CODEWITHSUNDEEP
bash
noway
noway

Reputation: 2585

Delete all text before a matching text in a file with Bash

I have the following text in a file, and I would like to find the word fox and would like to delete the text before the word fox from the begining of the text. (including fox)

I want to keep the whole text if there is no match.

The quick 
brown fox 
jumps over 
the lazy dog

The requested outcome will be:

jumps over 
the lazy dog

The tool to be used is not important, but I'm with Msys on Windows, and not all the fancy tools are implemented here.

Things I tried:

var=$(echo "${raw}" | awk "/replaceme/{i++}i")
var=$(echo "${raw}" | sed -n "/replaceme/,${p}")
var=$(echo "${raw}" | sed -r -n -e "/replaceme/,${p}")
var=$(echo "${raw}" | sed "/replaceme/,$!d")

Thanks

Upvotes: 1

Views: 80

Answers (3)

anubhava
anubhava

Reputation: 785196

You can do this easily in awk:

awk -v s="fox" '$0~s{re="^.*" s "[[:space:]]*"; sub(re, ""); }1' RS= file
jumps over
the lazy dog

When pattern is not found:

awk -v s="box" '$0~s{re="^.*" s "[[:space:]]*"; sub(re, ""); }1' RS= file
The quick
brown fox
jumps over
the lazy dog

EDIT: If you want to delete the text AFTER the match (including the line of match):

awk -v s="over" '$0~s{re="\n[^\n]*" s ".*$"; sub(re, ""); }1' RS= file
The quick
brown fox

Upvotes: 2

konsolebox
konsolebox

Reputation: 75488

If it's only up to the first occurrence of fox you can have:

sed '1,/fox/d' file

Add -i as an option if you want to modify the file.

To delete lines up to the last occurrence:

awk 'BEGIN{start=1}{a[NR]=$0}/fox/{start=NR+1}END{for(i=start;i<=NR;++i)print a[i]}' file

To modify:

awk 'BEGIN{start=1;file=ARGV[1]}{a[NR]=$0}/fox/{start=NR+1}END{printf("")>file;for(i=start;i<=NR;++i)print a[i]}>file' file

Upvotes: 0

merours
merours

Reputation: 4106

You can go down the regex path with /^.*fox(.+)|.+/s.

This way, it will take everything from the text after the last occurence of fox and, if none found, everything.

Upvotes: 0

Related Questions