Storing set of commands output to a shell variable throws error

Question

In My shell script, I have following lines...

name=`grep -i "client" fin${i}.txt`
jobplan_name=`${name} | cut -d "=" -f4 | cut -d " " -f1`
echo ${jobplan_name}

Output:

<JOBP:  not found

Please correct me where the problem is

Geeb · Accepted Answer · 2014-01-27 14:13:19Z

3

I think you need an echo:

jobplan_name=`echo ${name} | cut -d "=" -f4 | cut -d " " -f1`

answered Jan 27, 2014 at 14:13

Geeb

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2 Comments

chepner Over a year ago

Without the echo, you are trying to execute a command named by name and piping its output to cut, rather than piping the value of name itself to cut.

Geeb Over a year ago

Because the backticks say "give me the output of the command contained within". The contents of the name variable is not a command recognized by your system: Hence "not found". The output of the echo command is what you want: the string that the name variable contains.

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