How to make a script to make multiple grep's over a file? - bash

Question

I want to make a script that can do the following automatically:
grep 'string1' file.txt | grep 'string2' | grep 'string3' ... | grep 'stringN'
The idea is that the script can be run like this:
myScript.sh file.txt string1 string2 string3 ... stringN
and the script has to return all the lines of file.txt that contain all the strings.
For instance, if file.txt looks like this:
hello world
hello world run
hello planet world
And I can make a grep like this:
grep hello file.txt | grep world
and I get:
hello world
hello world run
hello planet world
I want to make a script that makes this automatically, with an undefined number of strings as parameters.
I found that it is hard to achieve this, since the number of strings can be variable. First, I tried to create an array called args like this in myScript.sh:
#!/bin/bash
args=("$#")
with the purpose of storing the arguments. I know that the ${args[0]} is going to be my file.txt and the rest are the strings that I need to use in the distinct greps, but I don't know how to proceed and if this is the best approach to solve the problem. I would appreciate any suggestion about how to program this.

Answer 1

sed is capable of doing this perfectly with a single process, and avoids these eval shenanigans. The resulting script is actually quite simple.
#!/bin/sh
file=$1
shift
printf '\\?%s?!d\n' "$#" |
sed -f - "$file"
We generate a line of sed script for each expression; if the expression is not (!) found, we delete (d) this input line, and start over with the next one.
This assumes your sed accepts - as the argument to -f to read the script from standard input. This is not completely portable; you would perhaps need to store the generated script in a temporary file instead if this is a problem.
This uses ? as the internal regex separator. If you need a literal ? in one of the patterns, you will need to backslash-escape it. In the general case, creating a script which finds an alternative separator which is in none of the search expressions would perhaps be possible, but at that point, I'd move to a proper scripting language (Python would be my preference) instead.

Answer 2

You can generate the pattern of operation and save it in a variable:
pattern="$(printf 'grep %s file.txt' "$1"; printf ' | grep %s' "${#:2}" ; printf '\n')"
and then
eval "$pattern"
Example:
% cat file.txt
foo bar
bar spam
egg
% grep_gen () { pattern="$(printf 'grep %s file.txt' "$1"; printf ' | grep %s' "${#:2}" ; printf '\n')"; eval "$pattern" ;}
% grep_gen foo bar
foo bar

Answer 3

You can create the command in a loop and then use eval to evaluate it.
This is using cat so you can group all the grep.
#! /bin/bash
file="$1"
shift
args=( "$#" )
cmd="cat '$file'"
for a in "${args[#]}"
do
cmd+=' | '
cmd+="grep '$a'"
done
eval $cmd

Answer 4

An eval-free alternative:
#!/bin/bash
temp1="$(mktemp)"
temp2="$(mktemp)"
grep "$2" "$1" > temp1
for arg in "${#:3}"; do
grep "$arg" temp1 > temp2
mv temp2 temp1
done
cat temp1
rm temp1
mktemp generates a temporary file with a unique name and returns its name; it should be widely available.
The loop then executes grep for each argument and renames the second temp file for the next loop.

Answer 5

This is the optimization of Diego Torres Milano's code and the answer to my original question:
#! /bin/bash
file=$1
shift
cmd="cat '$file'"
for 'a' in "$#"
do
cmd+=" | grep '$a'"
done
eval $cmd