[SOLVED] Find matching pattern and replace variable value in next line

Issue

I am trying to replace the value of a variable "Tag" next to the searched pattern, like:

If name in images is "repo.url/app1" Replace Tag: NEW_VALUE

Below is the yaml sample file where I need to replace the values.

namespace: dev

resources:
- file.yaml
- ../../..

images:
- name: repo.url/app1
  Tag: xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
- name: repo.url/app2
  Tag: xxxxxxxxxxxxxxxxxxxxxxxxxxxxx

So far I tried sed -i "/\-\ name\:\ repo.url\/app1/{n;n;s/\(Tag\).*/\1: $NEW_VALUE/}" file which doesn't do anything and doesn't throw any error. I am not sure, if there's a better way to handle such substitutions.

Solution

Simplest way with sed is just to match the line and then perform the substitution on the next (\w used to match all word-characters at the end of the line for replacement), e.g.

sed -E '/repo\.url\/app1$/{n;s/\w+$/yyyyyyyyyyyyyyyyyyyyyy/}' file.yaml

Example Use/Output

sed -E '/repo\.url\/app1$/{n;s/\w+$/yyyyyyyyyyyyyyyyyyyyyy/}' file.yaml
namespace: dev

resources:
- file.yaml
- ../../..

images:
- name: repo.url/app1
  Tag: yyyyyyyyyyyyyyyyyyyyyy
- name: repo.url/app2
  Tag: xxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Using awk

Another way is to use awk and a flag to indicate replace in next line (rep below). Then use sub() to replace the 'x's with whatever you need, e.g.

awk '
  rep { sub(/\w+$/,"yyyyyyyyyyyyyyyyyyyyyyyy"); rep = 0 } 
  $NF == "repo.url/app1" { rep=1 }
  1
' file.yaml

Where 1 is just shorthand for print current record.

Example Use/Output

awk 'rep { sub(/\w+$/,"yyyyyyyyyyyyyyyyyyyyyyyy"); rep = 0 } $NF == "repo.url/app1" { rep=1 }1' file.yaml
namespace: dev

resources:
- file.yaml
- ../../..

images:
- name: repo.url/app1
  Tag: yyyyyyyyyyyyyyyyyyyyyyyy
- name: repo.url/app2
  Tag: xxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Just redirect the output to a new file to save, e.g. awk '{...}' file.yaml > newfile.yaml.

Answered By - David C. Rankin

Answer Checked By - Clifford M. (WPSolving Volunteer)