Getting "command not found" while using cut on bash variable

Question

I have two variables in a bash script

hostname="ab78ascsoadp003.abc.com"
Loc=`$hostname | cut -c3,4`

I am getting an error ab78ascsoadp003.abc.com: command not found

I am trying to use cut command so that $Loc gets 78

`$hostname`'s content isn't a command, therefore it fails to be parsed as the first token of a pipeline. — Aaron, Sep 25 '18 at 08:43

score 3 · Answer 1 · answered Sep 25 '18 at 08:54

While you can use cut to achieve this, sometimes it is useful to stick to bash:

hostname="ab78ascsoadp003.abc.com"
Loc=${hostname:3:2}

${parameter:offset:length} Substring Expansion. Expands to up to length characters of parameter starting at the character specified by offset. If length is omitted, expands to the substring of parameter starting at the character specified by offset. length and offset are arithmetic expressions <snip>

_{source: man bash}

oguz ismail · Answer 2 · 2019-04-12T10:13:15.067

2

hostname="ab78ascsoadp003.abc.com"
Loc=$(cut -c3,4 <<<"$hostname")

edited Apr 12 '19 at 10:13

answered Sep 25 '18 at 08:42

oguz ismail

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score 1 · Answer 3 · answered Sep 25 '18 at 08:43

1

you are missing an echo

Loc=`echo $hostname | cut -c3,4`

answered Sep 25 '18 at 08:43

Eric

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Getting "command not found" while using cut on bash variable

3 Answers3

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