sed remove last 2 numerals

Question

I have the following.

oid=.1.3.6.1.4.1.795.2.5.0.xxx (these 0.xxx represent any random numeral)

I want to do the following

echo $oid | sed (some commands to remove 0.xxx)

and store the resulting string

.1.3.6.1.4.1.795.2.5

in oid.

I don't mind this being done with Grep, awk or sed anything.
Everything before 0.xxx can change only thing static will be 0.
Thanks for your help!

Answer 1

With bash's Parameter expansion:

oid=".1.3.6.1.4.1.795.2.5.0.xxx"
oid="${oid%.0.*}"
echo "$oid"

Output:

.1.3.6.1.4.1.795.2.5

Answer 2

As you said this is a general form, you want to remove the last 6 characters from your text(.0.xxx), so you can use this:

echo ".1.3.6.1.4.1.795.2.5.0.xxx" |sed 's/.\{6\}$//'

Then if you want use with variables:

oid=.1.3.6.1.4.1.795.2.5.0.xxx

Then:

echo $oid | sed 's/.\{6\}$//'

output is:

.1.3.6.1.4.1.795.2.5

Another solution

echo "${oid:0:${#oid}-6}"

Another solution using cut and '.' as delimeter

echo $oid | cut -d '.' -f -10

Answer 3

Using sed:

oid="$(echo -n .1.3.6.1.4.1.795.2.5.0.0 | sed -r 's/\.0\.[0-9]+$//')"

-r: makes sed interpret ERE (Extended Regular Expressions) patterns

sed command breakdown:

• s: asserts to perform a substitution
• /: starts the pattern
• \.: matches a . character
• 0: matches a 0 character
• \.: matches a . character
• [0-9]+: matches 1 or more digits
• $: matches the end of the line
• /: stops the pattern / starts the replacement string
• /: stops the replacement string / starts the modifiers

Answer 4

Using awk

oid=".1.3.6.1.4.1.795.2.5.0.xxx"
awk -F'.' '{for (i=2;i<=NF-2;i++) {printf "%s","."$i} }' <<< "$oid"

or all in one

awk -F'.' '{for (i=2;i<=NF-2;i++) {printf "%s","."$i} }' <<< ".1.3.6.1.4.1.795.2.5.0.xxx"

Output

.1.3.6.1.4.1.795.2.5

Short explanation

NF-2 – all elements without the last two elements

Answer 5

Using grep:

grep -Po '.*(?=\.0\.\d+$)'

Test :

$ oid='.1.3.6.1.4.1.795.2.5.0.908'

$ oid="$(grep -Po '.*(?=\.0\.\d+$)' <<<"$oid")"

$ echo "$oid"
.1.3.6.1.4.1.795.2.5

• grep -P will enable us to use PCRE

• grep -o will output only the matched portion

• .*(?=\.0\.\d+$) will match all characters prior to .0, followed by . and any number of digits at the end.