bash和grep：传递正则表达式参数

Question

I'm trying to write a bash script that helps solving crosswords. For example, the question is "Alcoholic Drink in German". I already have a 'B' at the first place, an 'R' at the last place and two gaps in between. So a regex would be $B..R^ Since I live in Switzerland, I'd like to use the ngerman dictionary (DICT=/usr/share/dict/ngerman). Here's how I'd do it directly on the shell:

    grep -i '^B...$' /usr/share/dict/ngerman

That works perfectly, and the word 'Bier' appears among three others. Since this syntax is cumbersome, I'd like to write a little batch script, that allows me to enter it like this:

    crosswords 'B..R'

Here's my approach:

#!/bin/bash

DICT=/usr/share/dict/ngerman

usage () {
    progname=$(basename $0)
    echo "usage: $progname regex"
}

if [ $# -le 0 ]; then 
    usage
    exit 1
fi

regex="'^$1$'"
cmd="grep -i $regex $DICT"
echo $regex
echo $cmd
$($cmd) | while read word; do
    echo "$word"
done

But nothing appears, it doesn't work. I also output the $regex and the $cmd variable for debugging reasons. Here's what comes out:

'^B..R$'

grep -i '^B..R$' /usr/share/dict/ngerman

That's exactly what I need. If I copy/paste the command above, it works perfectly. But if i call it with $($cmd) , it fails. What is wrong?

Answer 1

Change regex="^'$1$'" to regex="^$1$" and $($cmd) to $cmd

Here is a fixed version:

#!/bin/bash

DICT=/usr/share/dict/ngerman

usage () {
    progname=$(basename "$0")
    echo "usage: $progname regex"
}

if [ $# -le 0 ]; then 
    usage
    exit 1
fi

regex="^$1$"
cmd="grep -i $regex $DICT"
echo "$regex"
echo "$cmd"
$cmd | while read -r word; do
    echo "$word"
done

But this script has potential problems. For example try running it as ./script 'asdads * ' . This will expand to all files in a directory and all of them are going to be passed to grep.

Here is a bit improved version of your code with correct quoting and also with bonus input validation:

#!/bin/bash

DICT=/usr/share/dict/ngerman

usage () {
    progname=$(basename "$0")
    echo "usage: $progname regex"
}

if [ $# -le 0 ]; then 
    usage
    exit 1
fi

if ! [[ $1 =~ ^[a-zA-Z\.]+$ ]]; then
    echo 'Wrong word. Please use only a-zA-Z characters and dots for unknown letters'
    exit 1
fi

grep -i "^$1$" "$DICT" | while read -r word; do
    echo "$word"
done

Answer 2

you do not need to put quotes around regex variable string. and $($cmd) should change to $cmd

so the correct code is :

#!/bin/bash

DICT=/usr/share/dict/ngerman

usage () {
    progname=$(basename $0)
    echo "usage: $progname regex"
}

if [ $# -le 0 ]; then 
    usage
    exit 1
fi

regex="^$1$"
cmd="grep -i $regex $DICT"
echo $regex
echo $cmd
$cmd | while read word; do
    echo "$word"
done

Answer 3

Oh, now I got it. When I do it manually, '' are expanded! Here's my test program in C (param-test.c):

#include <stdio.h>
int main(int argc, char *argv[]) {
    puts(argv[1]);
    return 0;
}

Then I call:

    param-test 'foo'

And I see:

    foo

That's the problem! grep doesn't really get 'B..R', but just B..R.