Return to Answer

git grep

Here is the syntax using git grep combining multiple patterns using Boolean expressions:

git grep -e "life" --and -e "happy" --and -e "horse"

^{The above command will print lines matching all the patterns at once.}

If the files aren't under version control, add --no-index param.

Search files in the current directory that is not managed by Git.

Check man git-grep for help.

grep

Normally grep with -f parameter will print lines with at least one pattern, e.g.

grep -f args.txt file.txt

In your case it won't work.

So to print lines which matches all patterns at the same time, you can try this command:

while read n text; do [ $n -eq $(wc -l < args.txt) ] && echo $text; done < <(while read patt; do grep "$patt" text.txt; done < args.txt | sort | uniq -c)

Explanation:

1. The inner while loop will print all lines which matches at least one pattern in text.txt using pattern list from args.txt file.
2. Then this list is sorted (sort) and counted for number of occurrences (uniq -c).
3. The outer while loop will print only lines which have the same number of occurrences that number of patterns in args.txt (which is 3).

Another approach would be to remove all lines which does not match at least one pattern.

Here is the solution using Ex/Vim editor changing the file in-place:

while read patt; do ex +"v/$patt/d" -scwq text.txt; done < args.txt

^{Note: This will remove the unneeded lines from the file it-self.}

Here is shorter version which will print the result on the screen only:

ex $(xargs -I% printf "+v/%/d " < args.txt) +%p -scq! text.txt

^{Change +%p -scq! to -scwq to save it in-place into the file.}

And here is the solution by defining a shell alias:

alias grep-all="</dev/stdin $(xargs printf '|grep "%s"' < args.txt)"

Sample usage:

grep-all file.txt

See also: How to run grep with multiple AND patterns? and How to grep for 2 words existing on the same line?

git grep

Here is the syntax using git grep combining multiple patterns using Boolean expressions:

git grep -e "life" --and -e "happy" --and -e "horse"

^{The above command will print lines matching all the patterns at once.}

If the files aren't under version control, add --no-index param.

Search files in the current directory that is not managed by Git.

Check man git-grep for help.

grep

Normally grep with -f parameter will print lines with at least one pattern, e.g.

grep -f args.txt file.txt

In your case it won't work.

So to print lines which matches all patterns at the same time, you can try this command:

while read n text; do [ $n -eq $(wc -l < args.txt) ] && echo $text; done < <(while read patt; do grep "$patt" text.txt; done < args.txt | sort | uniq -c)

Explanation:

1. The inner while loop will print all lines which matches at least one pattern in text.txt using pattern list from args.txt file.
2. Then this list is sorted (sort) and counted for number of occurrences (uniq -c).
3. The outer while loop will print only lines which have the same number of occurrences that number of patterns in args.txt (which is 3).

Another approach would be to remove all lines which does not match at least one pattern.

Here is the solution using Ex/Vim editor changing the file in-place:

while read patt; do ex +"v/$patt/d" -scwq text.txt; done < args.txt

^{Note: This will remove the unneeded lines from the file it-self.}

Here is shorter version which will print the result on the screen only:

ex $(xargs -I% printf "+v/%/d " < args.txt) +%p -scq! text.txt

^{Change +%p -scq! to -scwq to save it in-place into the file.}

And here is the solution by defining a shell alias:

alias grep-all="</dev/stdin $(xargs printf '|grep "%s"' < args.txt)"

Sample usage:

grep-all file.txt

Related:

• How to run grep with multiple AND patterns?
• How to grep for 2 words existing on the same line?
• Check if all of multiple strings or regexes exist in a file

git grep

Here is the syntax using git grep:

git grep -e "life" --and -e "happy" --and -e "horse"

If the files aren't under version control, add --no-index param.

Search files in the current directory that is not managed by Git.

Check man git-grep for help.

grep

Normally grep with -f parameter will print lines with at least one pattern, e.g.

grep -f args.txt file.txt

In your case it won't work.

So to print lines which matches all patterns at the same time, you can try this command:

while read n text; do [ $n -eq $(wc -l < args.txt) ] && echo $text; done < <(while read patt; do grep "$patt" text.txt; done < args.txt | sort | uniq -c)

Explanation:

1. The inner while loop will print all lines which matches at least one pattern in text.txt using pattern list from args.txt file.
2. Then this list is sorted (sort) and counted for number of occurrences (uniq -c).
3. The outer while loop will print only lines which have the same number of occurrences that number of patterns in args.txt (which is 3).

Another approach would be to remove all lines which does not match at least one pattern.

Here is the solution using Ex/Vim editor changing the file in-place:

while read patt; do ex +"v/$patt/d" -scwq text.txt; done < args.txt

^{Note: This will remove the unneeded lines from the file it-self.}

Here is shorter version which will print the result on the screen only:

ex $(xargs -I% printf "+v/%/d " < args.txt) +%p -scq! text.txt

^{Change +%p -scq! to -scwq to save it in-place into the file.}

And here is the solution by defining a shell alias:

alias grep-all="</dev/stdin $(xargs printf '|grep "%s"' < args.txt)"

Sample usage:

grep-all file.txt

See also: How to run grep with multiple AND patterns? and How to grep for 2 words existing on the same line?

git grep

Here is the syntax using git grep combining multiple patterns using Boolean expressions:

git grep -e "life" --and -e "happy" --and -e "horse"

^{The above command will print lines matching all the patterns at once.}

If the files aren't under version control, add --no-index param.

