I have following data in my table
id nml
-- -----------------
1 Temora sepanil
2 Human Mixtard
3 stlliot vergratob
I need to get the result by extracting first word in column nml and get its last 3 characters with reverse order
That means output should be like
nml reverse
----------------- -------
Temora sepanil aro
Human Mixtard nam
stlliot vergratob toi
You use PostgreSQL's string functions to achieve desired output
in this case am using split_part,right,reverse function
select reverse(right(split_part('Temora sepanil',' ',1),3))
output:
aro
so you can write your query in following format
select nml
,reverse(right(split_part(nml,' ',1),3)) "Reverse"
from tbl
Split nml using regexp_split_to_array(string text, pattern text [, flags text ]) refer Postgres Doc for more info.
Use reverse(str) (refer Postgres Doc) to reverse the first word form previous split.
Use substr(string, from [, count]) (refer Postgres Doc) to select first three letters of the reversed test
Query
SELECT
nml,
substr(reverse(regexp_split_to_array(nml, E'\\s+')[0]),3) as reverse
FROM
MyTable
You can use the SUBSTRING, CHARINDEX, RIGHT and REVERSE function
here's the syntax
REVERSE(RIGHT(SUBSTRING(nml , 1, CHARINDEX(' ', nml) - 1),3))
sample:
SELECT REVERSE(RIGHT(SUBSTRING(nml , 1, CHARINDEX(' ', nml) - 1),3)) AS 'Reverse'
FROM TableNameHere
Related
I have a column (mycolumn) in my snowflake table (mytable) whose content has this pattern :
JohnDoe - Client Number One
MaryJane - Client Number Two
I would need to extract the first portion on the left of the string (JohnDoe,MaryJane - with no whitespace behind).
I tried to use the following approach, but I got stucked because I could only remove the first two block of words to the right, but not the - (dash) and the white spaces.
select substring(mycolumn,1,length(mycolumn)- CHARINDEX(' ', REVERSE(mycolumn))- CHARINDEX(' ', REVERSE(mycolumn))) from mytable
You can use regexp_substr():
select regexp_substr(mycolumn, '^[^ ]+')
from mytable;
Situation:
I have a column decoded from Hex to varchar where values look like this:
{"something":"example"}
Objective:
I would like to extract the second word between the quotes.
What i tried:
I started with a couple of substring and charindex functions but my code looks more complicated than it should be.
SELECT SUBSTRING(
SUBSTRING(
'{"something":"example"}',
charindex(':"','{"something":"example"}')+2,
LEN('{"something":"example"}')-charindex(':"','{"something":"example"}')+2),
0,
CHARINDEX('"',SUBSTRING(
'{"something":"example"}',
charindex(':"','{"something":"example"}')+2,
LEN('{"something":"example"}')-charindex(':"','{"something":"example"}')+2))
)
Any ideas?
If you're on SQL Server 2016+, you can use OPENJSON:
SELECT [value]
FROM OPENJSON('{"something":"example"}');
db<>fiddle
So against a table:
SELECT [value]
FROM (VALUES('{"something":"example"}'),
('{"another":"sample"}'))V(S)
CROSS APPLY OPENJSON(V.S);
This returns:
value
-------
example
sample
Can anyone help me, I have a problem regarding on how can I get the below result of data. refer to below sample data. So the logic for this is first I want delete the letters before the number and if i get that same thing goes on , I will delete the numbers before the letter so I can get my desired result.
Table:
SALV3000640PIX32BLU
SALV3334470A9CARBONGRY
TP3000620PIXL128BLK
Desired Output:
PIX32BLU
A9CARBONGRY
PIXL128BLK
You need to use a combination of the SUBSTRING and PATINDEX Functions
SELECT
SUBSTRING(SUBSTRING(fielda,PATINDEX('%[^a-z]%',fielda),99),PATINDEX('%[^0-9]%',SUBSTRING(fielda,PATINDEX('%[^a-z]%',fielda),99)),99) AS youroutput
FROM yourtable
Input
yourtable
fielda
SALV3000640PIX32BLU
SALV3334470A9CARBONGRY
TP3000620PIXL128BLK
Output
youroutput
PIX32BLU
A9CARBONGRY
PIXL128BLK
SQL Fiddle:http://sqlfiddle.com/#!6/5722b6/29/0
To do this you can use
PATINDEX('%[0-9]%',FieldName)
which will give you the position of the first number, then trim off any letters before this using SUBSTRING or other string functions. (You need to trim away the first letters before continuing with the next step because unlike CHARINDEX there is no starting point parameter in the PATINDEX function).
Then on the remaining string use
PATINDEX('%[a-z]%',FieldName)
to find the position of the first letter in the remaining string. Now trim off the numbers in front using SUBSTRING etc.
