CODEWITHSUNDEEP
bashsedls
user5259983
user5259983

Reputation:

Removing 2 last characters from a string

I'm trying to make a list of directories from two sources. The other directory has entries that can have -1 or -2 after them and I want to get rid of them.

ls output example:

something.something.anything-1
stuff.stuff.stuff.morestuff-2
st.uf.f

The code as how I have it now:

echo -e "\tDATA\t\t | \t\tSYSTEM"
for SYSTEM in `ls /system/app`; do
    for DATA in `ls /data/app | sed 's/-1*$//' | sed 's/-2*$//'`; do
        echo -n "$DATA | " 
    done
    echo "$SYSTEM"
done

It works just fine, but I'm currious if there's a better way of doing it, as I have to use sed twice. And I noticed there isn't really many good posts here to remove characters from strings using commands, this could be a good place to share your ideas.

UPDATE:

The updated code:

echo -e "\tDATA\t\t | \t\tSYSTEM"
for SYSTEM in `ls /system/app`; do
    for DATA in `ls /data/app | sed 's/-[[:digit:]]*$//'`; do
        echo -n "$DATA | " 
    done
    echo "$SYSTEM"
done

Wich works perfectly!

Upvotes: 0

Views: 172

Answers (4)

Casimir et Hippolyte
Casimir et Hippolyte

Reputation: 89639

using a parameter expansion:

toto="abcd-1"
echo ${toto/%-[12]}

Upvotes: 0

Kent
Kent

Reputation: 195289

since you have stored your string in a variable, you can just use bash's built-in way to handle it:

kent$  x="1234567"    
kent$  echo ${x: :-2}
12345

Upvotes: 1

ergonaut
ergonaut

Reputation: 7057

If you want to remove the last two chars always:

echo "abcdefg" | sed 's/..$//g'
> abcde

Or you can use a tighter regex for all digits

echo "abcdefg-2" | sed 's/-[[:digit:]]$//g'
> abcdefg

Or just those two:

echo "abcdefg-2" | sed 's/-[12]$//g'
> abcdefg

Upvotes: 2

hashier
hashier

Reputation: 4750

If you just want to remove the last 2 digits of any string use this:

$ echo -e "Hello-1\nHello-2" | sed 's/..$//'
Hello
Hello

Upvotes: 0

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