A bash script

jeffwang66 · 11-12-2013, 10:52 AM

Hello,

I want to write a bash script to delete the content after '#'. However, if '#' appears in a string with "", ignore this. For example,

input file:

test #delete
"test #not delete"

Output file:

test
"test #not delete"

Does anyone know how to write this script?

Thanks

druuna · 11-12-2013, 11:57 AM

Your input example is rather small, but give this a try:

Code:

sed '/".*"/!s/#.*//' input

If the result is what you want/need you can change the command to make sure it is done in-place (inside the input file), this also makes a copy of the original:

Code:

sed -i.bak '/".*"/!s/#.*//' input

jeffwang66 · 11-12-2013, 01:38 PM

Thanks for your help. But if the input file is:

input file:

test #delete
"test #not delete" #delete

I want the output is:

test
"test #not delete"

Firerat · 11-12-2013, 02:14 PM

Code:

sed s/#[a-Z]*$// input

it works for your sample input
But will it work for your real data?

TB0ne · 11-12-2013, 03:37 PM

Quote:

Originally Posted by jeffwang66

Thanks for your help. But if the input file is:

input file:

test #delete
"test #not delete" #delete

I want the output is:

test
"test #not delete"

Ok...so since you posted two identical threads, why haven't you also posted what YOU have done/tried so far??? We'll be glad to help you, but you have to show some effort of your own. Asking people to write scripts for you isn't a good thing.

Habitual · 11-12-2013, 05:17 PM

works here, Five by Five:

Code:

cat input && sed -i.bak '/".*"/!s/#.*//' input && cat input
test #delete
"test #not delete"
test 
"test #not delete"

Slackware-approved GNU sed version 4.2.1

Firerat · 11-12-2013, 05:26 PM

Quote:

Originally Posted by Habitual

works here, Five by Five:

Code:

cat input && sed -i.bak '/".*"/!s/#.*//' input && cat input
test #delete
"test #not delete"
test 
"test #not delete"

Slackware-approved GNU sed version 4.2.1

but it doesn't work for the new criteria in post 3

input

Code:

test #delete
"test #not delete" #delete

this 'works'

Code:

sed s/#[a-Z]*$// input

but I guess, only by 'coincidence'
probably no good with 'real world' input data

Disillusionist · 11-20-2013, 12:12 PM

To cater for the stated requirements and also likely additional requirements:

Code:

cat input|sed 's/#[-=+a-zA-z0-9# ][-=+a-zA-z0-9#! ]*$//'|sed 's/#$//'

This runs two sed statements, the first to find # characters that are followed by one or more character in ranges a-z (lowercase alpha) A-Z (uppercase alpha) 0-9 (numeric) - a minus symbol = an equals symbol + a plus sign # a hash symbol a space character, where the second or later additional character can be ! an exclamation mark (removing the #!/bin/bash issues. The second sed statement removes a single hash at the end of a line.

There are many characters that could be added to the syntax, but this should meet most of the situations that I think you are trying to cater for.

Firerat · 11-20-2013, 02:04 PM

Code:

sed -e 's/#[-=+a-zA-z0-9# ][-=+a-zA-z0-9#! ]*$//' -e 's/#$//' input

no need for cat, or pipe

colucix · 11-21-2013, 03:45 AM

With this input:

Code:

$ cat file
test #delete
#delete "test #not delete" #delete "test again #not delete" test #delete me to the end
#delete

try

Code:

$ sed ':a;s/^\(\("[^"]*"\|[^"]*\)*\)#[^"]\+/\1/;ta;/^$/d' file
test 
"test #not delete" "test again #not delete" test

Hope this helps!

Disillusionist · 11-22-2013, 04:24 AM

Wow!

I really must read the Sed & Awk book on my shelf!

I have only read a small section, and not for quite a while