I'm sure this is super easy, but how would I go about converting LastName,FirstName to LastName,FirstInitial?
For example changing Smith,John to Smith,J or Johnson,John to Johnson,J etc.
Thank You!
In case of LastName and FirstName columns:
select LastName,substr(FirstName,1,1)
from mytable
;
In case of a fullname saved in a single column:
select substr(fullname,1,instr(fullname || ',',',')-1) || substr(fullname,instr(fullname || ',',','),2)
from mytable
;
or
select regexp_replace (fullname,'([^,]*,?)(.).*','\1\2')
from mytable
;
Here is one way, using just "standard" instr and substr. Assuming your input is a single string in the format 'Smith,John':
select substr(fullname, 1, instr(fullname, ',')+1) from yourtable;
yourtable is the name of the table, and fullname is the name of the column.
instr(fullname, ',') finds the position of the comma within the input string (it would be 6 in 'Smith,John'); thensubstrtakes the substring that begins at the first position (the1in the function call) and ends at the position calculated byinstr`, PLUS 1 (to get the first initial as well).
Related
I have a column in a table ident_nums that contains different types of ids. I need to remove special characters(e.g. [.,/#&$-]) from that column and replace them with space; however, if the special characters are found at the beginning of the string, I need to remove it without placing a space. I tried to do it in steps; first, I removed the special characters and replaced them with space (I used
REGEXP_REPLACE) then found the records that contain spaces at the beginning of the string and tried to use the TRIM function to remove the white space, but for some reason is not working that.
Here is what I have done
Select regexp_replace(id_num, '[:(),./#*&-]', ' ') from ident_nums
This part works for me, I remove all the unwanted characters from the column, however, if the string in the column starts with a character I don't want to have space in there, I would like to remove just the character, so I tried to use the built-in function TRIM.
update ident_nums
set id_num = TRIM(id_num)
I'm getting an error ORA-01407: can't update ident_nums.id_num to NULL
Any ideas what I am doing wrong here?
It does work if I add a where clause,
update ident_nums
set id_num = TRIM(id_num) where id = 123;
but I need to update all the rows with the white space at the beginning of the string.
Any suggestions are welcome.
Or if it can be done better.
The table has millions of records.
Thank you
Regexp can be slow sometimes so if you can do it by using built-in functions - consider it.
As #Abra suggested TRIM and TRANSLATE is a good choice, but maybe you would prefer LTRIM - removes only leading spaces from string (TRIM removes both - leading and trailing character ). If you want to remove "space" you can ommit defining the trim character parameter, space is default.
select
ltrim(translate('#kdjdj:', '[:(),./#*&-]', ' '))
from dual;
select
ltrim(translate(orginal_string, 'special_characters_to_remove', ' '))
from dual;
Combination of Oracle built-in functions TRANSLATE and TRIM worked for me.
select trim(' ' from translate('#$one,$2-zero...', '#$,-.',' ')) as RESULT
from DUAL
Refer to this dbfiddle
I think trim() is the key, but if you want to keep only alpha numerics, digits, and spaces, then:
select trim(' ' from regexp_replace(col, '[^a-zA-Z0-9 ]', ' ', 1, 0))
regexp_replace() makes it possible to specify only the characters you want to keep, which could be convenient.
Thanks, everyone, It this query worked for me
update update ident_nums
set id_num = LTRIM(REGEXP_REPLACE(id_num, '[:space:]+', ' ')
where REGEXP_LIKE(id_num, '^[ ?]')
this should work for you.
SELECT id_num, length(id_num) length_old, NEW_ID_NUM, length(NEW_ID_NUM) len_NEW_ID_NUM, ltrim(NEW_ID_NUM), length(ltrim(NEW_ID_NUM)) length_after_ltrim
FROM (
SELECT id_num, regexp_replace(id_num, '[:(),./#*&-#]', ' ') NEW_ID_NUM FROM
(
SELECT '1234$%45' as id_num from dual UNION
SELECT '#SHARMA' as id_num from dual UNION
SELECT 'JACK TEST' as id_num from dual UNION
SELECT 'XYZ#$' as id_num from dual UNION
SELECT '#ABCDE()' as id_num from dual -- THe 1st character is space
)
)
I am trying to separate first and last name . I have a column called 'Fullname' and it has first and last name and a comma all in one column. I've tried the below but I get an error " its not a valid number". When I remove the comma it works, so I am not sure how to incorporate a comma in the formula so it can work.
