Try to get from the value - "some kind of tex #123fpe2"
all characters after the # sign
Prepared the code below
but the problem is that not always the length of the value after the sign # 8 characters
select REVERSE(Left(Reverse(vValue), 8)),
(
select vValue from Runtime.dbo.Live where TagName = 'CurrentBaseRecipeName32000000io') as ValueForCheck,
from Runtime.dbo.Live where TagName = 'CurrentBaseRecipeName32000000io'
Getting error:
select REVERSE(Left(Reverse(vValue), POSITION('#' IN vValue))),
(
select vValue ...
select REVERSE(Left(Reverse(vValue), 8)),
(
select vValue from Runtime.dbo.Live where TagName = 'CurrentBaseRecipeName32000000io') as ValueForCheck,
from Runtime.dbo.Live where TagName = 'CurrentBaseRecipeName32000000io'
If you are sure that there is only 1 # in the string, or if there are more than 1 and you want the part of the string after the last occurrence of #, then you can use SUBSTRING_INDEX():
SELECT SUBSTRING_INDEX('some kind of tex #123fpe2', '#', -1);
will return:
123fpe2
I assume this is MySQL because of the use of POSITION in the question. Use INSTR to get the position of the '#' and then SUBSTRING to return everything after the index returned by INSTR + 1
SELECT SUBSTRING(vValue, INSTR(vValue, '#') + 1) FROM Runtime.dbo.Live
WHERE ...
Just in case, here is a version for SQL Server using RIGHT
SELECT RIGHT(vValue, LEN(vValue) - CHARINDEX('#', vValue)) FROM Runtime.dbo.Live
WHERE ...
Related
For example, I have got a string with table name and the schema like:
[dbo].[statistical]
How can fetch just the table name statistical out from this string?
This is what PARSENAME is used for:
SELECT PARSENAME('[dbo].[statistical]', 1)
SELECT PARSENAME('[adventureworks].[dbo].[statistical]', 1)
SELECT PARSENAME('[adventureworks]..[statistical]', 1)
SELECT PARSENAME('[statistical]', 1)
SELECT PARSENAME('dbo.statistical', 1)
-- all examples return 'statistical'
You could alternatively try this:
declare #s varchar(100) = 'asd.stadfa';
select reverse(substring(s, 1, charindex('.', s) - 1)) from (
select reverse(#s) s
) a
charindex returns first occurence of character, so you reverse initial string to make last dot first. Then you just use substring to extract first part of reversed string, which is what you are looking for. Finally, you need to apply reverse one more time to reverse back extracted string :)
I have a string as follows: first, last (123456) the expected result should be 123456. Could someone help me in which direction should I proceed using Oracle?
It will depend on the actual pattern you care about (I assume "first" and "last" aren't literal hard-coded strings), but you will probably want to use regexp_substr.
For example, this matches anything between two brackets (which will work for your example), but you might need more sophisticated criteria if your actual examples have multiple brackets or something.
SELECT regexp_substr(COLUMN_NAME, '\(([^\)]*)\)', 1, 1, 'i', 1)
FROM TABLE_NAME
Your question is ambiguous and needs clarification. Based on your comment it appears you want to select the six digits after the left bracket. You can use the Oracle instr function to find the position of a character in a string, and then feed that into the substr to select your text.
select substr(mycol, instr(mycol, '(') + 1, 6) from mytable
Or if there are a varying number of digits between the brackets:
select substr(mycol, instr(mycol, '(') + 1, instr(mycol, ')') - instr(mycol, '(') - 1) from mytable
Find the last ( and get the sub-string after without the trailing ) and convert that to a number:
SQL Fiddle
Oracle 11g R2 Schema Setup:
CREATE TABLE test ( str ) AS
SELECT 'first, last (123456)' FROM DUAL UNION ALL
SELECT 'john, doe (jr) (987654321)' FROM DUAL;
Query 1:
SELECT TO_NUMBER(
TRIM(
TRAILING ')' FROM
SUBSTR(
str,
INSTR( str, '(', -1 ) + 1
)
)
) AS value
FROM test
Results:
| VALUE |
|-----------|
| 123456 |
| 987654321 |
My query is the following:
SELECT id, category FROM table1
This returns the following rows:
ID|category
1 |{IN, SP}
2 |
3 |{VO}
Does anyone know how i can remove the first char and last char of the string in PostgreSQL, so it removes: {}?
Not sure, what you mean with "foreign column", but as the column is an array, the best way to deal with that is to use array_to_string()
SELECT id, array_to_string(category, ',') as category
FROM table1;
The curly braces are not part of the stored value. This is just the string representation of an array that is used to display it.
Either using multiple REPLACE functions.
SELECT id, REPLACE(REPLACE(category, '{', ''), '}', '')
FROM table1
Or using a combination of the SUBSTRING, LEFT & LENGTH functions
SELECT id, LEFT(SUBSTRING(category, 2, 999),LENGTH(SUBSTRING(category, 2, 999)) - 1)
FROM table1
Or just SUBSTRING and LENGTH
SELECT id, SUBSTRING(category, 2, LENGTH(category)-2)
FROM table1
You could replace the {} with an empty string
SELECT id, replace(replace(category, '{', ''), '}', '') FROM table1
select id,
substring(category,charindex('{',category,len(category))+2,len(category)-2)
from table1;
select id
,left(right(category,length(category)-1),length(category)-2) category
from boo
select id
,trim(both '{}' from category)
from boo
Trim():
Remove the longest string containing only the characters (a space by
default) from the start/end/both ends of the string
The syntax for the replace function in PostgreSQL is:
replace( string, from_substring, to_substring )
Parameters or Arguments
string
The source string.
from_substring
The substring to find. All occurrences of from_substring found within string are replaced with to_substring.
to_substring
The replacement substring. All occurrences of from_substring found within string are replaced with to_substring.
