如何从外壳中的“更新2f8b547d..eb94967a快速转发...”日志中获取“ 2f8b547d..eb94967a”字符串？

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我正在构建一个shell脚本。

该脚本获取git日志，例如：

"Updating 2f8b547d..eb94967a Fast-forward...."

but I want to get 2f8b547d..eb94967a snippet.

我是新手。因此，感谢您的帮助。

更新：

For the more, I want use the snippet as a param. Because I will excute git log 2f8b547d..eb94967a

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lmagni
lmagni

As an alternative to awk (don't get me wrong, awk is super for this job as well), you can simply use cut with a space delimiter extract the second field, e.g.

cut -d' ' -f2 yourgit.log

You can also pipe output to cut or redirect the input file to it using < as well. It essentially does the same as the awk command, it just being a different choice of which utility to use.

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渗透的葬礼
渗透的葬礼

You can pipe it to awk like so:

echo "Updating 2f8b547d..eb94967a Fast-forward...." | awk '{print $2}'

Your result will be 2f8b547d..eb94967a.

如果它是一个脚本（例如，abc.sh）具有这样的输出，则可以运行：

$> ./abc.sh | awk '{print $2}'

awk takes the output and splits the information by space. Updating is represented with $1. 2f8b547d..eb94967a is $2 and so on. In the above example, we ask awk to print out the 2nd item in the output.

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