,substr(Fullname,1,',') as Lastname
,substr(Fullname,',',' ') as Firstname
Column
Fullname
Brown,John N
Green,Julie T
Desired results
Lastname FirstName
Brown John
Green Julie
You can use regexp_substr():
select regexp_substr(name, '[^,]+', 1, 1) as lastname,
regexp_substr(name, '[^, ]+', 1, 2) as firstname
The second argument to SUBSTR() is the position of the substring, the third argument is the length of the substring. It will not automatically search for a delimiter if you use strings there instead of numbers. You can use LOCATE() to find the positions that you want.
SUBSTR(Fullname, 1, LOCATE(Fullname, ',')-1) AS Lastname,
SUBSTR(Fullname, LOCATE(Fullname, ',')+1) AS Firstname
Can be performed in Classical way by using instr inside substr function as the following case :
select substr(fullname,1,instr(fullname,',')-1) Firstname,
substr(fullname,instr(fullname,',')+1,length(fullname)) Lastname
from tab;
SQL Fiddle Demo
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
I want to select names from a table where the 'name' column contains '%' anywhere in the value. For example, I want to retrieve the name 'Approval for 20 % discount for parts'.
SELECT NAME FROM TABLE WHERE NAME ... ?
You can use like with escape. The default is a backslash in some databases (but not in Oracle), so:
select name
from table
where name like '%\%%' ESCAPE '\'
This is standard, and works in most databases. The Oracle documentation is here.
Of course, you could also use instr():
where instr(name, '%') > 0
One way to do it is using replace with an empty string and checking to see if the difference in length of the original string and modified string is > 0.
select name
from table
where length(name) - length(replace(name,'%','')) > 0
Make life easy on yourselves and just use REGEXP_LIKE( )!
SQL> with tbl(name) as (
select 'ABC' from dual
union
select 'E%FS' from dual
)
select name
from tbl
where regexp_like(name, '%');
NAME
----
E%FS
SQL>
I read the documentation mentioned by Gordon. The relevent sentence is:
An underscore (_) in the pattern matches exactly one character (as opposed to one byte in a multibyte character set) in the value
Here was my test:
select c
from (
select 'a%be' c
from dual) d
where c like '_%'
The value a%be was returned.
While the suggestions of using instr() or length in the other two answers will lead to the correct answer, they will do so slowly. Filtering on function results simply take longer than filtering on fields.
I need to compare the value of a column (LASTNAME) with a system variable (:VARIABLE), but the variable is an email address, so I need to trim off the "#email.com" and "firstname." Some things I've tried:
select *
from TABLENAME
where LASTNAME LIKE :VARIABLE
select *
from TABLENAME
where LASTNAME IN :VARIABLE
I've been able to trim off the #email.com, can't figure out how to trim off FIRSTNAME. at the same time.
Regular expressions can help:
SQL> SELECT LTRIM(regexp_substr('firstname.lastname#abc.com','\.[^#]*'),'.') last_name from dual;
LAST_NAME
---------
lastname
Your query could then look like:
SELECT *
FROM tablename
WHERE UPPER(LTRIM(regexp_substr(:VARIABLE,'\.[^#]*'),'.')) = UPPER(lastname);
You can use a combination of SUBSTR and INSTR.
The instr function returns the location of a substring in a string. With this one you can locate the "."or the "#" char.
Then use the substr to get the substring from the begining to the previous located position
SELECT SUBSTR('Take the first four characters', 1, 4)
Use:
SELECT t.*
FROM TABLE t
WHERE t.lastname LIKE '%' || SUBSTR(:VARIABLE,
INSTR(:VARIABLE, '.') +1,
INSTR(:VARIABLE, '#')) || '%'
Add UPPER or LOWER to both sides if you need to be sure of case insensitive matching.
Reference:
SUBSTR
INSTR