grep和awk，结合命令吗? [英] grep and awk, combine commands?

问题描述

我的文件看起来像:

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sftp> cd outbox
sftp> ls -ltr
-rw-------   1 0        0            1911 Jun 12 20:40 61N0584832_EDIP000749728818_MFC_20190612203409.txt
-rw-------   1 0        0            1878 Jun 13 06:01 613577165_EDIP000750181517_MFC_20190613055207.txt

我只想用一个命令打印.txt文件名.

I want to print only the .txt file names, ideally in one command.

我可以做到:

grep -e '^-' outfile.log > outfile.log2

..仅给出以-"开头的行.

..which gives only the lines that start with '-'.

-rw-------   1 0        0            1911 Jun 12 20:40 61N0584832_EDIP000749728818_MFC_20190612203409.txt
-rw-------   1 0        0            1878 Jun 13 06:01 613577165_EDIP000750181517_MFC_20190613055207.txt

然后:

awk '{print $9}' outfile.log2 > outfile.log3

..给出所需的输出:

..which gives the desired output:

61N0584832_EDIP000749728818_MFC_20190612203409.txt
613577165_EDIP000750181517_MFC_20190613055207.txt

所以问题是，这2个命令可以合并为1个吗?

And so the question is, can these 2 commands be combined into 1?

推荐答案

您可以使用一个awk:

awk '/^-/{ print $9 }' file > outputfile

或

awk '/^-/{ print $9 }' file > tmp && mv tmp file

它是这样的:

• /^-/-查找以-
开头的每一行
• { print $9 }-仅显示匹配行的字段9.

• /^-/ - finds each line starting with -
• { print $9 } - prints Field 9 of the matching lines only.

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