Regex INSTR -Snowflake - snowflake-cloud-data-platform

Question

I have a question related to a snowflake query:
Oracle query:
INSTR(CFileName, '/', -1) + 1)
Snowflake query
regexp_instr(CFileName, '/', -1) + 1
But when I run this in the Snowflake environment, I get this error:
Invalid parameter value: -3. Reason: Position must be positive
Can you please tell me how to read this below code? And how the output logic?
select regexp_instr('ygzff/afzfzh/yfzdaf/yftfrxdff/aa/ff', '/',3)+1
The answer for above is 7 in snowflake. how sf calculate the answer?

Answer 1

select
'ygzff/afzfzh/yfzdaf/yftfrxdff/aa/ff' as input,
regexp_instr(input, '/',1) as r1,
regexp_instr(input, '/',2) as r2,
regexp_instr(input, '/',3) as r3,
r3+1;
gives:
INPUT
R1
R2
R3
R3+1
ygzff/afzfzh/yfzdaf/yftfrxdff/aa/ff
6
6
6
7
the 1,2,3 in my example are the start offset, and result is 1 based offset the / which is the 6. thus the answer 7
The Oracle INSTR position -1 is saying count from the end, and thus is returning the position after the last instance of '/'
we can flip the string around, to find the tail, then minus if the token was found, and then use that like:
select
'ygzff/afzfzh/yfzdaf/yftfrxdff/aa/ff' as input,
length(input) as l,
regexp_instr(reverse(input), '/') as r,
iff(r=0,0,l-(r-2)) as tp,
substr(input, tp) as tail;
INPUT
L
R
TP
TAIL
ygzff/afzfzh/yfzdaf/yftfrxdff/aa/ff
35
3
34
ff
or you can also use RIGHT
select
column1 as input,
length(input) as l,
regexp_instr(reverse(input), '/') as r,
right(input, iff(r=0,l,r-1))
from values
('ygzff/afzfzh/yfzdaf/yftfrxdff/aa/ff'),
('no slashes here');
INPUT
L
R
RIGHT(INPUT, IFF(R=0,L,R-1))
ygzff/afzfzh/yfzdaf/yftfrxdff/aa/ff
35
3
ff
no slashes here
15
0
no slashes here