替换匹配行前面N行的内容

要求：sed 4.2.2+

https://unix.stackexchange.com/questions/206886/print-previous-line-after-a-pattern-match-using-sed

  1. cat test.txt
  3. status: done
  4. memory: 1G
  5. name: bwa_mem_S1
  6. test: hehe
  8. status: waiting
  9. memory: 1G
  10. name: bwa_mem_S2
  11. test: hehe
  13. status: waiting
  14. memory: 1G
  15. name: qc_S1
  16. test: hehe
  18. status: waiting
  19. memory: 1G
  20. name: qc_S2
  21. test: hehe
  24. cat test.txt | sed -zE 's#waiting(\n[^\n]*\nname: bwa_mem*)#done\1#g'
  • -z: 一次性读取全部文件
  • -E: 使用扩展的正则匹配模式，可以匹配换行符
  • \1: 表示子组()匹配到的第一个