Extract part of a string in BASH

Question

I have a puzzle to solve which I could not do on my own. I believe it may be a simple one for others. I was trying to solve in a one liner.

I have following strings:

Tree-Forest
Vegetable-fruit-Garden
Fish-Frog-Ponds
Chicken-Duck-Farm

I need to extract the part before last -, like (omitting the last - and the string after that):

Tree
Vegetable-fruit
Fish-Frog
Chicken-Duck

Please help.

Have you read the bash documentation on Parameter Expansion? It shows all the operations you can perform, including this one. — Barmar
– Barmar, Commented May 10, 2015 at 18:22

anubhava · Accepted Answer · 2015-05-10 18:21:01Z

1

Using BASH string manipulations:

s='Tree-Forest Vegetable-fruit-Garden Fish-Frog-Ponds Chicken-Duck-Farm'
echo "${s%-*}"
Tree-Forest Vegetable-fruit-Garden Fish-Frog-Ponds Chicken-Duck

After edited question it seems you're editing a file then you can use sed also:

sed 's/-[^-]*$//' file
Tree
Vegetable-fruit
Fish-Frog
Chicken-Duck

answered May 10, 2015 at 18:21

anubhava

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7 Comments

Barmar Over a year ago

Where does the question mention a file?

anubhava Over a year ago

Before editing I thought it is all in one single string but now it says I have following strings. Now difficult to see how strings are available in a block like this unless there is file. If not a file then sed 's/-[^-]*$//' <<< "$input"

Barmar Over a year ago

Maybe in a for or while loop that sets the variable to each string.

anubhava Over a year ago

OP says I was trying to solve in a one liner and doesn't provide details then.

user3783462 Over a year ago

Excellent, thanks a lot for solving this. Please explain what [^-]*$ means?

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