bash 提取与“foo”匹配的文件的最后 10 行
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Extracting last 10 lines of a file that match "foo"
提问by virtue
I want to write the last ten lines which contain a spesific word such as "foo" in a file to a new text file named for instance boo.txt.
我想将文件中包含特定单词（例如“foo”）的最后十行写入名为 example 的新文本文件boo.txt。
How can I achieve this in the command prompt of a unix terminal?
如何在 unix 终端的命令提示符下实现这一点？
回答by Mat
You can use grepand tail:
您可以使用grep和tail：
grep "foo" input.txt | tail -n 10 > boo.txt
The default number of lines printed by tailis 10, so you can omit the -n 10part if you always want that many.
打印的默认行数tail是 10，所以-n 10如果你总是想要那么多，你可以省略这部分。
The >redirection will create boo.txtif it didn't exist. If it did exist prior to running this, the file will be truncated (i.e. emptied) first. So boo.txtwill contain at most 10 lines of text in any case.
如果>重定向boo.txt不存在，则会创建。如果它在运行之前确实存在，则该文件将首先被截断（即清空）。所以boo.txt无论如何都将包含最多 10 行文本。
If you wanted to append to boo.txt, you should change the redirection to use >>.
如果您想附加到boo.txt，您应该将重定向更改为使用>>。
grep "bar" input.txt | tail -n 42 >> boo.txt
You might also be interested in headif you are looking for the first occurrences of the string.
head如果您正在寻找字符串的第一次出现，您可能也会感兴趣。
回答by sarnold
grep foo /path/to/input/file | tail > boo.txt
