Left selection up until a certain character SQL - sql-server

Question

My select statement returns two columns Column A is based on Column B BUT I removed the initial 17 characters. What I would then like to do is take all the characters in Column A up until it hits a \ (backslash). Can anyone help me achieve this please - current code below
SELECT distinct substring(Path,17,18) AS Detail, Path
FROM [DB].[dbo].[Projects]
Where [Path] like '\DATA%'
AND [Deleted] = '0'
Just to re-iterate as my example wasn't very clear in the comment below. I am trying to extract from the following result
\DATA\More Data\Even More Data\Data 1
To show
Even More Data
So I have removed the proceeding 17 characters until the next \

Answer 1

For ColumnA, if you only want to take out the first 17 characters, you should use
RIGHT(Path, LEN(Path) - 17)
As your current solution will not work correctly if Path is longer than 35 characters.
As for returning the string up to the first backslash, use:
SELECT LEFT(Detail, CHARINDEX('\', Detail)) FirstFolder, Detail, Path
FROM
(
SELECT distinct RIGHT(Path, LEN(Path) - 17) AS Detail, Path
FROM [DB].[dbo].[Projects]
Where [Path] like '\DATA%'
AND [Deleted] = '0'
) a
Or all in one:
SELECT DISTINCT SUBSTRING(Path, 18, CHARINDEX('\', Path, 18) - 18)
FROM [DB].[dbo].[Projects]
WHERE [Path] like '\DATA%'
AND [Deleted] = '0'
This says:
extract a substring from path
start at Character 18
The length of the string will be calculated by finding the position of the first backslash in path starting at Character 18, minus 18 (because we started the search on the 18th character, and we want it relative to the start of our search not the start of the original string)
Update:
As #etsa correctly points out, if you cannot guarantee that Path is at least 18 characters long and contains a backslash after character 18 for every row, you should use the following to return only the rows that do meet this criteria:
SELECT DISTINCT SUBSTRING(Path, 18, CHARINDEX('\', Path, 18) - 18)
FROM [DB].[dbo].[Projects]
WHERE [Path] like '\DATA%'
AND [Deleted] = '0'
AND CHARINDEX('\', Path, 18) > 0

Answer 2

Regardless of the length of your data, this gets the parent folder which seems to be what you want.
DECLARE #table TABLE ([Path] VARCHAR(256))
INSERT INTO #table VALUES
('\DATA\More Data\Even More Data\Data 1'),
('\\server\top folder\middle folder\bottom folder\file 1'),
('x\x\x\x\x\x\x\x\x\x\x\x\x\x\x\x\x\x\x\this is what we want\x')
SELECT
[Path]
,CHARINDEX('\',REVERSE([Path])) as [Reverse Position of last \]
,CHARINDEX('\',REVERSE([Path]),CHARINDEX('\',REVERSE([Path])) + 1) as [Reverse Postion of next to last \]
,REVERSE(
SUBSTRING(
REVERSE([Path]),
CHARINDEX('\',REVERSE([Path]))+1,
CHARINDEX('\',REVERSE([Path]),CHARINDEX('\',reverse([Path]))+1)-CHARINDEX('\',REVERSE([Path])) - 1)) as [Your Desired Results]
FROM
#table
RETURNS
+--------------------------------------------------------------+----------------------------+-----------------------------------+----------------------+
| Path | Reverse Position of last \ | Reverse Postion of next to last \ | Your Desired Results |
+--------------------------------------------------------------+----------------------------+-----------------------------------+----------------------+
| \DATA\More Data\Even More Data\Data 1 | 7 | 22 | Even More Data |
| \\server\top folder\middle folder\bottom folder\file 1 | 7 | 21 | bottom folder |
| x\x\x\x\x\x\x\x\x\x\x\x\x\x\x\x\x\x\x\this is what we want\x | 2 | 23 | this is what we want |
+--------------------------------------------------------------+----------------------------+-----------------------------------+----------------------+