从文件列表中取出一个单词作为grep before的结果

Question

我有一个命令对包含"TypeId: 0"的完整路径文件进行grep，下面是命令

grep -rnw /home/username/app/data/store/0/part/.mv -e "TypeId: 0" | awk -F ":" '{print $1}'

，结果如下:/home/username/app/data/store/0/part/.mv/521/1673332792072/segmentconfig.yaml /home/username/app/data/store/0/part/.mv/521/1673333077920/segmentconfig.yaml /home/username/app/data/store/0/part/.mv/521/1673333077920/segmentconfig.yaml.old /home/username/app/data/store/0/part/.mv/515/1672993850766/segmentconfig.yaml /home/username/app/data/store/0/part/.mv/515/1672993850766/segmentconfig.yaml.old /home/username/app/data/store/0/part/.mv/703/1672987004847/segmentconfig.yaml /home/username/app/data/store/0/part/.mv/703/1672987004847/segmentconfig.yaml.old

现在我困惑如何从该列表中的每个文件grep "numofvertice"。有人有办法解决这个问题吗?

Answer 1

你可以试试:

grep -rnw /home/username/app/data/store/0/part/.mv -e "TypeId: 0" | awk -F ":" '{print $1}'|xargs -I{} grep "numofvertice" {}

Answer 2

像这样(GNUgrep):

<STDIN> | grep -oP 'bS+.yaml' | xargs cat

或带ack:

cd /home/username/app/data/store/0/part/.mv
ack -wl  -e "TypeId: 0" | xargs cat

Fromack --help:

-l，——files-with-matches
只打印包含匹配项的文件名