RIGHT of CHARINDEX not selecting correctly - sql

Question

I am trying to parse out a last name field that may have two last names that are separated by either a blank space ' ' or a hyphen '-' or it may only have one name.
Here is what I'm using to do that:
select top 1000
BENE_FIRST_NAME,
BENE_LAST_NAME,
FirstNm =
case
when BENE_FIRST_NAME like '% %' then
left(BENE_FIRST_NAME, CHARINDEX(' ', BENE_FIRST_NAME))
when BENE_FIRST_NAME like '%-%' then
left(BENE_FIRST_NAME, CHARINDEX('-', BENE_FIRST_NAME))
else BENE_FIRST_NAME
end,
LastNm =
case
when BENE_LAST_NAME like '% %' then
right(BENE_LAST_NAME, CHARINDEX(' ', BENE_LAST_NAME))
when BENE_LAST_NAME like '%-%' then
right(BENE_LAST_NAME, CHARINDEX('-', BENE_LAST_NAME))
else BENE_LAST_NAME
end,
CharIndxDash = CHARINDEX('-', BENE_LAST_NAME),
CharIndxSpace = CHARINDEX(' ', BENE_LAST_NAME)
from xMIUR_Elig_Raw_v3
Here are some results:
BENE_FIRST_NAME
BENE_LAST_NAME
FirstNm
LastNm
CharIndxDash
CharIndxSpace
JUANA
PEREZ-MARTINEZ
JUANA
RTINEZ
6
0
EMILIANO
PICENO ESPINOZA
EMILIANO
SPINOZA
0
7
JULIAN
NIETO-CARRENO
JULIAN
ARRENO
6
0
EMILY
SALMERON TERRIQUEZ
EMILY
TERRIQUEZ
0
9
The CHARINDEX seems to be selecting the correct position but it is not bringing in all of the CHARs to the right of that position. Sometimes it works like in the last record. But sometimes it is off by 1. And sometimes it is off by 2. Any ideas?

Answer 1

If you need to select part of a last name after space/hyphen, you need to get right part of the string with length = total_lenght - space_position:
...
LastNm =
case
when BENE_LAST_NAME like '% %' then
right(BENE_LAST_NAME, LEN(BENE_LAST_NAME) - CHARINDEX(' ', BENE_LAST_NAME))
when BENE_LAST_NAME like '%-%' then
right(BENE_LAST_NAME, LEN(BENE_LAST_NAME) -CHARINDEX('-', BENE_LAST_NAME))
else BENE_LAST_NAME
end,
...

Answer 2

Your last name logic doesn't make sense..
RIGHT takes N chars from the right of the string
CHARINDEX gives the position of a char from the left of the string
You can't use it to find a position from left and then take that number of chars from the right of the string
Here's a name:
JOHN MALKOVICH
The space is at 5. If you take 5 chars from the right, you get OVICH. The shorter the name before the space and the longer the name after the space, the fewer chars you get from the last name
Perhaps you mean to put a LEN in there so you take the string length minus the index of the space.. You can also use it in a call to SUBSTRING as the start index, and tell SQLS to take 9999 chars (of any number longer than the remaining string) and it will take up to the end of the string
SUBSTRING(name, CHARINDEX(' ', name)+1, 9999)

Answer 3

I think you can simplify your code by a lot. Consider below with a different but representative sample data
with data (name) as
(select 'first-last' union select 'first last' union select 'firstlast'),
data_prepped (name, indx) as
(select name,coalesce(nullif(charindex(' ', name)+charindex('-', name),0),len(name))
from data)
select name,
left(name, indx-1) as part1,
right(name, indx) as part2
from data_prepped