Reputation: 1
Given this file, I would like to
print the first install:
line that comes after the @ bash
line. Please see
the file for complete input, but here is a sample
@ base-files ; few lines install: x86_64/release/base-files/base-files-4.2-3.tar.xz 46428 6372358800e589b ; couple of lines install: x86_64/release/base-files/base-files-4.1-1.tar.bz2 49519 f91ed6eab060c3 @ bash ; few lines install: x86_64/release/bash/bash-4.1.17-9.tar.xz 1107128 e49b8d67d59d8617dfa31c ; couple of lines install: x86_64/release/bash/bash-4.1.16-8.tar.xz 1106812 5aa652ddc0a5d65f4af1e2 source: x86_64/release/bash/bash-4.1.16-8-src.tar.xz 6614280 79bb3ddc67d8f0d3da6 @ bash-completion ; few lines install: x86_64/release/bash-completion/bash-completion-1.3-1.tar.bz2 117489 538 source: x86_64/release/bash-completion/bash-completion-1.3-1-src.tar.bz2 216503
Output should be
install: x86_64/release/bash/bash-4.1.17-9.tar.xz 1107128 e49b8d67d59d8617dfa...
I have this command, but it prints all lines inclusive of the two.
awk '/@ bash$/,/install:/' setup.ini
In addition to the answers I created this one
awk '$1=="@" {c=$2} $1=="install:" && c=="bash" {print;exit}'
Upvotes: 1
Views: 179
Reputation: 1
awk '$1=="@" {c=$2} $1=="install:" && c=="bash" {print;exit}'
@
, save field two to variable c
install:
and c
is bash
, print and exitUpvotes: -1
Reputation: 786359
You can use:
awk '/@ bash$/{p=1} p&&/install:/{print; exit}' setup.ini
install: x86_64/release/bash/bash-4.1.17-9.tar.xz 1107128 e49b8d67d59d8617dfa31c
Upvotes: 5
Reputation: 755114
awk '/^@ bash/ { p = 1 } /^install:/ && p == 1 { print; p = 2 }'
The variable p
defaults to 0; when the @ bash
line is read, set it to 1; when the next install line is read, print it and set p to 2 so no other install lines will be printed. You could also exit instead of setting p = 2
, I suppose.
Upvotes: 4