numeric variable in egrep regular expression bash script

Question

So I am trying to make a script that contains egrep and accepts a numeric variable

#!/bin/bash

var=$1
list="egrep "^.{$var}$ /usr/share/dict/words"
cat list

For example, if var is 5, I would like this script to print out every line with 5 characters. For some reason the script does not do that. Help would be greatly appreciated!

Avoid the usage of egrep. This is depricated. Make use of grep -E instead. (see man grep) — kvantour
– kvantour, Commented May 21, 2019 at 12:55
Copy/paste your code to shellcheck.net, fix the issues it tells you about, and then let us know if you still have a problem you need help with. — Ed Morton
– Ed Morton, Commented May 21, 2019 at 18:38

oguz ismail · Accepted Answer · 2019-05-21 12:42:55Z

3

Your script doesn't work because there are several problems with these lines:

list="egrep "^.{$var}$ /usr/share/dict/words"
cat list

The first line isn't complete, it's missing a closing quote,
Even if you fixed it, you're assigning a literal string to list, not the output of a command,
RE and filename should be separated
cat doesn't print a variable's content, echo does that.

So:

#!/bin/bash
var="$1"
list="$(egrep '^.{'"$var"'}$' /usr/share/dict/words)"
echo "$list"

should work.

Or even better, you can use just an awk command:

awk 'length==5' /usr/share/dict/words

with $1 or any other variable:

awk -v n="$1" 'length==n' /usr/share/dict/words

edited May 21, 2019 at 12:42

answered May 21, 2019 at 10:09

oguz ismail

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