Bad substitution error when passing decremented variable to "sed" command - arrays

Question

I like to use "sed" command to delete two consecutive lines from a file.
I can delete single line using following syntax where
variable "index" holds the line number:
sed -i "${index}d" "$PWD$DEBUG_DIR$DEBUG_MENU"
Since I like to delete consecutive lines I did test this
syntax which supposedly decrements / increments the variable but I am getting a bad substitution error.
Addendum
I am not sure where to post this , but during troubleshooting of
this issue I have discovered that this syntax does not work as expected
(( index++)) nor (( index-- ))

Answer 1

Using sed with "index" in single line deletion twice in a row / sequentially / works and resolves few issues.
#delete command line
sed -i "${index}d" "$PWD$DEBUG_DIR$DEBUG_MENU"
#delete description line
sed -i "${index}d" "$PWD$DEBUG_DIR$DEBUG_MENU"

Answer 2

sed is the right tool for doing s/old/new, that is all. What you're trying to do isn't that so why are you trying to coerce sed into doing something when there's far more appropriate tools can do the job far easier?
The right way to do what your script:
#retrieve first matching line number
index=$(sed -n "/$choice/=" "$PWD$DEBUG_DIR$DEBUG_MENU")
#delete matching line plus next line from file
sed -i "/$index[1], (( $index[1]++))/"
"$PWD$DEBUG_DIR$DEBUG_MENU"
seems to be trying to do is this:
awk -v choice="$choice" '$0~choice{skip=2} !(skip&&skip--)' "$PWD$DEBUG_DIR$DEBUG_MENU"
For example:
$ seq 20 | awk -v choice="4" '$0~choice{skip=2} !(skip&&skip--)'
1
2
3
6
7
8
9
10
11
12
13
16
17
18
19
20
if you only want to delete the first match:
$ seq 20 | awk -v choice="4" '$0~choice{skip=2;cnt++} !(cnt==1 && skip && skip--)'
1
2
3
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
or the 2nd:
$ seq 20 | awk -v choice="4" '$0~choice{skip=2;cnt++} !(cnt==2 && skip && skip--)'
1
2
3
4
5
6
7
8
9
10
11
12
13
16
17
18
19
20
and to skip 5 lines instead of 2:
$ seq 20 | awk -v choice="4" '$0~choice{skip=5} !(skip&&skip--)'
1
2
3
9
10
11
12
13
19
20
Just use the right tool for the right job, don't go digging holes to plant trees with a teaspoon.

Answer 3

If you just want the first one, then quit when you see it:
sed -n "/$choice/ {=;q}" file
But you look like you're processing this file multiple times. There must be a simpler way to do it, if you can describe your over-arching goal.
For example, if you just want to remove the matched line and the next line, but only the first time, you can use awk: here we see "4" and "5" are gone, but "14" and "15" remain:
$ seq 20 | awk '/4/ && !seen {getline; seen++; next} 1'
1
2
3
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
With GNU sed, you can use +1 as the second address in an address range:
index=4
seq 10 | sed "$index,+1 d"
Or if you want to use bash/ksh arithmetic expansion: use the post-increment operator
seq 10 | sed "$((index++)),$index d"
Note here that, due to the pipeline, sed is running in a subshell: even though the index value is now 5, this occurs in a subshell. After the sed command ends, index has value 4 in the current shell.