mjhjmtu

I have a problem. I'm doing a script which one will get as a parameter $1 some charters. I have file "drinks.txt" where there have written few lines like this:

Cocacola
juice
CocaPepsi

I'm going to pass as a parameter "co" for example, and this should return me Cocacola and CocaPepsi. I'm failing with IF condition. Here is my code:

searchingParameter=$1
for drink in `cat drinks.txt`;do
 if [ "$drink" == "$1*" ]; then
 echo "$drink"
 fi
done

So this should print every drink that found with $1 plus *.
What's the correct way to do the IF for this?

Thanks in advance.

Doing this with a shell loop is awkward and error prone. If your drinks.txt file contains a * on a line by itself (or surrounded by spaces), it would expand to the names of the files in the current directory, for example.

Instead, just use grep:

grep -iF 'co' drinks.txt

Here, we use grep with -i to do a case-insensitive match. The -F option means that grep should use the pattern as a string rather than as a regular expression (this does not matter here, but would matter if your pattern contained characters like * or . or other special regular expression symbols).

The command would return each complete line that contained the string co, Co, cO or CO.

As part of a script:

#!/bin/sh

grep -iF -e "$1" drinks.txt

I'm using -e here to say "the next argument is a pattern". If I didn't do that and $1 started with a dash, it would be taken as an option to grep.

To force a match at the start of a word, you would use

grep -i -e 'b'"$1" drinks.txt

or

grep -i -e '<'"$1" drinks.txt

Where b and < both would match at a word boundary. Note that we had to remove the -F option here, because our pattern is now a regular expression. This also means that any regular expression characters in $1 will be special.