CODEWITHSUNDEEP
linuxunixsedawk
Gan
Gan

Reputation: 207

Find and remove in unix

Below is my input file:

sample.txt:

3"
6-position
7' 4" to 10' 3-1/2"
4.8"
Adjustable from 99" to 111" - max 148

and in the output I only need 3, i.e.

output.txt:

3
4.8

So basically I need to print the numeric value for the " symbol, other non-numeric text needs to be removed entirely.

I tried to implement this with sed, but I was not able to get the desired result.

Is there any way to achieve this on UNIX?

Upvotes: 0

Views: 215

Answers (4)

potong
potong

Reputation: 58488

This might work for you (GNU sed):

sed '/^[0-9.]\+"/!d;s/".*//' file

Upvotes: 0

anubhava
anubhava

Reputation: 785721

awk is more suited to perform this type of task:

awk '/^ *[0-9]*(\.[0-9]+)?" *$/{sub(/"/, ""); print}' inFile

OUTPUT:

3
4.8

Upvotes: 1

Lev Levitsky
Lev Levitsky

Reputation: 65821

One way with sed:

sed -n 's/^\([0-9]\+\(\.[0-9]\+\)\?\)"$/\1/p' sample.txt > out.txt

or with GNU sed

sed -rn 's/^([0-9]+(\.[0-9]+)?)"$/\1/p' sample.txt > out.txt

or with GNU grep

grep -oP '^[0-9]+(\.[0-9]+)?(?="$)' > out.txt

Be sure to use the correct inch mark (” or "). Or you can match both with a character class [”"].

Edit: updated to work for floating point numbers.

Upvotes: 1

demure
demure

Reputation: 7627

I think you are asking for grep -o [0-9][0-9]*\" sample.txt Which will match one or more numbers followed my a '"', and print each occurrence separately and without surrounding text.

Upvotes: 0

Related Questions