Question

我需要帮助删除两个空白行之前的文本。

任何人都可以通过SED帮助解决这个问题吗？

示例:(我的日志文件）行：“%%%% Info | Date：...”

来源：

%%%% Info | Date: (11-02-18)-Day=042 - %%%%

OLD: android-902 (11-02-18 02:30)

Wi-Fi up - USB (11-02-18 03:21:21)

%%%% - Info | Date: (12-02-18)-Day=043 - %%%%

Wi-Fi up - USB (12-02-18 07:17:02)

Smart Connected (12-02-18 10:19)

%%%% - Info | Date: (13-02-18)-Day=044 - %%%%


%%%% - Info | Date: (14-02-18)-Day=045 - %%%%


%%%% - Info | Date: (15-02-18)-Day=046 - %%%%


%%%% - Info | Date: (16-02-18)-Day=047 - %%%%

Wi-Fi up - USB (16-02-18 06:32:24)

输出：

%%%% Info | Date: (11-02-18)-Day=042 - %%%%

OLD: android-902 (07-02-18 02:30)

Wi-Fi up - USB (07-02-18 03:21:21)

%%%% - Info | Date: (12-02-18)-Day=043 - %%%%

Wi-Fi up - USB (12-02-18 07:17:02)

Smart Connected (12-02-18 10:19)

%%%% - Info | Date: (16-02-18)-Day=047 - %%%%

Wi-Fi up - USB (16-02-18 06:32:24)

提前致谢。

Answer 1

awk是一个非常强大的工具，但它不是用于多行处理的最佳工具，只需使用以下$ awk 'BEGIN{RS="%%%%[^%]+%%%%\n\n\n"}{printf "%s",$0}' input.log %%%% Info | Date: (11-02-18)-Day=042 - %%%% OLD: android-902 (11-02-18 02:30) Wi-Fi up - USB (11-02-18 03:21:21) %%%% - Info | Date: (12-02-18)-Day=043 - %%%% Wi-Fi up - USB (12-02-18 07:17:02) Smart Connected (12-02-18 10:19) %%%% - Info | Date: (16-02-18)-Day=047 - %%%% Wi-Fi up - USB (16-02-18 06:32:24)解决方案即可完成任务。

RS="%%%%[^%]+%%%%\n\n\n"

说明：

awk是ORS将用于查找每条记录的记录分隔符，因为您使用默认\n这些字段只会从输出中删除并替换通过+-----------------+--------+ | Project | Budget | +-----------------+--------+ | P1200 | 150 | | P1400 | 200 | | L3 Program 3 | 350 | | P1100 | 250 | | P1300 | 150 | | L3 Program 2 | 400 | | L2 Initiative 2 | 750 | | P2200 | 300 | | P2400 | 200 | | P2600 | 300 | | L3 Program 1 | 800 | | L2 Initiative 1 | 800 | | L1 Division | 1550 | +-----------------+--------+

sed - 删除两个空白行之前的文本

1 个答案: