Not getting grep with a variable

Question

I am assigning a variable to echo command which generates a random number.

For eg: n=$(echo $(( (RANDOM%6) +1)). I have to grep this variable in /etc/passwd.

So, my syntax reads like:

n=$(echo $(( (RANDOM%6) +1)) | grep "$n" /etc/passwd

But I get the entire file content(/etc/passwd) as output. I do not know where the syntax is wrong.

It is important I get the output in one line only.

Hint 1: Count your parentheses. Hint 2: The $(echo part is pointless; n=$(( (RANDOM%6) +1)) is sufficient. Hint 3: A pipe | creates child processes that run essentially in parallel. Do you think it makes sense to execute these two commands in parallel? — Uwe, Mar 9 '15 at 19:54

Gilles 'SO- stop being evil' · Accepted Answer · 2015-03-09 23:09:14Z

1

I notice your separating two commands that run in sequence and not sharing Io with a pipe rather than a semicolon. Try

n=$(echo $(( (RANDOM%6) +1)) ; grep "$n" /etc/passwd

670k163 gold badges1397 silver badges1900 bronze badges

answered Mar 9 '15 at 21:00

Cosmo F

1591 silver badge6 bronze badges

2

The echo is still pointless. – roaima Mar 9 '15 at 21:58
grep -F "$((RANDOM%6+1))" /etc/passwd but (RANDOM%6)+1 == RANDOM%7 – Costas Mar 9 '15 at 22:05

add a comment |

Adam Katz · Accepted Answer · 2015-03-10 22:20:06Z

For eg: n=$(echo $(( (RANDOM%6) +1)). I have to grep this variable in /etc/passwd.

So you're trying to print all lines of /etc/passwd that contain a random number from one to six?

Wouldn't that just be (assuming bash or zsh or ksh for $RANDOM, which the base POSIX shell does not support):

grep $((RANDOM%6+1)) /etc/passwd

If you need that random number elsewhere:

n=$((RANDOM%6+1))
grep $n /etc/passwd

I don't think this is what you're trying to do, but you've oversimplified your question and lost that intent.

2 Answers 2