Search files in the current directory that is not managed by Git.

Check man git-grep for help.

grep

Normally grep with -f parameter will print lines with at least one pattern, e.g.

grep -f args.txt file.txt

In your case it won't work.

So to print lines which matches all patterns at the same time, you can try this command:

while read n text; do [ $n -eq $(wc -l < args.txt) ] && echo $text; done < <(while read patt; do grep "$patt" text.txt; done < args.txt | sort | uniq -c)

Explanation:

1. The inner while loop will print all lines which matches at least one pattern in text.txt using pattern list from args.txt file.
2. Then this list is sorted (sort) and counted for number of occurrences (uniq -c).
3. The outer while loop will print only lines which have the same number of occurrences that number of patterns in args.txt (which is 3).

Another approach would be to remove all lines which does not match at least one pattern.

Here is the solution using Ex/Vim editor changing the file in-place:

while read patt; do ex +"v/$patt/d" -scwq text.txt; done < args.txt

^{Note: This will remove the unneeded lines from the file it-self.}

Here is shorter version which will print the result on the screen only:

ex $(xargs -I% printf "+v/%/d " < args.txt) +%p -scq! text.txt

^{Change +%p -scq! to -scwq to save it in-place into the file.}

And here is the solution by defining a shell alias:

alias grep-all="</dev/stdin $(xargs printf '|grep "%s"' < args.txt)"

Sample usage:

grep-all file.txt

See also: How to run grep with multiple AND patterns? and How to grep for 2 words existing on the same line?

git grep

Here is the syntax using git grep:

git grep -e "life" --and -e "happy" --and -e "horse"

If the file isn't under version control, add it by:

git init
git add file.txt

Check man git-grep for help.

grep

Normally grep with -f parameter will print lines with at least one pattern, e.g.

grep -f args.txt file.txt

In your case it won't work.

So to print lines which matches all patterns at the same time, you can try this command:

while read n text; do [ $n -eq $(wc -l < args.txt) ] && echo $text; done < <(while read patt; do grep "$patt" text.txt; done < args.txt | sort | uniq -c)

Explanation:

1. The inner while loop will print all lines which matches at least one pattern in text.txt using pattern list from args.txt file.
2. Then this list is sorted (sort) and counted for number of occurrences (uniq -c).
3. The outer while loop will print only lines which have the same number of occurrences that number of patterns in args.txt (which is 3).

Another approach would be to remove all lines which does not match at least one pattern.

Here is the solution using Ex/Vim editor changing the file in-place:

while read patt; do ex +"v/$patt/d" -scwq text.txt; done < args.txt

^{Note: This will remove the unneeded lines from the file it-self.}

Here is shorter version which will print the result on the screen only:

ex $(xargs -I% printf "+v/%/d " < args.txt) +%p -scq! text.txt

^{Change +%p -scq! to -scwq to save it in-place into the file.}

And here is the solution by defining a shell alias:

alias grep-all="</dev/stdin $(xargs printf '|grep "%s"' < args.txt)"

Sample usage:

grep-all file.txt

See also: How to run grep with multiple AND patterns? and How to grep for 2 words existing on the same line?

git grep

Here is the syntax using git grep:

git grep -e "life" --and -e "happy" --and -e "horse"

If the files aren't under version control, add --no-index param.

Search files in the current directory that is not managed by Git.

Check man git-grep for help.

grep

Normally grep with -f parameter will print lines with at least one pattern, e.g.

grep -f args.txt file.txt

In your case it won't work.

So to print lines which matches all patterns at the same time, you can try this command:

while read n text; do [ $n -eq $(wc -l < args.txt) ] && echo $text; done < <(while read patt; do grep "$patt" text.txt; done < args.txt | sort | uniq -c)

Explanation:

1. The inner while loop will print all lines which matches at least one pattern in text.txt using pattern list from args.txt file.
2. Then this list is sorted (sort) and counted for number of occurrences (uniq -c).
3. The outer while loop will print only lines which have the same number of occurrences that number of patterns in args.txt (which is 3).

Another approach would be to remove all lines which does not match at least one pattern.

Here is the solution using Ex/Vim editor changing the file in-place:

while read patt; do ex +"v/$patt/d" -scwq text.txt; done < args.txt

^{Note: This will remove the unneeded lines from the file it-self.}

Here is shorter version which will print the result on the screen only:

ex $(xargs -I% printf "+v/%/d " < args.txt) +%p -scq! text.txt

^{Change +%p -scq! to -scwq to save it in-place into the file.}

And here is the solution by defining a shell alias:

alias grep-all="</dev/stdin $(xargs printf '|grep "%s"' < args.txt)"

Sample usage:

grep-all file.txt

See also: How to run grep with multiple AND patterns? and How to grep for 2 words existing on the same line?