I have defined the following alias in ~/.bashrc:
alias fg='find . -name $1 | xargs grep --color $2'
in order to write
fg "*.txt" " my_text "
and find all file that have extension .txt and contain " my_text " but it does not work. Why?
3 Answers
Aliases in bash do not take parameters (as already pointed out), so when you need something like that you can use bash functions (like the one provided by @l0b0).
But what you are trying to achieve here, can be done in a better way by using only grep.
grep -r --color --include="*.txt" " my_text " ./
BTW, fg is a shell built in command, an important one. You should avoid using it as a name for aliases or functions.
EDIT: in a function
$ ffg() { rgrep --color --include="$1" "$2" ./; }
$ ffg "*.txt" " my_text "
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1
find ./ -name "$1" -exec grep -l "$2" {} \; should do the trick.
answered Mar 18, 2011 at 9:58
Shadur-don't-feed-the-AIShadur-don't-feed-the-AI
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$i should be $1, and $1 and $2 should be quoted to allow blanks and such in filenames and search-word. Commented Mar 26, 2011 at 12:07
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This works!
function fndg()
{
find . -name "$1" | xargs grep -rn --color "$2"
}
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findcan call grep without xargs invocation, see-execdir, but in this case, the called programgrepcan iterate itself. Commented Mar 26, 2011 at 12:06 -
xargs will be more efficient, as it will call grep with several filenames at once, instead of one at a time using -exec Commented Mar 26, 2011 at 23:25
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gnu findcan handle mutliple filenames at once too:find . -name "$1" -exec grep --color $2 -- {} +if ended with + instead of ";" Commented Mar 27, 2011 at 19:59
fgto a function instead.fg is a shell builtin, I recommend you chose a different name whether you implement this as an alias or (better still) a function.