将sed输出通过xargs输出到grep的问题

Question

I am having problems piping sed output to be used with grep through xargs . I think it is a problem of variable expansion, but I am not sure.

The scenario is as follows. Current directory has these files:

file1.log
file1.log.67654
file2.log
file2.log.66876
file3.log
file3.log.56768
...

From these files, only some of the ones ending in .log contain a certain string so that:

$ grep -l "certainString"
file1.log
file3.log

What I want is to find how many of these files OR the ones that start like these but have additional .numbers as extension (files file1.log , file1.log.67654 , file3.log and file3.log.56768 in this example) contain another string.

I thought that this command would do it:

grep -l "certainString" | sed "s/log/log*/g" | xargs grep -l "anotherString"

But after I replace log with log* with sed , xargs does not expand the * but inteprets it literally, therefore:

$ grep -l "certainString" | sed "s/log/log*/g" | xargs grep -l "anotherString"
grep: ./file1.log*: No such file or directory
grep: ./file3.log*: No such file or directory

I have read about this and people suggest using something like:

grep -l "certainString" | sed "s/log/log*/g" | xargs -I@ sh -c 'grep -l "anotherString"'

But I can´t make it work (I do not fully understand it, to be honest). I feel I am just there but I am missing a detail. Any help?

Answer 1

in case file names doesn't contain spaces or globbing metacharater ( * [ ] )

Just add @ , because -I@ means that @ is a token to be replaced by argument

grep -l "certainString" | sed "s/log/log*/g" | xargs -I@ sh -c 'grep -l "anotherString" @'