shell script to find a word in a list of files, all of them given as parameters

Question

I need a simple shell program which has to do something like this:

script.sh word_to_find file1 file2 file3 .... fileN

which will display

word_to_find 3 - if word_to_find appears in 3 files

or

word_to_find 5 - if word_to_find appears in 5 files 

This is what I've tried

#!/bin/bash

count=0
for i in $@; do
  if [ grep '$1' $i ];then
     ((count++))
  fi
done
echo "$1 $count"

But this message appears:

syntax error: "then" unexpected (expecting "done").

Before this the error was

[: grep: unexpected operator.

Answer 1

The code you showed is:

#!/bin/bash

count=0
for i in $@; do
  if [ grep '$1' $i ];then
     ((count++))
  fi
done
echo "$1 $count"

When I run it, I get the error:

script.sh: line 5: [: $1: binary operator expected

This is reasonable, but it is not the same as either of the errors reported in the question. There are multiple problems in the code.

The for i in $@; do should be for i in "$@"; do. Always use "$@" so that any spaces in the arguments are preserved. If none of your file names contain spaces or tabs, it is not critical, but it is a good habit to get into. (See How to iterate over arguments in bash script for more information.)

The if operations runs the [ (aka test) command, which is actually a shell built-in as well as a binary in /bin or /usr/bin. The use of single quotes around '$1' means that the value is not expanded, and the command sees its arguments as:

[
grep
$1
current-file-name
]

where the first is the command name, or argv[0] in C, or $0 in shell. The error I got is because the test command expects an operator such as = or -lt at the point where $1 appears (that is, it expects a binary operator, not $1, hence the message).

You actually want to test whether grep found the word in $1 in each file (the names listed after $1). You probably want to code it like this, then:

#!/bin/bash

word="$1"
shift
count=0

for file in "$@"
do
    if grep -l "$word" "$file" >/dev/null 2>&1
    then ((count++))
    fi
done
echo "$word $count"

We can negotiate on the options and I/O redirections used with grep. The POSIX grep options -q and/or -s options provide varying degrees of silence and -q could be used in place of -l. The -l option simply lists the file name if the word is found, and stops scanning on the first occurrence. The I/O redirection ensures that errors are thrown away, but the test ensures that successful matches are counted.

---

Incorrect output claimed

It has been claimed that the code above does not produce the correct answer. Here's the test I performed:

$ echo "This country is young" > young.iii
$ echo "This country is little" > little.iii
$ echo "This fruit is fresh" > fresh.txt
$ bash findit.sh country young.iii fresh.txt little.iii
country 2
$ bash -x findit.sh country young.iii fresh.txt little.iii
+ '[' -f /etc/bashrc ']'
+ . /etc/bashrc
++ '[' -z '' ']'
++ return
+ alias 'r=fc -e -'
+ word=country
+ shift
+ count=0
+ for file in '"$@"'
+ grep -l country young.iii
+ (( count++ ))
+ for file in '"$@"'
+ grep -l country fresh.txt
+ for file in '"$@"'
+ grep -l country little.iii
+ (( count++ ))
+ echo 'country 2'
country 2
$

This shows that for the given files, the output is correct on my machine (Mac OS X 10.10.2; GNU bash, version 3.2.57(1)-release (x86_64-apple-darwin14)). If the equivalent test works differently on your machine, then (a) please identify the machine and the version of Bash (bash --version), and (b) please update the question with the output you see from bash -x findit.sh country young.iii fresh.txt little.iii. You may want to create a sub-directory (such as junk), and copy findit.sh into that directory before creating the files as shown, etc.

You could also bolster your case by showing the output of:

$ grep country young.iii fresh.txt little.iii
young.iii:This country is young
little.iii:This country is little
$

Answer 2

Try this:

#!/bin/sh
printf '%s %d\n' "$1" $(grep -hm1 "$@" | wc -l)

Notice how all the script's arguments are passed verbatim to grep -- the first is the search expression, the rest are filenames.

The output from grep -hm1 is a list of matches, one per file with a match, and wc -l counts them.

I originally posted this answer with grep -l but that would require filenames to never contain a newline, which is a rather pesky limitation.

Maybe add an -F option if regular expression search is not desired (i.e. only search literal text).

Answer 3

#!/usr/bin/perl
use strict;
use warnings;
my $wordtofind = shift(@ARGV);
my $regex = qr/\Q$wordtofind/s;
my @file = ();
my $count = 0;
my $filescount = scalar(@ARGV);
for my $file(@ARGV)
{
    if(-e $file)
    {
        eval { open(FH,'<' . $file) or die "can't open file $file "; };
        unless($@)
        {
            for(<FH>)
            {
                if(/$regex/)
                {
                     $count++;
                     last;
                }
            }
            close(FH);
        }
    }
}
print "$wordtofind $count\n";

Answer 4

You could use an Awk script:

#!/usr/bin/env awk -f

BEGIN {
    n=0
} $0 ~ w {
    n++
} END {
    print w,n
}

and run it like this:

./script.awk w=word_to_find file1 file2 file3 ... fileN

or if you don't want to worry about assigning a variable (w) on the command line:

BEGIN {
    n=0
    w=ARGV[1]
    delete ARGV[1]
} $0 ~ w {
    n++
} END {
    print w,n
}