UPDATE dbo.table1
SET category = REPLACE(category, '{', '')
WHERE ID <=3
UPDATE dbo.table1
SET category = REPLACE(category, '}', '')
WHERE ID <=3
I am working with SQL server 2008, and facing problem about character replacement.
If I use
SELECT REPLACE(MYWORD,0,1) FROM MYTABLE
It is replacing all 0 into 1, I just want to replace Last character Like MYWORD = "ERMN0" so it will be MYWORD = "ERMN1"
using STUFF, which, IMO, ends up being most readable:
DECLARE #MyWORD VARCHAR(20) = 'ABCDEF123'
SELECT STUFF(#MyWORD, LEN(#MyWORD), 1, '2')
output:
ABCDEF122
You may use combination of LEFT, RIGHT, and CASE.
You need to use CASE to check the most RIGHT character whether it's a 0 or not and replace it with 1. And at last, combine it with the LEFT part (after being separated from the last character) of the MYWORD string.
However, depending on your requirement, it may have a drawback.
When there is a word ending with 10, it would also be replaced.
SELECT LEFT(MYWORD,LEN(MYWORD)-1) + CASE RIGHT(MYWORD,1) WHEN '0' THEN '1' ELSE RIGHT(MYWORD,1) END
Try this.
SELECT LEFT('ERMN0', Len('ERMN0')-1)
+ Replace(RIGHT('ERMN0', 1), 0, 1)
OUTPUT : ERMN1
In your case
SELECT LEFT(MYWORD, Len(MYWORD)-1)
+ Replace(RIGHT(MYWORD, 1), 0, 1) as [REPLACED] FROM MYTABLE
Try this
SELECT SUBSTRING(MYWORD, 1, LEN(MYWORD) - 1) +
REPLACE(SUBSTRING(MYWORD, LEN(MYWORD), LEN(MYWORD)), 0, 1) FROM MYTABLE
This will work
SELECT LEFT ('ERMN0' , Len('ERMN0') -1 ) + REPLACE(Right('ERMN0', 1), '0','1')
Or in your case
SELECT LEFT (MYWORD , Len(MYWORD) -1 ) + REPLACE(Right(MYWORD, 1), '0','1') AS MYWORD FROM MYTABLE
this is also use full to replace letters from end
It is used from replacing characters from end 1,2 or N
Declare #Name nvarchar(20) = 'Bollywood'
select #Name = REPLACE(#Name, SUBSTRING(#Name, len(#Name) - 1, 2), 'as')
SELECT #Name
output is "Bollywoas"
Here best part is you can repalce as many character from last you needed.
I have below string:
ThisSentence.ShouldBe.SplitAfterLastPeriod.Sentence
So I want to select Sentence since it is the string after the last period. How can I do this?
Just for completeness' sake, here's a solution using regular expressions (not very complicated IMHO :-) ):
select regexp_substr(
'ThisSentence.ShouldBe.SplitAfterLastPeriod.Sentence',
'[^.]+$')
from dual
The regex
uses a negated character class to match anything except for a dot [^.]
adds a quantifier + to match one or more of these
uses an anchor $ to restrict matches to the end of the string
You can probably do this with complicated regular expressions. I like the following method:
select substr(str, - instr(reverse(str), '.') + 1)
Nothing like testing to see that this doesn't work when the string is at the end. Something about - 0 = 0. Here is an improvement:
select (case when str like '%.' then ''
else substr(str, - instr(reverse(str), ';') + 1)
end)
EDIT:
Your example works, both when I run it on my local Oracle and in SQL Fiddle.
I am running this code:
select (case when str like '%.' then ''
else substr(str, - instr(reverse(str), '.') + 1)
end)
from (select 'ThisSentence.ShouldBe.SplitAfterLastPeriod.Sentence' as str from dual) t
And yet another way.
Not sure from a performance standpoint which would be best...
The difference here is that we use -1 to count backwards to find the last . when doing the instr.
With CTE as
(Select 'ThisSentence.ShouldBe.SplitAfterLastPeriod.Sentence' str, length('ThisSentence.ShouldBe.SplitAfterLastPeriod.Sentence') len from dual)
Select substr(str,instr(str,'.',-1)+1,len-instr(str,'.',-1)+1) from cte;
select
substr(
'ThisSentence.ShouldBe.SplitAfterLastPeriod.Sentence',
INSTR(
'ThisSentence.ShouldBe.SplitAfterLastPeriod.Sentence',
'.',
-1
)+1
)
from dual;
The INSTR function accepts a third parameter, the occurrence. It defaults to 1 (the first occurrence), but also accepts negative numbers (meaning counting from the last occurrence backwards).
select substr(str, instr(str, '.', -1) + 1)
from (
select 'ThisSentence.ShouldBe.SplitAfterLastPeriod.Sentence'
as str
from dual);
Sentence
how many dots in a string?
select length(str) - length(replace(str, '.', '') number_of_dots from ...
get substring after last dot:
select substr(str, instr(str, '.', 1, number_of_dots)+1) from ...