如何使用 sed 替换 shell 脚本中的 cron？

Question

我需要使用 sed 或 awk 替换文件中的 cron 条目。

试过这个：没用

sed -i 's/0 0 * * 0/0 1 * * 1/g' script.sh

脚本.sh

#!/bin/bash

mkdir -p .github/workflows

cd .github/workflows
touch semgrep.yml

cat << EOF > semgrep.yml
name: Semgrep
on:
  pull_request: {}
  push:
    branches:
      - master
      - main
    paths:
      - .github/workflows/semgrep.yml
  schedule:
    - cron: '0 0 * * 0'
jobs:
  semgrep:
    name: Static Analysis Scan
    runs-on: ubuntu-18-04

请帮助我。

Answer 1

使用mikefarah/yq就地编辑文件 ( -i )：

yq -i '.on.schedule[].cron = "0 1 * * 1"' semgrep.yml

会变成一个semgrep.yml包含

name: Semgrep
on:
  pull_request: {}
  push:
    branches:
      - master
      - main
    paths:
      - .github/workflows/semgrep.yml
  schedule:
    - cron: '0 0 * * 0'
jobs:
  semgrep:
    name: Static Analysis Scan
    runs-on: ubuntu-18-04

成一个包含

name: Semgrep
on:
  pull_request: {}
  push:
    branches:
      - master
      - main
    paths:
      - .github/workflows/semgrep.yml
  schedule:
    - cron: '0 1 * * 1'
jobs:
  semgrep:
    name: Static Analysis Scan
    runs-on: ubuntu-18-04