You may find this other solution helpful
SQL to find first non-numeric character in a string
Try this it may helps you
;With cte (Data)
AS
(
SELECT 'SALV3000640PIX32BLU' UNION ALL
SELECT 'SALV3334470A9CARBONGRY' UNION ALL
SELECT 'SALV3334470A9CARBONGRY' UNION ALL
SELECT 'SALV3334470B9CARBONGRY' UNION ALL
SELECT 'SALV3334470D9CARBONGRY' UNION ALL
SELECT 'TP3000620PIXL128BLK'
)
SELECT * , CASE WHEN CHARINDEX('PIX',Data)>0 THEN SUBSTRING(Data,CHARINDEX('PIX',Data),LEN(Data))
WHEN CHARINDEX('A9C',Data)>0 THEN SUBSTRING(Data,CHARINDEX('A9C',Data),LEN(Data))
ELSE NULL END AS DesiredResult FROM cte
Result
Data DesiredResult
-------------------------------------
SALV3000640PIX32BLU PIX32BLU
SALV3334470A9CARBONGRY A9CARBONGRY
SALV3334470A9CARBONGRY A9CARBONGRY
SALV3334470B9CARBONGRY NULL
SALV3334470D9CARBONGRY NULL
TP3000620PIXL128BLK PIXL128BLK
I would like to convert a string containing dates in SQL select from Oracle 11g database.
Original string (CLOB) example:
"1.12.2011 - event 1
2.2.2012 - event 2
13.3.2012 - event 44"
Desired output:
"20111201 - event 1
20120202 - event 2
20120313 - event 44"
Is there a better (faster) way than using 4 separate replacements?
regexp_replace(regexp_replace(regexp_replace(regexp_replace(my_string,
'(\d\d)\.(\d\d)\.(20\d\d)', '\3\2\1'),
'(\d\d)\.(\d)\.(20\d\d)', '\30\2\1'),
'(\d)\.(\d\d)\.(20\d\d)', '\3\20\1'),
'(\d)\.(\d)\.(20\d\d)', '\30\20\1')
Especially if you're using clobs you have to be careful unless you're certain of the data in there.
However, if your clob only looks like that then you need threeregexp_replace in order for this to work; it'll also be much more dynamic. Just explicitly specify digits using [[:digit:]] then specify a minimum and maximum number of times these digits could be there using {1,2}.
Then the following would work:
select regexp_replace(
regexp_replace(
regexp_replace( my_string
, '([[:digit:]]{1,2})\.([[:digit:]]{1,2})\.(20[[:digit:]]{2})'
, '\3-\2-\1')
, '-([[:digit:]]{1}(-|$))'
, '0\1' )
, ('-')
, '')
from dual
This means:
match ( group 1 ) 1 or 2 digits
match a full stop.
match ( group 2 ) 1 or 2 digits
match a full stop
match ( group 3 ) 20 + 2 digits.
Then take out only groups 1, 2 and 3, i.e. ignoring the full stops and return then in the order 3, 2, 1 padded with a hyphen
Then replace any [digit] that is followed by either a hyphen or the end of the string, i.e. the number of digits is only 1 with -0[digit].
Lastly replace all the hyphens.
Separately from that I agree with tbone. It would make a lot more sense to store this data in a separate table (event_id number, event_date date). Any string transformations are easy with no chance of getting it wrong, unlike in this situation, and the data is easy to query and compare.
there are no better options (both correct and readable) with better performance - or if there are, no one cares..
i prefer a 2-level regexp_replace for date part:
select regexp_replace(
regexp_replace( my_string,
'([[:digit:]]{1,2})\.([[:digit:]]{1,2})\.(20[[:digit:]]{2})',
'\3-0\2-0\1' ),
'(20[[:digit:]]{2})-0?([[:digit:]]{2})-0?([[:digit:]]{2})',
'\3\2\1' )
from dual;
Demo
Maybe try doing:
select to_char(to_date('13.3.2011', 'DD.MM.YYYY'),'YYYYMMDD') from dual;
Using T-SQL, how would I go about getting the last 3 characters of a varchar column?
So the column text is IDS_ENUM_Change_262147_190 and I need 190
SELECT RIGHT(column, 3)
That's all you need.
You can also do LEFT() in the same way.
Bear in mind if you are using this in a WHERE clause that the RIGHT() can't use any indexes.
You can use either way:
SELECT RIGHT(RTRIM(columnName), 3)
OR
SELECT SUBSTRING(columnName, LEN(columnName)-2, 3)
Because more ways to think about it are always good:
select reverse(substring(reverse(columnName), 1, 3))
declare #newdata varchar(30)
set #newdata='IDS_ENUM_Change_262147_190'
select REVERSE(substring(reverse(#newdata),0,charindex('_',reverse(#newdata))))
=== Explanation ===
I found it easier to read written like this:
SELECT
REVERSE( --4.
SUBSTRING( -- 3.
REVERSE(<field_name>),
0,
CHARINDEX( -- 2.
'<your char of choice>',
REVERSE(<field_name>) -- 1.
)
)
)
FROM
<table_name>
Reverse the text
Look for the first occurrence of a specif char (i.e. first occurrence FROM END of text). Gets the index of this char
Looks at the reversed text again. searches from index 0 to index of your char. This gives the string you are looking for, but in reverse
Reversed the reversed string to give you your desired substring
if you want to specifically find strings which ends with desired characters then this would help you...
select * from tablename where col_name like '%190'