How to use a variable in a command that's creating another variable

Question

I created my basic variable from a read command (I've done this manually and using a script):

read NAME

I then want to use that NAME variable to search a file and create another variable:

STUDENT=$(grep $NAME <students.dat | awk -F: '/$NAME/ {print $1}')

If I run the command manually with an actual name from that students.dat file (and not $NAME), it executes and displays what I want. However, when I run this command (manually or from the script using $NAME), it returns blank, and I'm not sure why.

RavinderSingh13 · Accepted Answer · 2017-02-20 23:15:54Z

2

@user1615415: Try:

cat script.ksh
echo "Enter name.."
read NAME
STUDENT=$(awk -vname="$NAME" -F: '($0 ~ name){print $3}' student.dat)

answered Feb 20, 2017 at 23:15

RavinderSingh13

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Comments

John Kugelman · Accepted Answer · 2017-02-20 23:13:49Z

1

Shell variables aren't interpolated in single quotes, only double quotes.

STUDENT=$(grep $NAME <students.dat | awk -F: "/$NAME/ {print \$1}")

$1 needs to be escaped to ensure it's not expanded by the shell, but by awk.

answered Feb 20, 2017 at 23:13

John Kugelman

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3 Comments

chepner Over a year ago

Passing the value of $NAME with the -v option is safer.

SuperheroSmith Over a year ago

Beautiful! That did it. Thanks!

codeforester Over a year ago

It's better to enclose $NAME in double quotes while passing to grep. Also, use -- to prevent failures caused by any leading - in $NAME. So, grep -- "$NAME" is much better.

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