Value not assigned to variable in shell script

Question

I don't know what I do wrong, but this code just doesn't work for me:

mode="-P"
select mmode in "default" "fast"; do
    case $mmode in
        default ) break;;
        fast ) mode="-T 2C -P" break;;
    esac
done

echo $mode

The output is always "-P". Somebody please tell me where's the problem, thank you in advance!

devnull · Accepted Answer · 2014-02-10 13:57:11Z

3

A missing ; causes it. Say:

fast ) mode="-T 2C -P"; break;;

answered Feb 10, 2014 at 13:57

devnull

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6 Comments

kennyevo Over a year ago

Oh god I knew it will be a little thing like that :D Thank you for the very fast solution!

devnull Over a year ago

@kennyevo ; is not a little thing. It is a command separator :)

Edouard Thiel Over a year ago

var=value cmd is correct but different from var=value; cmd : in the first case, var will only exist for cmd. It is very useful for things like IFS=: read ... or LANG=C file ...

devnull Over a year ago

@EdouardThiel You are talking about something completely different.

Edouard Thiel Over a year ago

@devnull in what sense ?

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