I want to trim a this character consist of a chinese character.
Select '10樓 10 /F'
Select '7樓 7/F'
Bad Result:
10樓 10 /F
7樓 7/F
I want to this:
10/F
7/F
You can use CHARINDEX to find the position of the chinese character.
Then use SUBSTRING to get the part you need based on that position.
And REPLACE all the spaces to nothing from that result.
For example:
select * , replace(substring(value, charindex(N'樓',value)+1, len(value)),' ','') as value2
from (values
(N'10樓 10 /F'),
(N'7樓 7/F')
) v(value);
Note that putting the N before the strings marks them as NVARCHAR.
Since the chinese character is a unicode character.
Try to do this:
with cte as (
select '10樓 10 /F' as t
union all
select '7樓 7/F'
)
select rtrim(ltrim(substring(t, charindex('樓', t) + 1, len(t)))) as t
from cte
Related
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 have a column that contains data like:
AB-123 XYZ
ABCD-456 AAA
BCD-789 BBB
ZZZ-963
Y-85
and this is what i need from those string:
123
456
789
963
85
I need the characters from the left after the dash('-') character, then ends before the space character is read.
Thank You guys.
Note: Original tag on this question was Oracle and this answer is based on that tag. Now that, tag is updated to SqlServer, this answer is no longer valid, if somebody looking for Oracle solution, this may help.
Use regular expression to arrive at sub string.
select trim(substr(regexp_substr('ABCD-456 AAA','-[0-9]+ '),2)) from dual
'-[0-9]+ ' will grab any string pattern which starts with dash has one or more digits and ends with a ' ' and returns number with dash
substr will remove '-' from above output
trim will remove any trailing ' '
Check This.
Using Substring and PatIndex.
select
SUBSTRING(colnm, PATINDEX('%[0-9]%',colnm),
PATINDEX('%[^0-9]%',ltrim(RIGHT(colnm,LEN(colnm)-CHARINDEX('-',colnm)))))
from
(
select 'AB-123 XYZ' colnm union
select 'ABCD-456 AAA' union
select 'BCD-789 BBB' union
select 'ZX- 23 BBB'
)a
OutPut :
Try this
http://rextester.com/YTBPQD69134
CREATE TABLE Table1 ([col] varchar(12));
INSERT INTO Table1
([col])
VALUES
('AB-123 XYZ'),
('ABCD-456 AAA'),
('BCD-789 BBB');
select substring
(col,
charindex('-',col,1)+1,
charindex(' ',col,1)-charindex('-',col,1)
) from table1;
Assume all values has '-' and followed by a space ' '. Below solution will not tolerant to exception case:
SELECT
*,
SUBSTRING(Value, StartingIndex, Length) AS Result
FROM
-- You can replace this section of code with your table name
(VALUES
('AB-123 XYZ'),
('ABCD-456 AAA'),
('BCD-789 BBB')
) t(Value)
-- Use APPLY instead of sub-query is for debugging,
-- you can view the actual parameters in the select
CROSS APPLY
(
SELECT
-- Get the first index of character '-'
CHARINDEX('-', Value) + 1 AS StartingIndex,
-- Get the first index of character ' ', then calculate the length
CHARINDEX(' ', Value) - CHARINDEX('-', Value) - 1 AS Length
) b
Getting Examples from similar Stack Overflow threads,
Remove all characters after a specific character in PL/SQL
and
How to Select a substring in Oracle SQL up to a specific character?
I would want to retrieve only the first characters before the occurrence of a string.
Example:
STRING_EXAMPLE
TREE_OF_APPLES
The Resulting Data set should only show only STRING_EXAM and TREE_OF_AP because PLE is my delimiter
Whenever i use the below REGEXP_SUBSTR, It gets only STRING_ because REGEXP_SUBSTR treats PLE as separate expressions (P, L and E), not as a single expression (PLE).
SELECT REGEXP_SUBSTR('STRING_EXAMPLE','[^PLE]+',1,1) from dual;
How can i do this without using numerous INSTRs and SUBSTRs?
Thank you.
The problem with your query is that if you use [^PLE] it would match any characters other than P or L or E. You are looking for an occurence of PLE consecutively. So, use
select REGEXP_SUBSTR(colname,'(.+)PLE',1,1,null,1)
from tablename
This returns the substring up to the last occurrence of PLE in the string.
If the string contains multiple instances of PLE and only the substring up to the first occurrence needs to be extracted, use
select REGEXP_SUBSTR(colname,'(.+?)PLE',1,1,null,1)
from tablename
Why use regular expressions for this?
select substr(colname, 1, instr(colname, 'PLE')-1) from...
would be more efficient.
with
inputs( colname ) as (
select 'FIRST_EXAMPLE' from dual union all
select 'IMPLEMENTATION' from dual union all
select 'PARIS' from dual union all
select 'PLEONASM' from dual
)
select colname, substr(colname, 1, instr(colname, 'PLE')-1) as result
from inputs
;
COLNAME RESULT
-------------- ----------
FIRST_EXAMPLE FIRST_EXAM
IMPLEMENTATION IM
PARIS
PLEONASM
I am using an Oracle regular expression to extract the first letter of each word in a string. The results are returned in a single cell, with spaces representing hard breaks. Here is an example...
input:
'I hope that some kind person
browsing stack overflow
can help me'
output:
ihtskp bso chm
What I am trying to do next is count the length of each "word" in my output, like this:
6 3 3
Alternatively, a count of the words in each line of the original string would be acceptable, as it would yield the same result.
Thanks!
Count the number of spaces and add one:
select (length(your_col) - length(replace(your_col, ' '))+1) from your_table;
It will give you the number of words per line. From there you can get all counts on one line by using listagg function:
select LISTAGG(cnt,' ') within group (order by null) from (
select (length(a)-length(replace(a,' '))+1) cnt from (
select 'apa bpa bv' a from dual
union all
select 'n bb gg' a from dual
union all
select 'ff ff rr gg' a from dual))
group by null;
Perhaps you also need to split the strings if they contain newlines or are they split already?
I tried to edit my original post but it hasn't appeared, but I figured out a way to solve my issue. I just decided to break the words into rows, since I know how to character count rows, and then reassembled the character counts into a single cell using listagg:
with my_string as (
select regexp_substr (words,'[0-9]+|[a-z]+|[A-Z]+',1,lvl) parsed
from (
select words, level lvl
from letters connect by level <= length(words) - length(replace(words,' ')) + 1)
)
select listagg(length(parsed),' ') within group (order by parsed) word_count
from my_string
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'