBM:
The TRANSLATE scalar function does not allow replacement of a character by another character which is encoded using a different number of bytes. The second and third arguments of the TRANSLATE scalar function must end with correctly formed characters.
Is there anyway to get around this?
Edit: #Paul Vernon's solution seems to be working:
· 6005308 ??6005308
–6009908 ?6009908
–6011177 ?6011177
��6011183�� ??6011183??
Try regexp_replace(c,'[^\w\d]','') or regexp_replace(c,'[^a-zA-Z\d]','')
E.g.
select regexp_replace(c,'[^a-zA-Z\d]','') from table(values('AB_- C$£abc�$123£')) t(c)
which returns
1
---------
ABCabc123
BTW Note that the allowed regular expression patterns are listed on this page Regular expression control characters
Outside of a set, the following must be preceded with a backslash to be treated as a literal
* ? + [ ( ) { } ^ $ | \ . /
Inside a set, the follow must be preceded with a backslash to be treated as a literal
Characters that must be quoted to be treated as literals are [ ] \
Characters that might need to be quoted, depending on the context are - &
I find PostgreSQL similar to operator works little strange. I accidentally checked for space in below query but surprised with the result.
select 'Device Reprocessing' similar to '%( )%' --return true
select 'Device Reprocessing' similar to '%()%' --return true
select 'DeviceReprocessing' similar to '%()%' --return true
Why 2nd and the 3rd query returns true? Is empty pattern always return true?
What I understand about SIMILAR TO operator is returns true or false depending on whether its pattern matches the given string.
You have defined a group with nothing in it, meaning anything will match. I think you will find any string matches %()%, even an empty string.
Normally you would use this grouping to list options so:
select 'DeviceReprocessing' similar to '%(Davinci|Dog)%'
Would return false since it contains neither "Davinci" nor "Dog", but this:
select 'DeviceReprocessing' similar to '%(vice|Dog)%'
would return true since it does contain at least one of the options.
Your first condition is true because the expression does contain a space.
I actually prefer the Regular Expression notation that does not require the % wildcards:
select 'DeviceReprocessing' ~ 'vice|Dog'
I want to list the trigger no system ending with "_BI" in firebird database,
but no result with this
select * from rdb$triggers
where
rdb$trigger_source is not null
and (coalesce(rdb$system_flag,0) = 0)
and (rdb$trigger_source not starting with 'CHECK' )
and (rdb$trigger_name like '%BI')
but with this syntaxs it gives me a "_bi" and "_BI0U" and "_BI0U" ending result
and (rdb$trigger_name like '%BI%')
but with this syntaxs it gives me null result
and (rdb$trigger_name like '%#_BI')
thank you beforehand
The problem is that the Firebird system tables use CHAR(31) for object names, this means that they are padded with spaces up to the declared length. As a result, use of like '%BI') will not yield results, unless BI are the 30th and 31st character.
There are several solutions
For example you can trim the name before checking
trim(rdb$trigger_name) like '%BI'
or you can require that the name is followed by at least one space
rdb$trigger_name || ' ' like '%BI %'
On a related note, if you want to check if your trigger name ends in _BI, then you should also include the underscore in your condition. And as an underscore in like is a single character matcher, you need to escape it:
trim(rdb$trigger_name) like '%\_BI' escape '\'
Alternatively you could also try to use a regular expressions, as you won't need to trim or otherwise mangle the lefthand side of the expression:
rdb$trigger_name similar to '%\_BI[[:SPACE:]]*' escape '\'
What is wrong with this statement that it is still giving me spaces after the field. This makes me think that the syntax combining the WHEN statements is off. My boss wants them combined in one statement. What am I doing wrong?
Case WHEN LTRIM(RTRIM(cSHortName))= '' Then NULL
WHEN cShortname is NOT NULL THEN
REPLACE (cShortName,SUBSTRING,(cShortName,PATINDEX('%A-Za-z0-9""},1,) ''_
end AS SHORT_NAME
Judging from the code, it seems that you may be trying to strip spaces and non-alphanumeric characters from the beginning and ending of the string.
If so, would this work for you?
I think it provides the substring from the first alphanumeric occurrence to the last.
SELECT
SUBSTRING(
cShortName,
PATINDEX('%A-Za-z0-9',cShortName),
( LEN(cShortName)
-PATINDEX('%A-Za-z0-9',REVERSE(cShortName))
-PATINDEX('%A-Za-z0-9',cShortName)
)
) AS SHORTNAME
Replace TRIM with LTRIM.
You can also test LEN(cShortName) = 0
Ummm there seems to be some problems in this script, but try this.
Case
WHEN LTRIM(RTRIM(cSHortName))= '' Then NULL
WHEN cShortname is NOT NULL THEN REPLACE(cShortName, SUBSTRING(cShortName, PATINDEX('%A-Za-z0-9', 1) , ''), '')
end AS SHORT_NAME
Why do you think it is supposed not to give you spaces after the field?
Edit:
As far as I understand you are trying to remove any characters from the string that do not match this regular expression range [a-zA-Z0-9] (add any other characters that you want to preserve).
I see no clean way to do that in Microsoft SQL Server (you are using Microsoft SQL Server it seems) using the built-in functions. There are some examples on the web that use a temporary table and a while loop, but this is ugly. I would either return the strings as is and process them on the caller side, or write a function that does that using the CLR and invoke it from the select statement.
I hope this helps.