[英]How to replace a cron in a shell script using sed?
我需要使用 sed 或 awk 替換文件中的 cron 條目。
試過這個:沒用
sed -i 's/0 0 * * 0/0 1 * * 1/g' script.sh
腳本.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
請幫助我。
使用mikefarah/yq就地編輯文件 ( -i
):
yq -i '.on.schedule[].cron = "0 1 * * 1"' semgrep.yml
會變成一個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
成一個包含
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
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