显示匹配行的前后 N 行? [英] Show the previous and after N lines each for the matched lines?

问题描述

以下命令将显示与模式匹配的行.

The following command will show the lines match the pattern.

cat file.txt | select-string -pattern "..."

如何显示匹配行的前三行和后三行?(就像grep的-C3参数.)也可以打印出行号吗?

How to show the previous and after 3 lines each for the matched lines? (Like the -C3 parameter of grep.) Also is it possible to print out the line number?

顺便说一句，这个命令比 grep 慢很多?

BTW, the command is much slower comparing with grep?

推荐答案

在Select-String的-Path参数中指定文件名，并使用-Context 开关指定前后包含的行数.

Specify the filename in the -Path parameter of Select-String, and use the -Context switch to specify the number of lines before and after to include.

select-string -path file.txt -pattern "..." -Context 1,3

文件名和行号将包含在输出中

The file name and line number will be included in the output

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